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Programming in C MCQ - Find the Output

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

int x = 5, sum = 0;
do {
    sum += x;
} while (x-- > 2);
printf("%d %d", sum, x);

The body adds 5+4+3+2=14; the last comparison tests 2>2 and decrements x to 1.

42 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;
do {
    i++;
    if (i % 3 == 0)
        continue;
    sum += i;
} while (i < 8);
printf("%d", sum);

Values 3 and 6 are skipped; 1+2+4+5+7+8=27.

43 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;
do {
    j = 1;
    do {
        if (i + j > 5)
            break;
        count += j;
        j++;
    } while (j <= 4);
    i++;
} while (i <= 4);
printf("%d", count);

The four outer passes add 10, 6, 3, and 1, totaling 20.

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

int x = 25, count = 0;
do {
    count += x % 4;
} while ((x = x / 2) > 2);
printf("%d %d", count, x);

The body sees 25,12,6,3; remainders 1+0+2+3=6. The final division leaves x=1.

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

int i, j, count = 0;
for (i = 1; i <= 3; i++) {
    j = i;
    do {
        count += j;
        j--;
    } while (j > 0);
}
printf("%d", count);

The three outer passes add 1, then 2+1, then 3+2+1, totaling 10.

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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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