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..
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
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I actually don't know how to use the optimization. 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..
#include <stdio.h>
int main(void) { char *name[2]; name[0] = "/bin/sh"; name[1] = NULL; execve(name[0], name, NULL); return(0); }
best regards Chungwei _______________________________________________ [EMAIL PROTECTED] mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-hackers To unsubscribe, send any mail to "[EMAIL PROTECTED]"
