When you don't pass parameters (ie: f(1)), you must add 4 of more in
addition to pointing to the return address. (even if you have 2, 3 or more
of parameters, it's alway 4)
Here the code:
void
f()
{
char a[4];
int *b;
b = a + 12;
*b += 0x8;
}
void
main()
{
int x;
x = 0;
f();
x = 1;
printf("%d\n", x);
}
To know why, read the dissassembler code from gdb, the answer is in here
:-)
Eric
On Mon, 8 Apr 2002, darko wrote:
> Hi all,
>
> I've started to study buffer overflows. I wrote the following code:
>
> void f() {
> char a[4];
> int *b;
> b = a + 0x8;
> (*b) += 0x8;
> }
>
> main() {
> int x;
> x = 0;
> f();
> x = 1;
> printf("%d\n", x);
> }
>
> I want, after the call to f(), the program jump to printf() so the value of x
> should remain 0, not 1. I always get segmentation faults, bus errors, etc.
> and never that fuc*ing "x = 0" !!
> Tested on a Celeron 433, red hat 7.2, gcc 2.96.
>
> byez
> darko
>
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