How to do the modulo operation in ARM assembly?

Question

I'm trying to add the values in two registers, and modulo them by 8. So, in C code, it would be like this

a = a + b;
c = a % 8;

how to do this above operation in ARM assembly.

Solution

Not all ARM processors have a direct instruction for division or modulo, so in most cases, a call to the modulo operation would end up as a function call to e.g. ___modsi3.

In this particular case, when doing modulo for 8, if the values can be assumed to be nonnegative, you can do the % 8 part as & 7. In that case, the assembly for your case would be:

add rA, rA, rB
and rC, rA, #7

Comments

[InfinitelyManic]

What case is this assuming since it fails for certain values; e.g., 71 mod 15 = 6 test=11. Unless I'm missing something.

[old_timer]

modulo 8 is an and with 7, because 8 is a power of 2. but modulo 15 is a completely different thing you cant use base 2 and operations you have to actually divide.

[InfinitelyManic]

@old_timer, yeah I assumed that may have been the case. I have used a subtraction based mod function; however, it's much slower than a div (where available) based mod function due to the loops.