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.. thank you very much for the reply
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.
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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hello greg.. yes.. it does the difference.. thanks a lot for your help..
Chungwei
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