Question 1

What will be the output of the program?

#include<stdio.h>
int main()
{
    int k, num = 30;
    k = (num < 10) ? 100 : 200;
    printf("%d\n", num);
    return 0;
}

Answer

Answer: B

No answer description is available. Let's discuss.

Question 2

What will be the output of the program?

#include<stdio.h>
int main()
{
    int a=0, b=1, c=3;
    *((a) ? &b : &a) = a ? b : c;
    printf("%d, %d, %d\n", a, b, c);
    return 0;
}

Answer

Answer: C

Step 1: int a=0, b=1, c=3; here variable a, b, and c are declared as integer type and initialized to 0, 1, 3 respectively.

Step 2: *((a) ? &b : &a) = a ? b : c; The right side of the expression(a?b:c) becomes (0?1:3). Hence it return the value '3'.

The left side of the expression *((a) ? &b : &a) becomes *((0) ? &b : &a). Hence this contains the address of the variable a *(&a).

Step 3: *((a) ? &b : &a) = a ? b : c; Finally this statement becomes *(&a)=3. Hence the variable a has the value '3'.

Step 4: printf("%d, %d, %d\n", a, b, c); It prints "3, 1, 3".

Question 3

What will be the output of the program?

#include<stdio.h>
int main()
{
    float a = 0.7;
    if(0.7 > a)
        printf("Hi\n");
    else
        printf("Hello\n");
    return 0;
}

Answer

Answer: A

if(0.7 > a) here a is a float variable and 0.7 is a double constant. The double constant 0.7 is greater than the float variable a. Hence the if condition is satisfied and it prints 'Hi'
Example:

#include<stdio.h>
int main()
{
    float a=0.7;
    printf("%.10f %.10f\n",0.7, a);
    return 0;
}

Output:
0.7000000000 0.6999999881

Question 4

What will be the output of the program?

#include<stdio.h>
int main()
{
    unsigned int i = 65536; /* Assume 2 byte integer*/
    while(i != 0)
        printf("%d",++i);
    printf("\n");
    return 0;
}

Answer

Answer: D

Here unsigned int size is 2 bytes. It varies from 0,1,2,3, ... to 65535.

Step 1:unsigned int i = 65536; here variable i becomes '0'(zero). because unsigned int varies from 0 to 65535.

Step 2: while(i != 0) this statement becomes while(0 != 0). Hence the while(FALSE) condition is not satisfied. So, the inside the statements of while loop will not get executed.

Hence there is no output.

Note: Don't forget that the size of int should be 2 bytes. If you run the above program in GCC it may run infinite loop, because in Linux platform the size of the integer is 4 bytes.

Question 5

What will be the output of the program?

#include<stdio.h>
int main()
{
    char ch;
    if(ch = printf(""))
        printf("It matters\n");
    else
        printf("It doesn't matters\n");
    return 0;
}

Answer

Answer: B

printf() returns the number of charecters printed on the console.

Step 1: if(ch = printf("")) here printf() does not print anything, so it returns '0'(zero).
Step 2: if(ch = 0) here variable ch has the value '0'(zero).
Step 3: if(0) Hence the if condition is not satisfied. So it prints the else statements.
Hence the output is "It doesn't matters".

Note: Compiler shows a warning "possibly incorrect assinment".

Questions for: Control Instructions