Modulo operation with negative numbers

Question

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:

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

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

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

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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.

I don't think there isn't any need to check if the number is negative.

A simple function to find the positive modulo would be this -

Edit: Assuming N > 0 and N + N - 1 <= INT_MAX

int modulo(int x,int N){
    return (x % N + N) %N;
}

This will work for both positive and negative values of x.

Original P.S: also as pointed out by @chux, If your x and N may reach something like INT_MAX-1 and INT_MAX respectively, just replace int with long long int.

And If they are crossing limits of long long as well (i.e. near LLONG_MAX), then you shall handle positive and negative cases separately as described in other answers here.