Fastest way to calculate a 128-bit integer modulo a 64-bit integer

Question

I have a 128-bit unsigned integer A and a 64-bit unsigned integer B. What\'s the fastest way to calculate A % B - that is the (64-bit) remainder from dividing A

Answer 1

You can use the division version of Russian Peasant Multiplication.

To find the remainder, execute (in pseudo-code):

X = B;

while (X <= A/2)
{
    X <<= 1;
}

while (A >= B)
{
    if (A >= X)
        A -= X;
    X >>= 1;
}

The modulus is left in A.

You'll need to implement the shifts, comparisons and subtractions to operate on values made up of a pair of 64 bit numbers, but that's fairly trivial (likely you should implement the left-shift-by-1 as X + X).

This will loop at most 255 times (with a 128 bit A). Of course you need to do a pre-check for a zero divisor.