Technical questions with answer in C programming language

1)

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.

      

2)

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. 

3)

int i;

     main()

{

int t;

for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))

             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

          

5)

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.

5)

 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. 

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