ARM __moddi3

When I was debugging an app on iPhone 4 (armv7), I found a function call:

Since ARM does not have Division/Modulo operators, compilers use other ways making your code work on devices like iPhone.

I followed the __moddi3 on iPhone 4, the moddi3() is provided by llvm’s compiler-rt (lib/builtins/moddi3.c).

Parameters a and b are 64-bit long and returned value is also 64-bit.

But in ARMv7, registers are 32-bit long. I’m trying to find out which registers are used.

I wrote a simple C file and compile to armv7 and x86_64

Compile to x86_64

Disassembe x86_64

x86_64 uses idivq, in this program:

Compile to armv7

Disassemble armv7

In armv7:

If both dividend and divisor are less than 0x100000000.

$r2 may be negative, like 0xfffffff3. ($r3 == 0).

means $r0 – $r2 == divisor

If signed integer $r2 is smaller than 0x80000000,
((long long) $r1 << 32 | $r0) - ((long long ) $r3 << 32 | $r2) == divisor.

ARM __moddi3 by @sskaje: