On Friday, 5 March 2004 at 18:43:11 -0500, Chungwei Hsiung wrote: > Greg 'groggy' Lehey wrote: > >> On Friday, 5 March 2004 at 13:43:04 -0500, Chungwei Hsiung wrote: >> >> >>> Hello.. >>> I am super new to this list, and I have a simple question that I don't >>> know why it does that. I have a simple test program. I compile it, and >>> gdb to disassemble main. I got the following.. >>> >>> 0x8048201 <main+9>: mov $0x0,%eax >>> 0x8048206 <main+14>: sub %eax,%esp >>> ... >>> >>> I don't know if at line 5, we move zero to %eax. why do we need to sub >>>> eax, %esp? why do we need to substract 0 from the stack pointer?? >>> Any help is really appreciated. >> >> This is probably because you didn't optimize the output. You'd be >> surprised how many redundant instructions the compiler puts in under >> these circumstances. Try optimizing and see what the code looks like. >> >> If this *was* done with optimization, let's see the source code. > > Hello.. thank you very much for the reply > I actually don't know how to use the optimization.
Use the gcc command line options. See below.
>I just compile it with gcc 3.2.2, and use gdb to disassemble main to
>get this assembly. Is it possible I can get the non-redundent output?
>here is the code I compile..
>
> ...
The best way to look at the assembly output is to generate it directly
from the compiler. I get:
$ cc -O -pipe -mcpu=pentiumpro -S exec.c
$ cat exec.s
.LC0:
.string "/bin/sh"
...
main:
pushl %ebp
movl %esp, %ebp
subl $24, %esp
andl $-16, %esp
movl $.LC0, -8(%ebp)
leal -8(%ebp), %edx
movl $0, 4(%edx)
movl -8(%ebp), %eax
movl %eax, (%esp)
movl %edx, 4(%esp)
movl $0, 8(%esp)
call execve
movl $0, %eax
movl %ebp, %esp
popl %ebp
ret
This doesn't look that much like your code. Without the -O (optimize)
flag I get:
$ cc -pipe -mcpu=pentiumpro -S exec.c
$ cat exec.s
...
main:
pushl %ebp
movl %esp, %ebp
subl $24, %esp
andl $-16, %esp
movl $0, %eax
subl %eax, %esp
movl $.LC0, -8(%ebp)
So yes, it looks as if you're not optimizing.
Greg
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