Why does (x-(x/y)*y) evaluate to the same as x%y?

Question

#include <stdio.h>

int main() {
    int x = 9;
    int y = 2;
    int z = x - (x / y) * y;
    printf("%d", z);
    return 0;
}

Why does this code print the value of x % y ?

From a strictly mathematical standpoint, (x/y)*y is the same as x , so one might expect 0 to be printed if looked at in this way.

Answer 1

The reason has to due with how the / operator works with particular types.

The division operation x/y is performed as integer division because both operands are integer types. So the resulting value has the fractional portion of the division truncated.

Multiplying this result by y will therefore not necessarily be the same as x due to the truncation of the result value. The difference between this result and x is x%y .

From section 6.5.5 of the C standard :

6 When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded. If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a ; otherwise, the behavior of both a/b and a%b is undefined.

So the standard explicitly states that this equality holds.

If either operand of the division was a floating point type, then x-(x/y)*y would always be 0 or a value very close to 0 due to the inexact nature of floating point operations.