Modulo operation with negative numbers

C99 requires that when a/b is representable:

(a/b) * b + a%b shall equal a

This makes sense, logically. Right?

Let's see what this leads to:

Example A. 5/(-3) is -1

=> (-1) * (-3) + 5%(-3) = 5

This can only happen if 5%(-3) is 2.

Example B. (-5)/3 is -1

=> (-1) * 3 + (-5)%3 = -5

This can only happen if (-5)%3 is -2

The % operator in C is not the modulo operator but the remainder operator.

Modulo and remainder operators differ with respect to negative values.

With a remainder operator, the sign of the result is the same as the sign of the dividend (numerator) while with a modulo operator the sign of the result is the same as the divisor (denominator).

C defines the % operation for a % b as:

  a == (a / b * b) + a % b

with / the integer division with truncation towards 0. That's the truncation that is done towards 0 (and not towards negative inifinity) that defines the % as a remainder operator rather than a modulo operator.

Based on the C99 Specification: a == (a / b) * b + a % b

We can write a function to calculate (a % b) == a - (a / b) * b!

int remainder(int a, int b){    return a - (a / b) * b;}

For modulo operation, we can have the following function (assuming b > 0)

int mod(int a, int b){    int r = a % b;    return r < 0 ? r + b : r;}

My conclusion is that a % b in C is a remainder operation and NOT a modulo operation.