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C Aptitude Questions and Answers



Predict the output or error(s) for the following:
25.   main()
{
printf("%p",main);
}

Answer:
                        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.

26.       main()
{
clrscr();
}
clrscr();
           
Answer:
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).

27.       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.

28.       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

29.       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.

30.       main()
{
 char *p;
 p="Hello";
 printf("%c\n",*&*p);
}

Answer:
H

Explanation:
* 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.

31.       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.

32.       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.

33.     void main()
{
            int i=5;
            printf("%d",i++ + ++i);
}

Answer:
Output Cannot be predicted  exactly.

Explanation:
Side effects are involved in the evaluation of   i

34.       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.

35.       #include
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.

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