I'm trying to understand how my computer works ;-)
Now I have a lot of questions...
if I compile the following function (with -O2):
func(char *sm){
char buffer[256];
int i;
for(i=0; i<=255; i++) buffer[i] = sm[i];
}
it gives me this assembler code:
0x08048358 <func+0>: push %ebp
0x08048359 <func+1>: mov %esp,%ebp
0x0804835b <func+3>: sub $0x108,%esp
0x08048361 <func+9>: mov 0x8(%ebp),%ecx
0x08048364 <func+12>: xor %edx,%edx
0x08048366 <func+14>: mov %esi,%esi
0x08048368 <func+16>: mov (%edx,%ecx,1),%al
0x0804836b <func+19>: mov %al,0xfffffef8(%edx,%ebp,1)
0x08048372 <func+26>: inc %edx
0x08048373 <func+27>: cmp $0xff,%edx
0x08048379 <func+33>: jle 0x8048368 <func+16>
0x0804837b <func+35>: leave
0x0804837c <func+36>: ret
0x0804837d <func+37>: lea 0x0(%esi),%esi
gcc allocates 264 (sub $0x108,%esp) bytes on the stack. But only the lower 256
are used. Without the gcc parameter "-O2" even 280 Bytes on the stack are
reserved.
For what are those unused bytes?
And what does "mov %esi,%esi" do? Nothing?
When will the "lea 0x0(%esi),%esi" instuction be executed?
There's an other small program:
long getesp() {
__asm__("movl %esp,%eax");
}
void main() {
printf("%08X\n",getesp());
}
Everytime I execute it, it gives me a slightly different value. Shouldn't the
esp register be the same value everytime?
I'm sorry to bore you with my stupid questions, I know, I should buy my self a
good assembly book. But the bookstores have closed these days ;-)
By the way, who knows a good book (x86, AT&T,linux/unix)? I don't even find an
explanation of the assembly commands (what does "lea" do?) in the internet.
Sorry for my miserable english. I'm still learning...
stefan mueller
