Hi Guy,
guy keren wrote:
AFAIK Valgrind does not detect neither stack nor static buffer
overflows at all.
[EMAIL PROTECTED]:~$ cat c2.c
#include <stdio.h>
void f(char* p_i )
{
char i[1024];
f(i);
}
int main()
{
f((char*)NULL);
return 0;
}
[EMAIL PROTECTED]:~$ gcc -Wall c2.c
[EMAIL PROTECTED]:~$ ./a.out
Segmentation fault (core dumped)
[EMAIL PROTECTED]:~$ valgrind
valgrind valgrind.bin valgrind-listener
[EMAIL PROTECTED]:~$ valgrind ./a.out
==5741== Memcheck, a memory error detector.
==5741== Copyright (C) 2002-2006, and GNU GPL'd, by Julian Seward et al.
==5741== Using LibVEX rev 1658, a library for dynamic binary translation.
==5741== Copyright (C) 2004-2006, and GNU GPL'd, by OpenWorks LLP.
==5741== Using valgrind-3.2.1-Debian, a dynamic binary instrumentation
framework.
==5741== Copyright (C) 2000-2006, and GNU GPL'd, by Julian Seward et al.
==5741== For more details, rerun with: -v
==5741==
==5741== Stack overflow in thread 1: can't grow stack to 0xBE68BF44
=
how do you interpret these 'Stack overflow in thread 1: can't grow
stack to 0xBE68BF44' messages?
Your program does a stack overflow. What Valgrind doesn't detect are
stack *buffer* overflows, which is something completely different.
Try (stolen from the Wikiperdia article on Valgrind btw):
int Static[5];
int func(void)
{
int Stack[5];
Static[5] = 0; /* boom! Static[0] to Static[4] exist, Static[5] is out of
bounds */
Stack [5] = 0; /* bang! Stack[0] to Stack[4] exist, Stack[5] is out of
bounds */
return 0;
}
regarding static buffers - a test program shows that indeed valgrind
does not report such overflows. even worse - it seems to hide errors
of writing into read-only global variables (apparently it allocates
global const buffers in read/write memory, while when loading the
program without valgrind, ld.so (or whoever) loads them into read-only
memory, and writes into them causes a crash.
Indeed.
Gilad
--guy.
--
Gilad Ben-Yossef
Chief Coffee Drinker
Codefidence Ltd.
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