C-language

C  Interview Questions

Note :  All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment,
The underlying machine is an x86 system,
Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this
assumptions (for example sizeof(int) == 2 may be assumed).
Predict the output or error(s) for the following:


1. void main()
{
 int  const * p=5;
 printf("%d",++(*p));
}
Answer:
   Compiler error: Cannot modify a constant value.
Explanation:  
p is a pointer to a "constant integer". But we tried to change the
value of the "constant integer".

2. main()
{
 char s[ ]="man";
 int i;
 for(i=0;s[ i ];i++)
 printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer:
             mmmm
                        aaaa
                        nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the
same idea. Generally  array name is the base address for that array. Here s is
the base address. i is the index number/displacement from the base address. So,
indirecting it with * is same as s[i]. i[s] may be surprising. But in the  case of  C
it is same as s[i].

3. main()
{
 float me = 1.1;
 double you = 1.1;
 if(me==you)
printf("I love U");
else
  printf("I hate U");
}
Answer:
I hate U
2
Explanation:
For floating point numbers (float, double, long double) the values
cannot be predicted exactly. Depending on the number of bytes, the precession
with of the value  represented varies. Float takes 4 bytes and long double takes
10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point
numbers with relational operators (== , >, <, <=, >=,!= ) .

4. main()
 {
 static int var = 5;
 printf("%d ",var--);
 if(var)
  main();
 }
Answer:
5 4 3 2 1
          Explanation:
When static storage class is given, it is initialized once. The change
in the value of a static variable is retained even between the function calls. Main
is also treated like any other ordinary function, which can be called recursively.

5. main()
{
  int c[ ]={2.8,3.4,4,6.7,5};
  int j,*p=c,*q=c;
  for(j=0;j<5;j++) {
  printf(" %d ",*c);
     ++q;   }
  for(j=0;j<5;j++){
printf(" %d ",*p);
++p;   }
}
Answer:
             2 2 2 2 2 2 3 4 6 5
          Explanation:
Initially pointer c is assigned to both  p and  q. In the first loop,
since only q is incremented and not c , the value 2 will be printed 5 times. In
second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

6. main()
{
 extern int i;
 i=20;
printf("%d",i);
}
Answer:
Linker Error : Undefined symbol '_i'
Explanation:
              extern storage class in the following declaration, http://www.exforsys.com Page 3 6/26/2004
3
                          extern int i;
specifies to the compiler that the memory for  i is allocated in some other
program and that address will be given to the current program at the time of
linking. But linker finds that no other variable of name i is available in any other
program with memory space allocated for it. Hence a linker error has occurred .

7. main()
{
 int i=-1,j=-1,k=0,l=2,m;
 m=i++&&j++&&k++||l++;
 printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer:
             0 0 1 3 1
Explanation :
Logical operations always give a result of  1 or 0  . And also the
logical AND (&&) operator has higher priority over the logical OR (||) operator. So
the expression  ‘i++ && j++ && k++’ is executed first. The result of this
expression is 0    (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which
evaluates to 1 (because OR operator always gives 1  except for ‘0 || 0’
combination- for which it gives 0). So the value of m is 1. The values of other
variables are also incremented by 1.

8. main()
{
 char *p;
 printf("%d %d ",sizeof(*p),sizeof(p));
}
Answer:
             1 2
Explanation:
The sizeof() operator gives the number of bytes taken by its
operand. P is a character pointer, which needs one byte for storing its value (a
character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store
the address of the character pointer sizeof(p) gives 2.

9. main()
{
 int i=3;
 switch(i)
  {
     default:printf("zero");
     case 1: printf("one");
     break;
    case 2:printf("two");
    break;
   case 3: printf("three");
    break;
   }
}
Answer :
three http://www.exforsys.com Page 4 6/26/2004
4
Explanation :
The default case can be placed anywhere inside the  loop. It is
executed only when all other cases doesn't match.

10.main()
{
   printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times
the least significant 4 bits are filled with 0's.The %x format specifier specifies that
the integer value be printed as a hexadecimal value.

11.main()
{
            char string[]="Hello World";
 display(string);
}
void display(char *string)
{
printf("%s",string);
}
           Answer:
Compiler Error : Type mismatch in redeclaration of function display
           Explanation :
In third line, when the function  display is encountered, the
compiler doesn't know anything about the function display. It assumes the
arguments and return types to be integers, (which is the default type). When it
sees the actual function display, the arguments and type contradicts with what it
has assumed previously. Hence a compile time error occurs.

12.main()
{
 int c=- -2;
 printf("c=%d",c);
}
Answer:
               c=2;
           Explanation:
Here unary minus (or negation) operator is used twice. Same
maths  rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can  only be
applied to variables as a decrement operator (eg., i--). 2 is a constant and not a
variable.

13. #define int char
main()
{
 int i=65;
 printf("sizeof(i)=%d",sizeof(i)); http://www.exforsys.com Page 5 6/26/2004
5
}
Answer:
            sizeof(i)=1
Explanation:
Since the #define replaces the string  int by the macro char

14.main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}
Answer:
i=0
   Explanation:
In the expression  !i>14 , NOT (!) operator has more precedence
than ‘ >’ symbol.  ! is a unary logical operator. !i (!10) is 0 (not of true is false).
0>14 is false (zero).

15. #include<stdio.h>
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is
pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10,
which is then incremented to 11. The value of ++*p  is 11. ++*str1, str1 is
pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is
98.
 Now performing (11 + 98 – 32), we get 77("M");
 So we get the output 77 :: "M" (Ascii is 77).
16. #include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8}  };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1 http://www.exforsys.com Page 6 6/26/2004
6
Explanation:
p=&a[2][2][2]  you declare only two 2D arrays, but you are trying
to access the third 2D(which you are not declared) it will print garbage values.
*q=***a starting address of a is assigned integer pointer. Now q is pointing to
starting address of a. If you print *q, it will print first element of 3D array.

17. #include<stdio.h>
main()
{
struct xx
{
      int x=3;
      char name[]="hello";
 };
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration
18. #include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
 struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence, the elements
are of yy are to be accessed through the instance of structure xx, which needs an
instance of yy to be known. If the instance is created after defining the structure
the compiler will not know about the instance relative to xx. Hence for nested
structure yy you have to declare member.
19.main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer: http://www.exforsys.com Page 7 6/26/2004
7
hai
Explanation:
\n  - newline
\b  - backspace
\r  - linefeed
20.main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from left
to right. The evaluation is by popping out from the stack. and the  evaluation is
from right to left, hence the result.
21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
the macro call square(4) will substituted by 4*4 so the expression
becomes i = 64/4*4 . Since / and * has equal priority the expression will be
evaluated as (64/4)*4 i.e. 16*4 = 64
 
22.main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s   %s",p,p1);
}
Answer:
ibj!gsjfoet
 Explanation:
   ++*p++ will be parse in the given order
*p that is value at the location currently pointed by p will be taken
++*p the retrieved value will be incremented
when ; is encountered the location will be incremented that is p++ will be
executed                                                                                                                                                                        
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by
executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’
and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p
becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot
print anything.  http://www.exforsys.com Page 8 6/26/2004
8
23. #include <stdio.h>
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in the
program. So the most recently assigned value will be taken.
24. #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of
the compiler. So textual replacement of clrscr() to 100 occurs.The input  program
to compiler looks like this :
  main()
  {
       100;
       printf("%d\n",100);
  }
Note:
100; is an executable statement but with no action. So it doesn't
give any problem
25.main()
{
41printf("%p",main);
}8Answer:
  Some address will be printed.
Explanation:
  Function names are just addresses (just like array names are
addresses).
main() is also a function. So the address of function main will be printed. %p in
printf specifies that the argument is an address. They are printed as hexadecimal
numbers.
27) main()
{
clrscr();
}
clrscr();

Answer: http://www.exforsys.com Page 9 6/26/2004
9
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function
call. In the second clrscr(); is a function declaration (because it is
not inside any function).
28) enum colors {BLACK,BLUE,GREEN}
 main()
{
 
 printf("%d..%d..%d",BLACK,BLUE,GREEN);
 
 return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.
29) void main()
{
 char far *farther,*farthest;
 
 printf("%d..%d",sizeof(farther),sizeof(farthest));
 
 }
Answer:
4..2
Explanation:
  the second pointer is of char type and not a far pointer
30) main()
{
 int i=400,j=300;
 printf("%d..%d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program.
Any number of printf's may be given. All of them take only the first
two values. If more number of assignments given in  the
program,then printf will take garbage values.
31)  main()
{
 char *p;
 p="Hello";
 printf("%c\n",*&*p);
}
Answer:
H
Explanation: http://www.exforsys.com Page 10 6/26/2004
10
* is a dereference operator & is a reference  operator. They can be  
applied any number of times provided it is meaningful. Here  p
points to  the first character in the string "Hello". *p dereferences it
and so its value is H. Again  & references it to an address and *
dereferences it to the value H.
32)  main()
{
    int i=1;
    while (i<=5)
    {
       printf("%d",i);
       if (i>2)
   goto here;
       i++;
    }
}
fun()
{
   here:
     printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words the scope of the labels
is limited to functions. The label 'here' is available in function fun()
Hence it is not visible in function main.
33)  main()
{
   static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
    int i;
    char *t;
    t=names[3];
    names[3]=names[4];
    names[4]=t;
    for (i=0;i<=4;i++)
     printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.
34) void main()
{
 int i=5;
 printf("%d",i++ + ++i);
}
Answer:
Output Cannot be predicted  exactly.
Explanation: http://www.exforsys.com Page 11 6/26/2004
11
Side effects are involved in the evaluation of   i
35) void main()
{
 int i=5;
 printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an
illegal combination of operators.
 
36) #include<stdio.h>
main()
{
int i=1,j=2;
switch(i)
 {
 case 1:  printf("GOOD");
     break;
 case j:  printf("BAD");
      break;
 }
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this
implies that we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.
37) main()
{
int i;
printf("%d",scanf("%d",&i));  // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0.  Here
10 is given as input which should have been scanned successfully.
So number of items read is 1.
38) #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
 }
Answer:
100  http://www.exforsys.com Page 12 6/26/2004
12
39) main()
{
int i=0;

for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is
"evaluated". Here it evaluates to 0 (false) and comes out of the
loop, and i is incremented (note the semicolon after the for loop).
40) #include<stdio.h>
main()
{
  char s[]={'a','b','c','\n','c','\0'};
  char *p,*str,*str1;
  p=&s[3];
  str=p;
  str1=s;
  printf("%d",++*p + ++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p
meAnswer:"p is pointing to '\n' and that is incremented by one."
the ASCII value of '\n' is 10. then it is incremented to 11. the value
of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is
incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both
11 and 98 is added and result is subtracted from 32.
i.e. (11+98-32)=77("M");

41) #include<stdio.h>
main()
{
  struct xx
   {
      int x=3;
      char name[]="hello";
   };
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the
structure declaration http://www.exforsys.com Page 13 6/26/2004
13
42) #include<stdio.h>
main()
{
struct xx
 {
   int x;
   struct yy
    {
      char s;
      struct xx *p;
    };
              struct yy *q;
             };
           }
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.
43) main()
{
 extern int i;
 i=20;
 printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined
somewhere else. The compiler passes the external variable to be
resolved by the linker. So compiler doesn't find an error. During
linking the linker searches for the definition of i. Since it is not
found the linker flags an error.
44) main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of
declaration. Even though a is a global variable, it is not available
for main. Hence an error.
45) main()
{
 extern out;
 printf("%d", out);
}
 int out=100; http://www.exforsys.com Page 14 6/26/2004
14
Answer:
100
Explanation:
This is the correct way of writing the previous program.
     
46) main()
{
 show();
}
void show()
{
 printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything
about it. So the default return type (ie, int) is assumed. But when
compiler sees the actual definition of show mismatch occurs since it
is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().

47) main( )
{
  int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
  printf(“%u %u %u %d \n”,a,*a,**a,***a);
   printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
   }
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
  The given array is a 3-D one. It can also be viewed as a 1-D array.
         
                                                                                                                           
2 4 7 8 3 4 2 2 2 3 3 4
   100  102  104  106 108   110  112  114  116   118   120   122
thus, for the first printf statement a, *a, **a  give address of  first
element . since the indirection ***a gives the value. Hence, the
first line of the output.
for the second printf a+1 increases in the third dimension thus
points to value at 114, *a+1 increments in second dimension thus
points to 104, **a +1 increments the first dimension thus points to
102 and ***a+1 first gets the value at first location and then
increments it by 1. Hence, the output.
48) main( )
{
  int a[ ] = {10,20,30,40,50},j,*p; http://www.exforsys.com Page 15 6/26/2004
15
  for(j=0; j<5; j++)
    {
printf(“%d” ,*a);
a++;
    }
    p = a;
   for(j=0; j<5; j++)
      {
printf(“%d ” ,*p);
p++;
      }
 }
Answer:
Compiler error: lvalue required.
 
Explanation:
Error is in line with statement a++. The operand must be an lvalue
and may be of any of scalar type for the any operator, array name
only when subscripted is an lvalue. Simply array name is a nonmodifiable lvalue.
**49) main( )
{
 static int  a[ ]   = {0,1,2,3,4};
 int  *p[ ] = {a,a+1,a+2,a+3,a+4};
 int  **ptr =  p;
 ptr++;
 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
 *ptr++;
 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
 *++ptr;
 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
 ++*ptr;
  printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
111
 222
 333
 344
Explanation:
Let us consider the array and the two pointers with some address
a
0 1 2 3 4
   100      102      104      106      108
          p
100 102 104 106 108
        1000    1002    1004    1006    1008
      ptr
1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002,
if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – http://www.exforsys.com Page 16 6/26/2004
16
starting location of array p, (1002 – 1000) / (scaling factor) = 1,
*ptr – a = value at address pointed by ptr – starting value of array
a, 1002 has a value 102  so the value is (102 – 100)/(scaling
factor) = 1,  **ptr is the value stored in the location pointed by
the pointer of ptr = value pointed by value pointed by 1002 =
value pointed by 102 = 1. Hence the output of the firs printf is  1,
1, 1.
After execution of *ptr++ increments value of the value in ptr by
scaling factor, so it becomes1004. Hence, the outputs for the
second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by
scaling factor, so it becomes1004. Hence, the outputs for the third
printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value
pointed by the value is incremented by the scaling factor. So the
value in array p at location 1006 changes from 106  10 108,.
Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000
= 3, *ptr – a = 108 – 100 = 4, **ptr = 4.
50) main( )
{
 char  *q;
 int  j;
 for (j=0; j<3; j++) scanf(“%s” ,(q+j));
 for (j=0; j<3; j++) printf(“%c” ,*(q+j));
 for (j=0; j<3; j++) printf(“%s” ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take input
in the same pointer thus we keep writing over in the same location,
each time shifting the pointer value by 1. Suppose the inputs are
MOUSE,  TRACK and VIRTUAL. Then for the first input suppose the
pointer starts at location 100 then the input one is stored as
M O U S E \0
When the second input is given the pointer is incremented as j
value becomes 1, so the input is filled in memory starting from
101.
M T R A C K \0
The third input  starts filling from the location 102
M T V I R T U A L \0
This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2  =
M T V
The second printf prints three strings starting from locations q,
q+1, q+2
 i.e  MTVIRTUAL, TVIRTUAL and VIRTUAL.
 
51) main( )
{
 void *vp;
 char ch = ‘g’, *cp = “goofy”;
 int j = 20; http://www.exforsys.com Page 17 6/26/2004
17
 vp = &ch;
 printf(“%c”, *(char *)vp);
 vp = &j;
 printf(“%d”,*(int *)vp);
 vp = cp;
 printf(“%s”,(char *)vp + 3);
}
Answer:
 g20fy
Explanation:
Since a void pointer is used it can be type casted to any  other type
pointer. vp = &ch  stores address of char ch and the next
statement prints the value stored in vp after type casting it to the
proper data type pointer. the output is ‘g’. Similarly  the output
from second printf is ‘20’. The third printf statement type casts it to
print the string from the 4
th
 value hence the output is ‘fy’.
52) main ( )
{
 static char *s[ ]  = {“black”, “white”, “yellow”, “violet”};
 char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
 p = ptr;
 **++p;
 printf(“%s”,*--*++p + 3);
}
Answer:
 ck
Explanation:
In this problem we have an array of char pointers pointing to start
of 4 strings. Then we have ptr which is a pointer to a pointer of
type char and a variable p which is a pointer to a  pointer to a
pointer of type char. p hold the initial value of ptr, i.e. p = s+3.
The next statement increment value in p by 1 , thus now value of p
=  s+2. In the printf statement the expression is evaluated *++p
causes gets value s+1 then the pre decrement is executed and we
get s+1 – 1 = s . the indirection operator now gets the value from
the array of s and adds 3 to the starting address.  The string is
printed starting from this position. Thus, the output is ‘ck’.
53) main()
{
 int  i, n;
 char *x = “girl”;
 n = strlen(x);
 *x = x[n];
 for(i=0; i<n; ++i)
   {
printf(“%s\n”,x);
x++;
   }
 }
Answer:
(blank space) http://www.exforsys.com Page 18 6/26/2004
18
irl
rl
l
Explanation:
Here a string (a pointer to char) is initialized with a value “girl”.
The strlen function returns the length of the string, thus n has a
value 4. The next statement assigns value at the nth location (‘\0’)
to the first location. Now the string becomes “\0irl” . Now the printf
statement prints the string after each iteration it increments it
starting position.  Loop starts from 0 to 4. The first time x[0] = ‘\0’
hence it prints nothing and pointer value is incremented. The
second time it prints from x[1] i.e “irl” and the third time it prints
“rl” and the last time it prints “l” and the loop terminates.
54) int i,j;
 for(i=0;i<=10;i++)
 {
 j+=5;
 assert(i<5);
 }
Answer:
Runtime error: Abnormal program termination.
   assert failed (i<5), <file name>,<line number>
Explanation:
asserts are used during debugging to make sure that certain
conditions are satisfied. If assertion fails, the program will
terminate reporting the same. After debugging use,
 #undef NDEBUG
and this will disable all the assertions from the source code.
Assertion
is a good debugging tool to make use of.
 
55) main()
 {
 int i=-1;
 +i;
 printf("i = %d, +i = %d \n",i,+i);
 }
Answer:
 i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes
you can just ignore it just because it has no effect in the
expressions (hence the name dummy operator).
56) What are the files which are automatically opened when a C file is
executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard
error).
57) what will be the position of the file marker?
 a: fseek(ptr,0,SEEK_SET); http://www.exforsys.com Page 19 6/26/2004
19
 b: fseek(ptr,0,SEEK_CUR);
Answer :
 a: The SEEK_SET sets the file position marker to the starting of the
file.
  b: The SEEK_CUR sets the file position marker to  the current
position
 of the file.
58) main()
 {
 char name[10],s[12];
 scanf(" \"%[^\"]\"",s);
 }
 How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it
matches with a quotation mark and then it  reads all character upto
another quotation mark.
59) What is the problem with the following code segment?
 while ((fgets(receiving array,50,file_ptr)) != EOF)
   ;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking
for != NULL.
60) main()
 {
 main();
 }
Answer:
 Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the function is
called its return address is stored in the call stack. Since there is
no condition to terminate the function call, the call stack overflows
at runtime. So it terminates the program and results in an error.
61) main()
 {
 char *cptr,c;
 void *vptr,v;
 c=10;  v=0;
 cptr=&c; vptr=&v;
 printf("%c%v",c,v);
 }
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since
void is an empty type. In the second line you are creating variable
vptr of type void * and v of type void hence an error. http://www.exforsys.com Page 20 6/26/2004
20
62) main()
 {
 char *str1="abcd";
 char str2[]="abcd";
 printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
 }
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of
the pointer variable. In second sizeof the name str2 indicates the
name of the array whose size is 5 (including the '\0' termination
character). The third sizeof is similar to the second one.
63) main()
 {
 char not;
 not=!2;
 printf("%d",not);
 }
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the
boolean value FALSE, and any non-zero value is considered to be
the boolean value TRUE. Here 2 is a non-zero value  so TRUE.
!TRUE is FALSE (0) so it prints 0.
64) #define FALSE -1
 #define TRUE   1
 #define NULL   0
 main() {
    if(NULL)
  puts("NULL");
    else if(FALSE)
  puts("TRUE");
    else
  puts("FALSE");
    }
Answer:
TRUE
Explanation:
The input program to the compiler after processing  by the
preprocessor is,
 main(){
  if(0)
   puts("NULL");
 else if(-1)
   puts("TRUE");
 else
   puts("FALSE");
  } http://www.exforsys.com Page 21 6/26/2004
21
Preprocessor doesn't replace the values given inside the double
quotes. The check by if condition is boolean value false so it goes
to else. In second if -1 is boolean value true hence "TRUE" is
printed.
65) main()
 {
 int k=1;
 printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
 }
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated  by whitespace) they are concatenated (this is called as "stringization"
operation). So the string is as if it is given as "%d==1 is %s". The
conditional operator( ?: ) evaluates to "TRUE".
66) main()
 {
 int y;
 scanf("%d",&y); // input given is 2000
 if( (y%4==0 && y%100 != 0) || y%100 == 0 )
      printf("%d is a leap year");
 else
      printf("%d is not a leap year");
 }
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.
67)   #define max 5
 #define int arr1[max]
 main()
 {
 typedef char arr2[max];
 arr1 list={0,1,2,3,4};
 arr2 name="name";
 printf("%d %s",list[0],name);
 }
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be
used to declare the variable name of the type arr2. But it is not the
case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are
used for declaring new types.
68) int i=10;
 main() http://www.exforsys.com Page 22 6/26/2004
22
 {
   extern int i;
             {
      int i=20;
  {
   const volatile unsigned i=30;
   printf("%d",i);
  }
       printf("%d",i);
    }
 printf("%d",i);
 }
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost
block i is declared as,
 const volatile unsigned
which is a valid declaration. i is assumed of type  int. So printf
prints 30. In the next block, i has value 20 and so printf prints 20.
In the outermost block, i is declared as extern, so no storage space
is allocated for it. After compilation is over the linker resolves it to
global variable i (since it is the only variable visible there). So it
prints i's value as 10.
69) main()
 {
     int *j;
     {
      int i=10;
      j=&i;
      }
      printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside
that block only. But the lifetime of i is lifetime of the function so it
lives upto the exit of main function. Since the i is still allocated
space, *j prints the value stored in i since j points i.
70) main()
 {
 int i=-1;
 -i;
 printf("i = %d, -i = %d \n",i,-i);
 }
Answer:
i = -1, -i = 1
Explanation: http://www.exforsys.com Page 23 6/26/2004
23
-i is executed and this execution doesn't affect the value of i. In
printf first you just print the value of i. After that the value of the
expression -i = -(-1) is printed.
71) #include<stdio.h>
main()
 {
   const int i=4;
   float j;
   j = ++i;
   printf("%d  %f", i,++j);
 }
Answer:
Compiler error
    Explanation:
i is a constant. you cannot change the value of constant
72) #include<stdio.h>
main()
{
  int a[2][2][2] = { {10,2,3,4}, {5,6,7,8}  };
  int *p,*q;
  p=&a[2][2][2];
  *q=***a;
  printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2]  you declare only two 2D arrays. but you are trying
to access the third 2D(which you are not declared)  it will print
garbage values. *q=***a starting address of a is assigned integer
pointer. now q is pointing to starting address of a.if you print *q
meAnswer:it will print first element of 3D array.
73) #include<stdio.h>
main()
  {
    register i=5;
    char j[]= "hello";                    
     printf("%s  %d",j,i);
}
Answer:
hello 5
Explanation:
if you declare i as register  compiler will treat it as ordinary integer
and it will take integer value. i value may be  stored  either in
register  or in memory.
74) main()
{
   int i=5,j=6,z;
   printf("%d",i+++j); http://www.exforsys.com Page 24 6/26/2004
24
  }
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)  
 
76) struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
 struct aaa abc,def,ghi,jkl;
 int x=100;
 abc.i=0;abc.prev=&jkl;
 abc.next=&def;
 def.i=1;def.prev=&abc;def.next=&ghi;
 ghi.i=2;ghi.prev=&def;
 ghi.next=&jkl;
 jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
 x=abc.next->next->prev->next->i;
 printf("%d",x);
}
Answer:
2
Explanation:
  above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.
77) struct point
 {
 int x;
 int y;
 };
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);
}

Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the
structure either with arrow mark or with indirection operator.
Note:
Since structure point  is globally declared x & y are initialized as
zeroes  http://www.exforsys.com Page 25 6/26/2004
25
 
78) main()
{
 int i=_l_abc(10);
   printf("%d\n",--i);
}
int _l_abc(int i)
{
 return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be
returned.
79) main()
{
 char *p;
 int *q;
 long *r;
 p=q=r=0;
 p++;
 q++;
 r++;
 printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator  when applied to pointers increments address
according to their corresponding data-types.
 80) main()
{
 char c=' ',x,convert(z);
 getc(c);
 if((c>='a') && (c<='z'))
 x=convert(c);
 printf("%c",x);
}
convert(z)
{
  return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.
81) main(int argc, char **argv)
{
 printf("enter the character"); http://www.exforsys.com Page 26 6/26/2004
26
 getchar();
 sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
 return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum
without converting it to integer values.
82) # include <stdio.h>
int one_d[]={1,2,3};
main()
{
 int *ptr;
 ptr=one_d;
 ptr+=3;
 printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.
83) # include<stdio.h>
aaa() {
  printf("hi");
 }
bbb(){
 printf("hello");
 }
ccc(){
 printf("bye");
 }
main()
{
  int (*ptr[3])();
  ptr[0]=aaa;
  ptr[1]=bbb;
  ptr[2]=ccc;
  ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is
assigned to address of the function aaa. Similarly ptr[1] and ptr[2]
for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(),
since ptr[2] points to ccc. http://www.exforsys.com Page 27 6/26/2004
27
85) #include<stdio.h>
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against
NULL.
86) main()
{
 int i =0;j=0;
 if(i && j++)
    printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine
the truth value of the boolean expression. So the statement
following the if statement is not executed.  The values of i and j
remain unchanged and get printed.
     
87) main()
{
 int i;
 i = abc();
 printf("%d",i);
}
abc()
{
 _AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the
information from the accumulator. Here _AH is the pseudo global
variable denoting the accumulator. Hence, the value of the
accumulator is set 1000 so the function returns value 1000.
88) int i;
         main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i)) http://www.exforsys.com Page 28 6/26/2004
28
              printf("%d--",t--);
             }
 // If the inputs are 0,1,2,3 find the o/p
Answer:
  4--0
  3--1
  2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during
execution
            will be,
          t        i       x
          4       0      -4
          3       1      -2
          2       2       0
         
89) main(){
  int a= 0;int b = 20;char x =1;char y =10;
  if(a,b,x,y)
        printf("hello");
 }
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the
rightmost value is returned and the other values are evaluated and
ignored. Thus the value of last variable y is returned to check in if.
Since it is a non zero value if becomes true so, "hello" will be
printed.
90) main(){
 unsigned int i;
 for(i=1;i>-2;i--)
          printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since
the both types doesn't match, signed is promoted to unsigned
value. The unsigned equivalent of -2 is a huge value so condition
becomes false and control comes out of the loop.
91) In the following pgm add a  stmt in the function  fun such that the address
of
'a' gets stored in 'j'.
main(){
  int * j;
  void fun(int **);
  fun(&j);
 }
 void fun(int **k) {
  int a =0;
  /* add a stmt here*/
 } http://www.exforsys.com Page 29 6/26/2004
29
Answer:
  *k = &a
Explanation:
         The argument of the function is a pointer to a pointer.
     
92) What are the following notations of defining functions known as?
i.      int abc(int a,float b)
          {
                 /* some code */
 }
ii.    int abc(a,b)
        int a; float b;
          {
            /* some code*/
          }
Answer:
i.  ANSI C notation
ii. Kernighan & Ritche notation
93) main()
{
char *p;
p="%d\n";
            p++;
            p++;
            printf(p-2,300);
}
Answer:
 300
Explanation:
The pointer points to % since it is incremented twice and again
decremented by 2, it points to '%d\n' and 300 is printed.
94) main(){
 char a[100];
 a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
 abc(a);
}
abc(char a[]){
 a++;
     printf("%c",*a);
 a++;
 printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a
points to 'b' then after incrementing to 'c' so bc will be printed.
               
95) func(a,b)
int a,b;
{
 return( a= (a==b) );
} http://www.exforsys.com Page 30 6/26/2004
30
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
 }
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another
function  2 and 3, integers. When this function is  invoked from
main, the following substitutions for formal parameters take place:
func for pf, 3 for val1 and 6 for val2. This function returns the
result of the operation performed by the function 'func'. The
function func has two integer parameters. The formal parameters
are substituted as 3 for a and 6 for b. since 3 is not equal to 6,
a==b returns 0. therefore the function returns 0 which in turn is
returned by the function 'process'.
96) void main()
{
 static int i=5;
 if(--i){
  main();
  printf("%d ",i);
 }
}
Answer:
 0 0 0 0
Explanation:
 The variable "I" is declared as static, hence memory for I will be
allocated for only once, as it encounters the statement. The function
main() will be called recursively unless I becomes  equal to 0, and since
main() is recursively called, so the value of static I ie., 0 will be printed
every time the control is returned.
97) void main()
{
 int k=ret(sizeof(float));
 printf("\n here value is %d",++k);
}
int ret(int ret)
{
 ret += 2.5;
 return(ret);
}
Answer:
 Here value is 7 http://www.exforsys.com Page 31 6/26/2004
31
Explanation:
The int ret(int ret), ie., the function name and the argument name
can be the same.
 Firstly, the function ret() is called in which the sizeof(float) ie., 4 is
passed,  after the first expression the value in ret will be 6, as ret is
integer hence the value stored in ret will have implicit type conversion
from float to int. The ret is returned in main() it is printed after and
preincrement.

98) void main()
{
 char a[]="12345\0";
 int i=strlen(a);
 printf("here in 3 %d\n",++i);
}
Answer:
here in 3 6
Explanation:
 The char array 'a' will hold the initialized string, whose length will
be counted from 0 till the null character. Hence the 'I' will hold the value
equal to 5, after the pre-increment in the printf statement, the 6 will be
printed.

99) void main()
{
 unsigned giveit=-1;
 int gotit;
 printf("%u ",++giveit);
 printf("%u \n",gotit=--giveit);
}
Answer:
 0 65535
Explanation:

100) void main()
{
 int i;
 char a[]="\0";
 if(printf("%s\n",a))
  printf("Ok here \n");
 else
  printf("Forget it\n");
}
Answer:
 Ok here
Explanation:
Printf will return how many characters does it print. Hence
printing a null character returns 1 which makes the if
statement true, thus "Ok here" is printed.

101) void main()
{
 void *v; http://www.exforsys.com Page 32 6/26/2004
32
 int integer=2;
 int *i=&integer;
 v=i;
 printf("%d",(int*)*v);
}
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
Void pointer is a generic pointer type. No pointer arithmetic can
be done on it. Void pointers are normally used for,
1. Passing generic pointers to functions and returning such
pointers.
2. As a intermediate pointer type.
3. Used when the exact pointer type will be known at a later
point of time.
102) void main()
{
 int i=i++,j=j++,k=k++;
printf(“%d%d%d”,i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is available to use in program code from the point of
its declaration.
So expressions such as  i = i++ are valid statements. The i, j and k
are automatic variables and so they contain some garbage value.
Garbage in is garbage out (GIGO).
103) void main()
{
 static int i=i++, j=j++, k=k++;
printf(“i = %d j = %d k = %d”, i, j, k);
}
Answer:
i = 1 j = 1 k = 1
Explanation:
Since static variables are initialized to zero by default.
104) void main()
{
 while(1){
  if(printf("%d",printf("%d")))
   break;
  else
   continue;
 }
}
Answer:
Garbage values
Explanation: http://www.exforsys.com Page 33 6/26/2004
33
The inner printf executes first to print some garbage value. The
printf returns no of characters printed and this value also cannot be
predicted. Still the outer printf  prints something and so returns a
non-zero value. So it encounters the break statement and comes
out of the while statement.
104) main()
{
 unsigned int i=10;
 while(i-->=0)
  printf("%u ",i);
}
Answer:
10 9 8 7 6 5 4 3 2 1 0 65535 65534…..
Explanation:
Since i is an unsigned integer it can never become negative. So the
expression i-- >=0  will always be true, leading to an infinite loop.

105) #include<conio.h>
main()
{
 int x,y=2,z,a;
 if(x=y%2) z=2;
 a=2;
 printf("%d %d ",z,x);
}
 Answer:
Garbage-value 0
Explanation:
The value of y%2 is 0. This value is assigned to x. The condition
reduces to if (x) or in other words if(0) and so z goes uninitialized.
Thumb Rule: Check all control paths to write bug free code.
106) main()
{
 int a[10];
 printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
 *a and -*a cancels out. The result is as simple as 1 + 3 = 4 !
107) #define prod(a,b) a*b
main()
{
 int x=3,y=4;
 printf("%d",prod(x+2,y-1));
}
Answer:
10 http://www.exforsys.com Page 34 6/26/2004
34
Explanation:
 The macro expands and evaluates to as:
 x+2*y-1 => x+(2*y)-1 => 10
108) main()
{
 unsigned int i=65000;
 while(i++!=0);
 printf("%d",i);
}
Answer:
 1
Explanation:
Note the semicolon after the while statement. When the value of i
becomes 0 it comes out of while loop. Due to post-increment on i
the value of i while printing is 1.
 
109) main()
{
 int i=0;
 while(+(+i--)!=0)
  i-=i++;
 printf("%d",i);
}
Answer:
-1
Explanation:
Unary + is the only dummy operator in C. So it has no effect on
the expression and now the while loop is,  while(i--!=0) which is
false and so breaks out of while loop. The value –1 is printed due
to the post-decrement operator.
 
113) main()
{
 float f=5,g=10;
 enum{i=10,j=20,k=50};
 printf("%d\n",++k);
 printf("%f\n",f<<2);
 printf("%lf\n",f%g);
 printf("%lf\n",fmod(f,g));
}
Answer:
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float
Explanation:
  Enumeration constants cannot be modified, so you  cannot apply
++.
  Bit-wise operators and % operators cannot be applied on float
values.
  fmod() is to find the modulus values for floats as % operator is for
ints.   http://www.exforsys.com Page 35 6/26/2004
35
110)   main()
{
 int i=10;
 void pascal f(int,int,int);
f(i++,i++,i++);
 printf(" %d",i);
}
void pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}
Answer:
Compiler error:  unknown type integer
Compiler error:  undeclared function write
Explanation:
Pascal keyword doesn’t mean that pascal code can be used. It
means that the function follows Pascal argument passing mechanism in
calling the functions.
111)  void pascal f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
void cdecl f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
main()
{
 int i=10;
f(i++,i++,i++);
 printf(" %d\n",i);
i=10;
f(i++,i++,i++);
printf(" %d",i);
}
Answer:
10 11 12 13
12 11 10 13
Explanation:
Pascal argument passing mechanism forces the arguments to be
called from left to right. cdecl is the normal C argument passing
mechanism where the arguments are passed from right to left.

112). What is the output of the program given below
main()
    {
       signed char i=0;
       for(;i>=0;i++) ;
       printf("%d\n",i);
    }
Answerhttp://www.exforsys.com Page 36 6/26/2004
36
  -128
Explanation
Notice the semicolon at the end of the for loop. THe initial
value of the i is set to 0. The inner loop executes to
increment the value from 0 to 127 (the positive range of
char) and then it rotates to the negative value of -128. The
condition in the for loop fails and so comes out of the for
loop. It prints the current value of i that is -128.
 
113) main()
    {
       unsigned char i=0;
       for(;i>=0;i++) ;
       printf("%d\n",i);
    }
Answer
 infinite loop
Explanation
The difference between the previous question and this one is that
the char is declared to be unsigned. So the i++ can never yield negative
value and i>=0 never becomes false so that it can come out of the for
loop.
114) main()
     {
       char i=0;
       for(;i>=0;i++) ;
       printf("%d\n",i);
       
 }
Answer:
  Behavior is implementation dependent.
Explanation:
The detail if the char is signed/unsigned by default is
implementation dependent. If the implementation treats the char
to be signed by default the program will print –128 and terminate.
On the other hand if it considers char to be unsigned by default, it
goes to infinite loop.
Rule:
You can write programs that have implementation
dependent behavior. But dont write programs that depend on such
behavior.
115) Is the following statement a declaration/definition. Find what does it
mean?
int (*x)[10];
Answer
  Definition.
 x is a pointer to array of(size 10) integers.
  Apply clock-wise rule to find the meaning of this definition. http://www.exforsys.com Page 37 6/26/2004
37
116). What is the output for the program given below
     typedef enum errorType{warning, error, exception,}error;
     main()
    {
        error g1;
        g1=1;
        printf("%d",g1);
     }
Answer
  Compiler error: Multiple declaration for error
Explanation
The name error is used in the two meanings. One means
that it is a enumerator constant with value 1. The another use is
that it is a type name (due to typedef) for enum errorType. Given a
situation the compiler cannot distinguish the meaning of error to
know in what sense the error is used:
 error g1;
g1=error;
 // which error it refers in each case?
When the compiler can distinguish between usages then it
will not issue error (in pure technical terms, names can only be
overloaded in different namespaces).
Note: the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for
programmer’s convenience.
 

117)          typedef struct error{int warning, error,
exception;}error;
     main()
    {
        error g1;
        g1.error =1;
        printf("%d",g1.error);
     }
Answer
  1
Explanation
The three usages of name errors can be distinguishable by the
compiler at any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as
in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by the
variable name as in :
g1.error =1;
         printf("%d",g1.error);
    typedef struct error{int warning, error, exception;}error; http://www.exforsys.com Page 38 6/26/2004
38
This can be used to define variables without using  the preceding struct
keyword as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it
is perfectly legal and valid.
Note
This code is given here to just explain the concept behind. In real
programming don’t use such overloading of names. It reduces the
readability of the code. Possible doesn’t mean that we should use it!
 
118) #ifdef something
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:
  Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation.
The name something is not already known to the compiler
making the declaration
int some = 0;
effectively removed from the source code.
119)  #if something == 0
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer
  0 0
Explanation
This code is to show that preprocessor expressions are not
the same as the ordinary expressions. If a name is  not
known the preprocessor treats it to be equal to zero.
120). What is the output for the following program
       main()
                            {
      int arr2D[3][3]; http://www.exforsys.com Page 39 6/26/2004
39
       printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0]))
);
               }
Answer
1
Explanation
This is due to the close relation between the arrays and
pointers. N dimensional arrays are made up of (N-1)
dimensional arrays.
 arr2D is made up of a 3 single arrays that contains 3
integers each .
The name arr2D refers to the beginning of all the 3 arrays.
*arr2D refers to the start of the first 1D array (of 3
integers) that is the same address as arr2D. So the
expression (arr2D == *arr2D) is true (1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero
doesn’t change the value/meaning. Again arr2D[0] is the
another way of telling *(arr2D + 0). So the expression
(*(arr2D + 0) == arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the
result is true(1) and the same is printed.
121) void main()
         {
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are represented
in memory”);
         }
 Answer
You can answer this if you know how values are represented
in memory
Explanation
~ (tilde operator or bit-wise negation operator) operates on
0 to produce all ones to fill the space for an integer. –1 is
represented in unsigned value as all 1’s and so both are
equal.
122) int swap(int *a,int *b)
{
 *a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
  int x=10,y=20;
 swap(&x,&y);
arr2D
arr2D[1]
arr2D[2]
arr2D[3] http://www.exforsys.com Page 40 6/26/2004
40
  printf("x= %d y = %d\n",x,y);
}
Answer
 x = 20 y = 10
Explanation
This is one way of swapping two values. Simple checking will help
understand this.
123)  main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b
124) main()
 {
  int i=5;
  printf("%d",++i++);
}
Answer:
  Compiler error: Lvalue required in function main
Explanation:
  ++i yields an rvalue.  For postfix ++ to operate an lvalue is
required.
125) main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++)
and ++*p++. Parenthesis just works as a visual clue for the
reader to see which expression is first evaluated.
126)
int aaa() {printf(“Hi”);}
int bbb(){printf(“hello”);}
iny ccc(){printf(“bye”);}
main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2](); http://www.exforsys.com Page 41 6/26/2004
41
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions
that takes no arguments and returns the type int. By the
assignment ptr[0] = aaa; it means that the first function pointer in
the array is initialized with the address of the function aaa.
Similarly, the other two array elements also get initialized with the
addresses of the functions bbb and ccc. Since ptr[2] contains the
address of the function ccc, the call to the function ptr[2]() is same
as calling ccc(). So it results in printing  "bye".
127)
main()
{
int i=5;
printf(“%d”,i=++i ==6);
}
Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because == is of
higher precedence than = operator. In the inner expression, ++i is
equal to 6 yielding true(1). Hence the result.
128) main()
{
   char p[ ]="%d\n";
p[1] = 'c';
printf(p,65);
}
Answer:
A
Explanation:
Due to the assignment p[1] = ‘c’ the string becomes, “%c\n”.
Since this string becomes the format string for printf and ASCII
value of 65 is ‘A’, the same gets printed.

129) void ( * abc( int, void ( *def) () ) ) ();
Answer::
 abc is a  ptr to a  function which takes 2 parameters .(a). an
integer variable.(b).        a ptrto a funtion which returns void. the
return type of the function is  void.
Explanation:
  Apply the clock-wise rule to find the result.
130) main()
{
while (strcmp(“some”,”some\0”))  http://www.exforsys.com Page 42 6/26/2004
42
printf(“Strings are not equal\n”);
 }
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference.
So “some” and “some\0” are equivalent. So, strcmp returns 0
(false) hence breaking out of the while loop.
131) main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with characters, ‘\0’ is not
appended automatically to the string. Since str1 doesn’t have null
termination, it treats whatever the values that are in the following
positions as part of the string until it randomly reaches a ‘\0’. So
str1 and str2 are not the same, hence the result.

132) main()
{
int i = 3;
for (;i++=0;) printf(“%d”,i);
}
Answer:
  Compiler Error: Lvalue required.
Explanation:
As we know that increment operators return rvalues  and
hence it cannot appear on the left hand side of an
assignment operation.

133) void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas
calloc returns the allocated memory space initialized to zeros. http://www.exforsys.com Page 43 6/26/2004
43

134) void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
 printf(“%d”, i);
}
Answer:
  32767
Explanation:
Since i is static it is initialized to 0. Inside the while loop the
conditional operator evaluates to false, executing  i--. This
continues till the integer value rotates to positive value (32767).
The while condition becomes false and hence, comes  out of the
while loop, printing the i value.
135) main()
{
  int i=10,j=20;
 j = i, j?(i,j)?i:j:j;
  printf("%d %d",i,j);
}
Answer:
10 10
Explanation:
  The Ternary operator ( ? : ) is equivalent for if-then-else
statement. So the question can be written as:
  if(i,j)
          {
if(i,j)
       j = i;
  else
      j = j;  
  }
    else
  j = j;
136) 1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.
Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a constant
char )
 *a='F'       : illegal
   a="Hi"       : legal
2. 'const' applies to 'a'  rather than to the value of a (constant
pointer to char ) http://www.exforsys.com Page 44 6/26/2004
44
 *a='F'       : legal
 a="Hi"       : illegal
3. Same as 1.
137) main()
{
  int i=5,j=10;
 i=i&=j&&10;
  printf("%d %d",i,j);
}
Answer:
1 10
Explanation:
The expression can be written as i=(i&=(j&&10)); The inner
expression (j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is
1. Hence the result.
138) main()
{
  int i=4,j=7;
 j = j || i++ && printf("YOU CAN");
  printf("%d %d", i, j);
}
Answer:
4 1
Explanation:
The boolean expression needs to be evaluated only till the truth
value of the expression is not known. j is not equal to zero itself
means that the expression’s truth value is 1. Because it is followed
by || and true || (anything) => true where (anything) will not be
evaluated. So the remaining expression is not evaluated and so the
value of i remains the same.
Similarly when && operator is involved in an expression, when any
of the operands become false, the whole expression’s truth value
becomes false and hence the remaining expression will not be
evaluated.    
false && (anything) => false where (anything) will  not be
evaluated.
139) main()
{
  register int a=2;
 printf("Address of a = %d",&a);
  printf("Value of a   = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
   & (address of ) operator cannot be applied on register
variables. http://www.exforsys.com Page 45 6/26/2004
45

140) main()
{
  float i=1.5;
 switch(i)
  {
  case 1: printf("1");
   case 2: printf("2");
   default : printf("0");
 }
}
Answer:
Compiler Error: switch expression not integral
Explanation:
   Switch statements can be applied only to integral types.
141) main()
{
  extern i;
 printf("%d\n",i);
  {
   int i=20;
  printf("%d\n",i);
  }
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and so using extern
has no use in resolving it.
142) main()
{
  int a=2,*f1,*f2;
 f1=f2=&a;
  *f2+=*f2+=a+=2.5;
 printf("\n%d %d %d",a,*f1,*f2);
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So changes
through f1 and f2 ultimately affects only the value of a.

143) main()
{
  char *p="GOOD";
 char a[ ]="GOOD";
printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d",
sizeof(p), sizeof(*p), strlen(p));
 printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
}
Answer: http://www.exforsys.com Page 46 6/26/2004
46
   sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
 sizeof(a) = 5, strlen(a) = 4
Explanation:
  sizeof(p) => sizeof(char*) => 2
 sizeof(*p) => sizeof(char) => 1
  Similarly,
 sizeof(a) => size of the character array => 5
When sizeof operator is applied to an array it returns the sizeof the
array and it is not the same as the sizeof the pointer variable. Here
the sizeof(a) where a is the character array and the size of the
array is 5 because the space necessary for the terminating NULL
character should also be taken into account.
144) #define DIM( array, type) sizeof(array)/sizeof(type)
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr, int));  
}
Answer:
10  
Explanation:
The size  of integer array of 10 elements is 10 * sizeof(int). The
macro expands to sizeof(arr)/sizeof(int) => 10 * sizeof(int) /
sizeof(int) => 10.
145) int DIM(int array[])
{
return sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr));  
}
Answer:
1  
Explanation:
Arrays cannot be passed to functions as arguments and only the
pointers can be passed. So the argument is equivalent to int *
array (this is one of the very few places where [] and * usage are
equivalent). The return statement becomes, sizeof(int *)/
sizeof(int) that happens to be equal in this case.
146) main()
{
 static int a[3][3]={1,2,3,4,5,6,7,8,9};
 int i,j;
 static *p[]={a,a+1,a+2};
  for(i=0;i<3;i++)
 {
   for(j=0;j<3;j++)
   printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j), http://www.exforsys.com Page 47 6/26/2004
47
   *(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
  }
}
Answer:
   1       1       1       1
   2       4       2       4
  3       7       3       7
   4       2       4       2
   5       5       5       5
  6       8       6       8
   7       3       7       3
   8       6       8       6
  9       9       9       9
Explanation:
  *(*(p+i)+j) is equivalent to p[i][j].
147) main()
{
  void swap();
 int x=10,y=8;    
  swap(&x,&y);
 printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
   *a ^= *b,  *b ^= *a, *a ^= *b;
}
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without using a
temporary variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that
may take any number of arguments (not no arguments) and
returns nothing. So this doesn’t issue a compiler error by the call
swap(&x,&y); that has two arguments.
This convention is historically due to pre-ANSI style (referred to as
Kernighan and Ritchie style) style of function declaration. In that
style, the swap function will be defined as follows,
void swap()
int *a, int *b
{
   *a ^= *b,  *b ^= *a, *a ^= *b;
}
where the arguments follow the (). So naturally the declaration for
swap will look like, void swap() which means the swap can take
any number of arguments.
148) main()
{
  int i = 257;
 int *iPtr = &i;
  printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) ); http://www.exforsys.com Page 48 6/26/2004
48
}
Answer:
  1 1
Explanation:
The integer value 257 is stored in the memory as, 00000001
00000001, so the individual bytes are taken by casting it to char *
and get printed.
149) main()
{
  int i = 258;
 int *iPtr = &i;
  printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
  2 1
Explanation:
The integer value 257 can be represented in binary as, 00000001
00000001. Remember that the INTEL machines are ‘small-endian’
machines.  Small-endian means that the lower order bytes are
stored in the higher memory addresses and the higher order bytes
are stored in lower addresses. The integer value 258 is stored in
memory as: 00000001 00000010.  
150) main()
{
  int i=300;
 char *ptr = &i;
  *++ptr=2;
 printf("%d",i);
}
Answer:
556
Explanation:
The integer value 300  in binary notation is: 00000001 00101100.
It is  stored in memory (small-endian) as: 00101100 00000001.
Result of the expression *++ptr = 2 makes the memory
representation as: 00101100 00000010. So the integer
corresponding to it  is  00000010 00101100 => 556.
151) #include <stdio.h>
main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
                  least = (*ptr<least ) ?*ptr :least;
printf("%d",least);
}
Answer:
0
Explanation:  http://www.exforsys.com Page 49 6/26/2004
49
After ‘ptr’ reaches the end of the string the value pointed by ‘str’ is
‘\0’. So the value of ‘str’ is less than that of ‘least’. So the value of
‘least’ finally is 0.
152) Declare an array of N pointers to functions returning pointers to functions
returning pointers to characters?
Answer:
  (char*(*)( )) (*ptr[N])( );
153) main()
{
struct student
{
char name[30];
struct date dob;
}stud;
struct date
        {
         int day,month,year;
         };
     scanf("%s%d%d%d", stud.rollno, &student.dob.day,
&student.dob.month,      &student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Inside the struct definition of ‘student’ the member of type struct
date is given. The compiler doesn’t have the definition of date
structure (forward  reference is not allowed in C in this case) so it
issues an error.
154) main()
{
struct date;
struct student
{
char name[30];
struct date dob;
}stud;
struct date
         {
         int day,month,year;
 };
scanf("%s%d%d%d", stud.rollno, &student.dob.day,
&student.dob.month, &student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Only declaration of struct date is available inside the structure
definition of ‘student’ but to have a variable of type struct date the
definition of the structure is required.  http://www.exforsys.com Page 50 6/26/2004
50
155) There were 10 records stored in “somefile.dat” but the following program
printed 11 names. What went wrong?
void main()
{
struct student
{
char name[30], rollno[6];
}stud;
FILE *fp = fopen(“somefile.dat”,”r”);
while(!feof(fp))
 {
       fread(&stud, sizeof(stud), 1 , fp);
puts(stud.name);
}
}
Explanation:
fread reads 10 records and prints the names successfully. It
will return EOF only when fread tries to read another record
and fails reading EOF (and returning EOF). So it prints the
last record again. After this only the condition feof(fp)
becomes false, hence comes out of the while loop.
156) Is there any difference between the two declarations,
1. int foo(int *arr[]) and
2. int foo(int *arr[2])
Answer:
No
Explanation:
Functions can only pass pointers and not arrays. The numbers that
are allowed inside the [] is just for more readability. So there is no
difference between the two declarations.
157) What is the subtle error in the following code segment?
void fun(int n, int arr[])
{
int *p=0;
int i=0;
while(i++<n)
  p = &arr[i];
*p = 0;
}
Answer & Explanation:
If the body of the loop never executes p is assigned no
address. So p remains NULL where *p =0 may result in
problem (may rise to runtime error “NULL pointer
assignment” and terminate the program).    
158) What is wrong with the following code?
int *foo()
{
int *s = malloc(sizeof(int)100);
assert(s != NULL); http://www.exforsys.com Page 51 6/26/2004
51
return s;
}
Answer & Explanation:
assert macro should be used for debugging and finding out bugs.
The check s != NULL is for error/exception handling and for that
assert shouldn’t be used. A plain if and the corresponding remedy
statement has to be given.
159) What is the hidden bug with the following  statement?
assert(val++ != 0);
Answer & Explanation:
Assert macro is used for debugging and removed in release
version. In assert, the experssion involves side-effects. So the
behavior of the code becomes different in case of debug version
and the release version thus leading to a subtle bug.
Rule to Remember:
Don’t use expressions that have side-effects in assert statements.
160) void main()
{
int *i = 0x400;  // i points to the address 400
*i = 0;  // set the value of memory location pointed by i;
}
Answer:
Undefined behavior
Explanation:
The second statement results in undefined behavior  because it
points to some location whose value may not be available for
modification.  This type of pointer in which the non-availability of
the implementation of the referenced location is known as
'incomplete type'.
161) #define assert(cond) if(!(cond)) \
  (fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\
 __FILE__,__LINE__), abort())
void main()
{
int i = 10;
if(i==0)
    assert(i < 100);
else
    printf("This statement becomes else for if in assert macro");
}
Answer:
No output
Explanation:
The else part in which the printf is there becomes the else for if in the
assert macro. Hence nothing is printed.
The solution is to use conditional operator instead of if statement,
#define assert(cond) ((cond)?(0): (fprintf (stderr, "assertion failed: \ %s,
file %s, line %d \n",#cond, __FILE__,__LINE__), abort())) http://www.exforsys.com Page 52 6/26/2004
52
Note:
However this problem of “matching with nearest else” cannot be
solved by the usual method of placing the if statement inside a
block like this,
#define assert(cond) { \
if(!(cond)) \
  (fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\
 __FILE__,__LINE__), abort()) \
}
162) Is the following code legal?
struct a
    {
int x;
 struct a b;
    }
Answer:
  No
Explanation:
Is it not legal for a structure to contain a member that is of the
same
type as in this case. Because this will cause the structure
declaration to be recursive without end.
163) Is the following code legal?
struct a
    {
int x;
            struct a *b;
    }
Answer:
Yes.
Explanation:
*b is a pointer to type struct a and so is legal. The compiler knows,
the size of the pointer to a structure even before the size of the
structure
is determined(as you know the pointer to any type is of same size).
This type of structures is known as ‘self-referencing’ structure.
164) Is the following code legal?
typedef struct a
    {
int x;
 aType *b;
    }aType
Answer:
  No
Explanation:
The typename aType is not known at the point of declaring the
structure (forward references are not made for typedefs).
165) Is the following code legal?
typedef struct a aType; http://www.exforsys.com Page 53 6/26/2004
53
struct a
{
int x;
aType *b;
};
Answer:
 Yes
Explanation:
The typename aType is known at the point of declaring the
structure, because it is already typedefined.
166) Is the following code legal?
void main()
{
typedef struct a aType;
aType someVariable;
struct a
{
int x;
      aType *b;
              };
}
Answer:
  No
Explanation:
  When the declaration,
typedef struct a aType;
is encountered body of struct a is not known. This  is known as
‘incomplete types’.

167) void main()
{
printf(“sizeof (void *) = %d \n“, sizeof( void *));
 printf(“sizeof (int *)    = %d \n”, sizeof(int *));
 printf(“sizeof (double *)  = %d \n”, sizeof(double *));
 printf(“sizeof(struct unknown *) = %d \n”, sizeof(struct unknown *));
 }
Answer :
sizeof (void *) = 2
sizeof (int *)    = 2
sizeof (double *)  =  2
sizeof(struct unknown *) =  2
Explanation:
The pointer to any type is of same size.
168) char inputString[100] = {0};
To get string input from the keyboard which one of the following is better?
 1) gets(inputString)
 2) fgets(inputString, sizeof(inputString), fp)
Answer & Explanation:
The second one is better because gets(inputString)  doesn't know
the size of the string passed and so, if a very big input (here, more
than 100 chars) the charactes will be written past the input string. http://www.exforsys.com Page 54 6/26/2004
54
When fgets is used with stdin performs the same operation as gets
but is safe.
169) Which version do you prefer of the following two,
1) printf(“%s”,str);  // or the more curt one
2) printf(str);
Answer & Explanation:
Prefer the first one. If the str contains any  format characters like
%d then it will result in a subtle bug.
170) void main()
{
int i=10, j=2;
int *ip= &i, *jp = &j;
int k = *ip/*jp;
printf(“%d”,k);
}
Answer:
Compiler Error: “Unexpected end of file in comment started in line
5”.
Explanation:
The programmer intended to divide two integers, but by the
“maximum munch” rule, the compiler treats the operator
sequence / and * as /* which happens to be the starting of
comment. To force what is intended by the programmer,
int k = *ip/ *jp;
// give space explicity separating / and *
//or
int k = *ip/(*jp);
// put braces to force the intention
will solve the problem.
171) void main()
{
char ch;
for(ch=0;ch<=127;ch++)
printf(“%c   %d \n“, ch, ch);
}
Answer:
 Implementaion dependent
Explanation:
The char type may be signed or unsigned by default. If it is signed
then ch++ is executed after ch reaches 127 and rotates back to -
128. Thus ch is always smaller than 127.
172) Is this code legal?
int *ptr;
ptr = (int *) 0x400;
Answer:
  Yes
Explanation: http://www.exforsys.com Page 55 6/26/2004
55
The pointer ptr will point at the integer in the memory location
0x400.
173) main()
{
char a[4]="HELLO";
printf("%s",a);
}
Answer:
  Compiler error: Too many initializers
Explanation:
The array a is of size 4 but the string constant requires 6 bytes to
get stored.
174) main()
{
char a[4]="HELL";
printf("%s",a);
}
Answer:
  HELL%@!~@!@???@~~!
Explanation:
The character array has the memory just enough to hold the string
“HELL” and doesnt have enough space to store the terminating null
character. So it prints the HELL correctly and continues to print
garbage values till it  accidentally comes across a NULL character.
175) main()
{
  int a=10,*j;
 void *k;
  j=k=&a;
   j++;
  k++;
   printf("\n %u %u ",j,k);
}
Answer:
  Compiler error: Cannot increment a void pointer
Explanation:
Void pointers are generic pointers and they can be used only when
the type is not known and as an intermediate address storage
type. No pointer arithmetic can be done on it and you cannot apply
indirection operator (*) on void pointers.
176) main()
  {
   extern int i;
  {  int i=20;
   {
     const volatile unsigned i=30; printf("%d",i);
   }
    printf("%d",i);
  } http://www.exforsys.com Page 56 6/26/2004
56
    printf("%d",i);
 }
 int i;
177) Printf can be implemented by using  __________ list.
Answer:
  Variable length argument lists
178) char *someFun()
  {
  char *temp = “string constant";
  return temp;
  }
  int main()
  {
  puts(someFun());
  }
Answer:
  string constant
Explanation:
  The program suffers no problem and gives the output correctly because
the character constants are stored in code/data area and not allocated in stack,
so this doesn’t lead to dangling pointers.
179)  char *someFun1()
  {
  char temp[ ] = “string";
  return temp;
  }
  char *someFun2()
  {
  char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};
  return temp;
  }
  int main()
  {
  puts(someFun1());
  puts(someFun2());
  }
Answer:
 Garbage values.
Explanation:
 Both the functions suffer from the problem of dangling pointers. In
someFun1() temp is a character array and so the space for it is allocated in heap
and is initialized with character string “string”. This is created dynamically as the
function is called, so is also deleted dynamically  on exiting the function so the
string data is not available in the calling function main() leading to print some
garbage values. The function someFun2() also suffers from the same problem but
the problem can be easily identified in this case.