My Ubuntu gives a "process exited" rather than segmentation fault

Asked Oct 17 '14 at 13:45

Active Oct 17 '14 at 21:46

Viewed 101 times

I'm currently studying return-oriented programming and I only have the basic knowledge of daily uses for linux. I am using Ubuntu 32-bit server 12.04. While I disabled aslr and am trying to get a segmentation fault(just by inserting like, 400 A's) on the program, it only shows me the following in gdb:

[Inferior 1 (process 1965) exited with code 015]

edit: the code is a simple one:

#include <stdio.h>
int main(int argc, char *argv[])
{
char buf[256];
memcpy(buf, argv[1],strlen(argv[1]));
printf(buf);
}

When I just do it on the command line, it doesn't show me anything. How do I make it so that it gives me a segmentation fault? Plus, does anyone know if this is the appropriate OS version for running ctf problems?

Thanks.

edited Oct 17 '14 at 21:46

asked Oct 17 '14 at 13:45

BunnyRabbit

Which program are you trying to debug? – Bart Friederichs Oct 17 '14 at 13:48
Take a look at: http://stackoverflow.com/questions/2045314/why-cant-i-cause-a-seg-fault – TeTeT Oct 17 '14 at 17:58
You don't get a segfault because this code doesn't segfault. Forcing a segfault can be as easy as `memcpy(NULL, argv[0])` (untested ;-)). – Bart Friederichs Oct 19 '14 at 16:58
@BartFriederichs It's supposed to segfault because if the argv[1] I give exceeds buf[]'s size, it's supposed to overwrite what it's not supposed to at memcpy. – BunnyRabbit Nov 02 '14 at 09:34

My Ubuntu gives a "process exited" rather than segmentation fault

0 Answers0