Strange instructions in compiler output (was: A simple question)

[Anthony Schneider]

try compiling with the -static flag the gcc.  then 'disassemble execve'.

-Anthony.

On Sat, Mar 06, 2004 at 02:26:51PM +0000, chungwei Hsiung wrote:
> Hello everyone
>   Thanks for fellows' previous helps. I actually have a further question. I read an article that it says if I compile the following program
> 
> #include <stdio.h>
> int main(){
>   char *name[2];
>   name[0] = "/bin/sh";
>   name[1] = NULL;
> 
>   execve(name[0],name,NULL);
>   return 0;
> }
> 
> by 
> gcc -o shellcode -ggdb -static shellcode.c
> 
> when i "disassemble execve" inside gdb, I should be able to see the assembly code for execve, but I can't see those codes for execve().
> Does anyone know how I can get the assembly code and see how the execve() works??
> 
> btw, I am using gcc3.2.2
> any help is really appreciated
> 
> best regards
> Chungwei
> 
> 
> On Sat, 6 Mar 2004 10:02:09 +1030
> Greg 'groggy' Lehey <grog at FreeBSD.org> 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..
> > >
> > > 0x80481f8 <main>:       push   %ebp
> > > 0x80481f9 <main+1>:     mov    %esp,%ebp
> > > 0x80481fb <main+3>:     sub    $0x8,%esp
> > > 0x80481fe <main+6>:     and    $0xfffffff0,%esp
> > > 0x8048201 <main+9>:     mov    $0x0,%eax
> > > 0x8048206 <main+14>:    sub    %eax,%esp
> > > 0x8048208 <main+16>:    movl   $0x804a6ce,0xfffffff8(%ebp)
> > > 0x804820f <main+23>:    movl   $0x0,0xfffffffc(%ebp)
> > > 0x8048216 <main+30>:    sub    $0x4,%esp
> > > 0x8048219 <main+33>:    push   $0x0
> > > 0x804821b <main+35>:    lea    0xfffffff8(%ebp),%eax
> > > 0x804821e <main+38>:    push   %eax
> > > 0x804821f <main+39>:    pushl  0xfffffff8(%ebp)
> > > 0x8048222 <main+42>:    call   0x804823c <execve>
> > > 0x8048227 <main+47>:    add    $0x10,%esp
> > > 0x804822a <main+50>:    mov    $0x0,%eax
> > > 0x804822f <main+55>:    leave
> > > 0x8048230 <main+56>:    ret
> > >
> > > 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.
> > 
> > Greg
> > --
> > Note: I discard all HTML mail unseen.
> > Finger grog at FreeBSD.org for PGP public key.
> > See complete headers for address and phone numbers.
> > 
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