I'm playing with one stack overflow example. This example looks like this:
void return_input (void){
char array[30];
gets (array);
printf("%s\n", array);
}
main() {
return_input();
return 0;
}
All this code is in the file called overflow.c. We have vulnerable function called return_input, particularly it's 30 byte char array. I compiled it and opened vulnerable function in gdb and got following output:
(gdb) disas return_input
0x08048464 <+0>: push %ebp
0x08048465 <+1>: mov %esp,%ebp
0x08048467 <+3>: sub $0x48,%esp
0x0804846a <+6>: mov %gs:0x14,%eax
0x08048470 <+12>: mov %eax,-0xc(%ebp)
0x08048473 <+15>: xor %eax,%eax
0x08048475 <+17>: lea -0x2a(%ebp),%eax
0x08048478 <+20>: mov %eax,(%esp)
0x0804847b <+23>: call 0x8048360 <gets@plt>
0x08048480 <+28>: lea -0x2a(%ebp),%eax
0x08048483 <+31>: mov %eax,(%esp)
0x08048486 <+34>: call 0x8048380 <puts@plt>
0x0804848b <+39>: mov -0xc(%ebp),%eax
0x0804848e <+42>: xor %gs:0x14,%eax
0x08048495 <+49>: je 0x804849c <return_input+56>
0x08048497 <+51>: call 0x8048370 <__stack_chk_fail@plt>
0x0804849c <+56>: leave
0x0804849d <+57>: ret
End of assembler dump.
As you see from the function prologue we reserved hex48(dec 72) bytes on the stack for local variables. First I was trying to find the address where our vulnerable array starts on the stack. I think it's -0x2a(%ebp), am I right? Hex2a is 42 decimal. As I understand it means that we can write safely 42 bytes before we start to overwrite EBP saved on the stack. But when I run this example it's enough to right only 37 bytes to get segmentation fault:
rustam@rustam-laptop:~/temp/ELF_reader$ ./overflow
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
Segmentation fault (core dumped)
How is 37 bytes enough to overflow buffer? If our local char array is -42 bytes from saved EBP