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C Aptitude Questions and Answers



Predict the output or error(s) for the following:
36.    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.

37.       #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
            }


Answer:
100

38.      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).

39.       #include
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");

40.       #include
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

41.       #include
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

42.       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.

43.       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.

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