p="%d\n";

    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

Blog Author: Vijay Kumar

Go to: Java Aptitude

for(i=1;i>-2;i--)

    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.

Blog Author: Vijay Kumar

Go to: Java Aptitude

if(a,b,x,y)

    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.

Blog Author: Vijay Kumar

Go to: Java Aptitude

i = abc();

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.

Blog Author: Vijay Kumar

Go to: Java Aptitude

if(i && j++)

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 i.e. false for if. 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.

Blog Author: Vijay Kumar

Go to: Java Aptitude

FILE *ptr;

    #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.

Blog Author: Vijay Kumar

Go to: Java Aptitude

int one_d[]={1,2,3};

    # 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.

Blog Author: Vijay Kumar

Go to: Java Aptitude

main(int argc, char **argv)

    main(int argc, char **argv)

{

 printf("enter the character");

 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. 

Blog Author: Vijay Kumar

Go to: Java Aptitude

int i=_l_abc(10);

    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.

Blog Author: Vijay Kumar

Go to: Java Aptitude

printf("%d",i+++j);

    main()

{

      int i=5,j=6,z;

      printf("%d",i+++j);

     }

Answer:

11

Explanation:

the expression i+++j is treated as (i++ + j)   

Blog Author: Vijay Kumar

Go to: Java Aptitude