Date: Sat, 6 Mar 2004 15:25:17 -0500 From: Anthony Schneider <anthony@x-anthony.com> To: chungwei Hsiung <chsiung2@buffalo.edu> Cc: freebsd-hackers@freebsd.org Subject: Re: Strange instructions in compiler output (was: A simple question) Message-ID: <20040306202517.GA8384@x-anthony.com> In-Reply-To: <20040306142651.314430be@bear.bflony.adelphia.net> References: <4048CA38.6040203@yahoo.com> <20040305233209.GO67801@wantadilla.lemis.com> <20040306142651.314430be@bear.bflony.adelphia.net>
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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
>=20
> #include <stdio.h>
> int main(){
> char *name[2];
> name[0] =3D "/bin/sh";
> name[1] =3D NULL;
>=20
> execve(name[0],name,NULL);
> return 0;
> }
>=20
> by=20
> gcc -o shellcode -ggdb -static shellcode.c
>=20
> when i "disassemble execve" inside gdb, I should be able to see the assem=
bly 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??
>=20
> btw, I am using gcc3.2.2
> any help is really appreciated
>=20
> best regards
> Chungwei
>=20
>=20
> On Sat, 6 Mar 2004 10:02:09 +1030
> Greg 'groggy' Lehey <grog@FreeBSD.org> wrote:
>=20
> > 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.
> >=20
> > 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.
> >=20
> > If this *was* done with optimization, let's see the source code.
> >=20
> > Greg
> > --
> > Note: I discard all HTML mail unseen.
> > Finger grog@FreeBSD.org for PGP public key.
> > See complete headers for address and phone numbers.
> >=20
> _______________________________________________
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