1. 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.

1. 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.

1. 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.

1. #define int char

main()
{
            int i=65;
            printf("sizeof(i)=%d",sizeof(i));
}
Answer:
                        sizeof(i)=1
Explanation:
Since the #define replaces the string  int by the macro char

1. 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).

1. #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).

1. #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
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.
           

1. #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

1. #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.

C Aptitude questions : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18

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