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Programming in C MCQ - Find the Output

1 Find the output of the following C code (assume #include and that the code is inside int main(void)):

int x = 5;

if (x = 0)
    printf("YES");
else
    printf("NO");

The condition uses assignment (=), not equality comparison (==). Assigning 0 makes the condition false, so the else branch prints NO.

2 Find the output of the following C code (assume #include and that the code is inside int main(void)):

int x = 10;

if (x > 5)
    printf("A");

if (x > 8)
    printf("B");
else
    printf("C");

The two if statements are independent. Both x > 5 and x > 8 are true. The else belongs only to the second if, so the output is AB.

3 Find the output of the following C code (assume #include and that the code is inside int main(void)):

int x = 4;

if (x > 2);
{
    printf("Hello");
}

The semicolon immediately after if forms an empty controlled statement. The following compound block is independent and prints Hello.

4 Find the output of the following C code (assume #include and that the code is inside int main(void)):

int a = -5;

if (a)
    printf("TRUE");
else
    printf("FALSE");

In C, zero is false and every nonzero scalar value is true. Because a is -5, TRUE is printed.

5 Find the output of the following C code (assume #include and that the code is inside int main(void)):

int a = 5, b = 10;

if (a > 3 && b < 8)
    printf("P");
else
    printf("Q");

a > 3 is true, but b < 8 is false. Logical AND requires both conditions to be true, so Q is printed.

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Question:

Find the output of the following C code (assume #include <stdio.h> and that the code is inside int main(void)):int i, j, sum = 0; for (i = 1;…

A. 24
B. 30
C. 32
D. 40

Correct answer: C

Explanation:

Only odd i+j pairs contribute: 6 for i=1, 8 for i=2, and 18 for i=3, totaling 32.

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Question

Find the output of the following C code (assume #include <stdio.h> and that the code is inside int main(void)):

int i, j, sum = 0;
for (i = 1; i <= 3; i++) {
    for (j = 1; j <= 4; j++) {
        if ((i + j) % 2 == 0)
            continue;
        sum += i * j;
    }
}
printf("%d", sum);

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