I\'m having problems with return-to-libc exploit. The problem is that nothing happens, but no segmentation fault (and yes I\'m actually overflowing the stack).
This is m
Let's do some asm before:
Code
$ cat gets.c
int main(int argc, char **argv) {
char array[512];
gets(array);
}
Asm
$ gcc gets.c -o getsA.s -S -fverbose-asm
$ cat gets.s
....
.globl main
.type main, @function
main:
leal 4(%esp), %ecx #,
andl $-16, %esp #,
pushl -4(%ecx) # (1)
pushl %ebp # 2
movl %esp, %ebp #,
pushl %ecx # 3
subl $516, %esp #,
leal -516(%ebp), %eax #, tmp60
movl %eax, (%esp) # tmp60,
call gets # << break here
addl $516, %esp #, << or here to see the stack picture
popl %ecx # (3')
popl %ebp # (2')
leal -4(%ecx), %esp # (1')
ret
.size main, .-main
The prologue and epilogue (these are with alignment code) is described in detail here Understanding the purpose of some assembly statements
Stack layout:
(char) array[0]
...
(char) array[511]
(32bit) $ecx - pushed by 3 - it was the address on the stack of the eip which main will return to
(32bit) $ebp - pushed by 2
(32bit) $esp - pushed by 1 - change the $esp to the original value
So, if you want to change a return address of main, you should not to change address in stack which will be used by ret, but also to repeat the values saved in stack by (1),(2),(3) pushes. Or you can embed a new return address in the array itself and overwrite only (3) by the your new stack address+4. (use 516 byte string)
I suggest you use this source code to hack it:
$ cat getss.c
f()
{
char array[512];
gets(array);
}
int main(int argc, char **argv) {
f();
}
because f have no problems with stack realignement
.globl f
.type f, @function
f:
pushl %ebp #
movl %esp, %ebp #,
subl $520, %esp #,
leal -512(%ebp), %eax #, tmp59
movl %eax, (%esp) # tmp59,
call gets #
leave
ret
.size f, .-f
Stack layout for f():
(char) array[0]
...
(char) array[511]
(32bit) old ebp
(32bit) return address
Breakpoint at ret instruction in f() with 520 bytes of "A"
(gdb) x/w $sp
0xXXXXXa3c: 0x41414141