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Operators

Arithmetic, comparison, logical and increment operators in C.

C has a rich set of operators. Most look the same as in other languages, but a few, such as ++ and the way integer division works, deserve special attention.

Arithmetic

C
Open in playground
#include <stdio.h>

int main(void) {
    int a = 17, b = 5;
    printf("a + b = %d\n", a + b);
    printf("a - b = %d\n", a - b);
    printf("a * b = %d\n", a * b);
    printf("a / b = %d\n", a / b);
    printf("a %% b = %d\n", a % b);
    printf("17.0 / 5 = %.1f\n", 17.0 / 5);
    return 0;
}
Output
a + b = 22
a - b = 12
a * b = 85
a / b = 3
a % b = 2
17.0 / 5 = 3.4

Remember: with two integers, / throws away the remainder.

Increment, decrement and compound assignment

x++ adds one to x, and x-- subtracts one. Compound operators such as += and *= update a variable in place.

C
Open in playground
#include <stdio.h>

int main(void) {
    int count = 10;
    count++;
    count += 5;
    count *= 2;
    count--;
    printf("count = %d\n", count);

    int i = 5;
    int x = i++;   /* x gets the old value, then i becomes 6 */
    int y = ++i;   /* i becomes 7 first, then y gets 7 */
    printf("x = %d, y = %d, i = %d\n", x, y, i);
    return 0;
}
Output
count = 31
x = 5, y = 7, i = 7

Comparison and logical operators

In C, a comparison gives 1 for true and 0 for false.

C
Open in playground
#include <stdio.h>

int main(void) {
    int age = 20;
    int has_id = 1;

    printf("%d\n", age >= 18);
    printf("%d\n", age == 21);
    printf("%d\n", age >= 18 && has_id);
    printf("%d\n", age < 13 || age > 65);
    printf("%d\n", !has_id);
    return 0;
}
Output
1
0
1
0
0

The ternary operator

condition ? a : b picks one of two values.

C
Open in playground
#include <stdio.h>

int main(void) {
    int n = 7;
    printf("%d is %s\n", n, n % 2 == 0 ? "even" : "odd");
    return 0;
}
Output
7 is odd

GATE-style practice questions

GATE-styleMCQ · one correct answer

What does the following program print?

#include <stdio.h>
int main(void) {
    int i = 5;
    int j = i++ + 10;
    int k = ++i * 2;
    printf("%d %d %d\n", i, j, k);
    return 0;
}
Show answer and explanationAnswer: B

i++ + 10 uses the old value 5 (j = 15) and then i becomes 6. ++i * 2 first makes i 7, then k = 14. Each statement modifies i only once, so the behaviour is well defined.

GATE-styleNAT · numerical answer

What does the following program print?

#include <stdio.h>
int main(void) {
    int a = 0, b = 5;
    if (a && ++b)
        printf("inside ");
    printf("%d\n", b);
    return 0;
}
Show answer and explanationAnswer: 5

&& short-circuits: since a is 0 (false), the right side ++b is never evaluated, so b stays 5 and "inside" is not printed.

GATE-styleNAT · numerical answer

What does the following program print?

#include <stdio.h>
int main(void) {
    int x = 10;
    int y = x > 5 ? x < 8 ? 1 : 2 : 3;
    printf("%d\n", y + (17 % 5) * 2 - 7 / 2);
    return 0;
}
Show answer and explanationAnswer: 3

The ternary groups right to left: x > 5 is true, then x < 8 is false, so y = 2. Then 17 % 5 = 2 and 7 / 2 = 3 (integer division): 2 + 2 × 2 − 3 = 3.