For remainder:
If (1 << (Bitwidth / 2)) % Divisor == 1, we can add the high and low halves
together and use a (Bitwidth / 2) urem. If (BitWidth /2) is a legal integer
type, this urem will be expand by DAGCombiner using multiply by magic
constant. We do have to take into account that adding high and low
together can produce a carry, making it a (BitWidth / 2)+1 bit number.
So we need to also add back in the carry from the first addition.
For division:
We can use the above trick to compute the remainder, subtract that
remainder from the dividend, then multiply by the multiplicative
inverse of the Divisor modulo (1 << BitWidth).
This is based on the section "Remainder by Summing Digits" in
Hacker's delight.
The remainder trick is similar to a trick you may have learned for
determining if a decimal number is divisible by 3. You can add all the
digits together and see if the sum is divisible by 3. If you're not sure
if the sum is divisible by 3, you can add its digits together. This
can be repeated until you have a single decimal digit. If that digit
is 3, 6, or 9, then the original number is divisible by 3. This works
because 10 % 3 == 1.
gcc already does this same trick. There are additional tricks gcc
does urem as well as srem, udiv, and sdiv that I plan to add in
future patches.
Reviewed By: RKSimon
Differential Revision: https://reviews.llvm.org/D130862
38ffa2bb96