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Programming in C MCQ - Find the Output

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

int a = 2, b = 20, count = 0;
while (a < b) {
    a += 3;
    b -= 2;
    count++;
}
printf("%d %d %d", a, b, count);

After four iterations, a=14 and b=12, so a<b becomes false.

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

int a = 1, b = 5, count = 0;
while (a++ < 4 && --b > 1)
    count += a + b;
printf("%d %d %d", a, b, count);

The body adds 6 on each of three successful tests. The final a++ fails, incrementing a to 5 without changing b.

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

int i = 1, j, count = 0;
while (i <= 4) {
    j = i;
    while (j <= 4) {
        count += i;
        j += 2;
    }
    i++;
}
printf("%d", count);

The contributions are 2 (i=1), 4 (i=2), 3 (i=3), and 4 (i=4): total 13.

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

int i = 0, sum = 0;
while (i < 10) {
    i += 2;
    if (i % 4 == 0)
        continue;
    sum += i;
}
printf("%d", sum);

The values 2, 6, and 10 are added; 4 and 8 are skipped.

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

int n = 4625, result = 0;
while (n > 0) {
    int digit = n % 10;
    if (digit % 2 == 0)
        result = result * 10 + digit;
    n /= 10;
}
printf("%d", result);

Digits are read right to left (5,2,6,4); the even digits form 264.

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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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