Question

I found the following assembly code and I have no idea what it is supposed to be doing (mainly because cmovg follows the movl instruction ):

pushl %ebp
movl %esp, %ebp
movl 8(%ebp), %edx
movl %edx, %eax
sarl $31, %eax
testl %edx, %edx
movl $1, %edx
cmovg %edx, %eax
popl %ebp
ret

So here is how I have interpreted it so far:
pushes onto stack

a new pointer (stack pointer) creates to point at the same location as base pointer

gets the input (let’s call it x)

copies x into register %eax (res = x)

res = res >> 31 sign extension

tests x

sets x = 1

if >, res = x

restores pointer

returns res

However, I am not sure what the significance of this subroutine is. To me it seems useless. I would appreciate it if you could point out what is being done here.

Answer 1

This code returns the sign of X. In C:

int sign(int x) {
    if (x>0)
        return 1;
    else if (x==0)
        return 0;
    else
        return -1;
}

The instruction sarl $31, %eax will put -1 in eax if it was negative, or 0 otherwise. Then the cmovg instruction will replace this value with 1 if x was positive.