1. Computer Security Buffer Overflow lab Eu-Jin Goh

2. Setting up Environment Demo

3. target2.c int foo( char *arg, char *out ) { strcpy( out, arg ); return 0; } int main( int argc, char *argv[] ) { char buf[64]; if ( argc != 2 ) { … } foo( argv[1], buf ); return 0; }

4. Stack in target2 – layout argv[1] == argv[0] == “/tmp/target2” argc $ra – to which main() will return $fp – for main’s stack frame buf[64] ptr to buf == “out”// args to foo() ptr to argv[1] == “arg”// args to foo() foo’s $ra

5. sploit2 Need: 1.Exploit string length –Need exploit str to write from buf[0] to main $ra –Find main $ra and &(buf[0]) and subtract –Why not foo’s $ra? 2.Address of the buffer (“buf” in target2) address we want program to jump to Caveat: buf adr depends exploit string size Why? because exploit str above buf on stack But once exploit str len fixed, adr of buf won’t change.

6. Details 1.Size of overflow buffer Buf addr = 0x9ffffdb0 reg ebp = 0x9ffffdf8 Difference is 0x48 = 72 Buffer size = 72 + 4 + 4 + 1 = 81 2.Find &buf again when str len is 81 Buf = 0x9ffffe60

7. Crafting the exploit string Want target to jump to start of buf, place shellcode (size 45 bytes) at the start of the string $ra exists at offset 76 need exploit string[76] to contain the addr target2 buf (0x9ffffe60)