Hello. I've started to learn assembly this week, and I found a example that I couldn't understand, so, if anyone could explain it to me, I would appreciate.
int increment(int x) {
x = x + 1;
return(x);
}and this got translated to:
pushl %ebp
movl %esp, %ebp
incl 8(%ebp)
movl 8(%ebp), %eax
popl %ebp
retSo my question is, why incl/movl 8, and not 4 ? doesnt esp "move it" 4 places ?
Thank you very much
Jump to PostThe function return address is in the first 8 bytes. The 8 is the offset into the stack pointer whose address is in ebp register.
Jump to PostI just told you. Before a function is called the program pushes the return address onto the stack, which is 8 bytes. Then it pushes each of the parameters. The ebp register contains the stack address of where the return address was stored. So to get to the first byte …
The function return address is in the first 8 bytes. The 8 is the offset into the stack pointer whose address is in ebp register.
But why 8?
I just told you. Before a function is called the program pushes the return address onto the stack, which is 8 bytes. Then it pushes each of the parameters. The ebp register contains the stack address of where the return address was stored. So to get to the first byte of the parameter it has to add 8 to the ebp register.
but i'm working a 32 bit CPU, shouldn't the return address be 4 bytes instead of 8 bytes ? (sorry for the first misunderstanding)
depends on the compiler. What compiler did you use to produce that assembly code?
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