Why does 5/2 give 2 in C? Integer division explained

The surprise

Many students write printf("%d", 5/2); and expect 2.5. The program prints 2. Nothing is broken. C is doing exactly what the rules say.

Key idea

When both numbers are integers, C does integer division. The decimal part is thrown away. It is not rounded.

Why it happens

C looks at the types of the two operands before it divides:

• int / int gives an int, so 5/2 is 2
• float / int or int / float gives a float, so 5.0/2 is 2.5

In maths terms, C gives you the quotient only:

5 = 2 × 2 + 1

Here the quotient is 2 and the remainder is 1. The remainder is what the % operator gives you.

How to get 2.5

Make at least one side a real number:

#include <stdio.h>

int main() {
    int a = 5, b = 2;
    printf("%d\n", a / b);          /* 2   */
    printf("%d\n", a % b);          /* 1   */
    printf("%.1f\n", (float)a / b); /* 2.5 */
    return 0;
}

The (float) before a is called type casting. It turns a into a float before the division happens.

Common mistake

Writing (float)(a / b) does not work. The division is done first, so you get 2 and then it becomes 2.0. Cast one operand, not the answer.

Quick practice

1. What does 7/2 print? What does -7/2 print?
2. What does 7 % 2 print?
3. Fix this so it prints 3.5: printf("%f", 7/2);

Exam tip

Board exams love this trap. If a program uses int variables for an average, say clearly in your answer that the decimal part is lost.

Remember: the types decide the answer, not the numbers.