Can somebody explain to me the following snippet of assembly:
mydiv:
pushl %ebp
movl %esp, %ebp
movl 8(%ebp), %edx ; get x
movl %edx, %eax
sarl $31, %edx ; divide by y
idivl 12(%ebp) ; return %eax
popl %ebp
ret
This is the equivalent to the following C function:
int mydiv( int x, int y )
{
return x / y;
}
The part which I'm having trouble understanding is the sarl instruction: why do you need to shift edx?
It's sign extending.
idivl has a 64-bit argument (edx:eax), so you need to ensure that the MSB contains the correct sign bits, based on the msb of eax.
So if eax is positive, it's msb will be 0, e.g. 5 -> 0000 ... 0101. If it's negative it's msb will be 1, e.g. -5 -> 1111 ... 1011. sarl performs an arithmetic right-shift, so edx will either be 0000 ... 0000 or 1111 ... 1111.