execute shellcode on a ret2ret exploit

Question

I am on Kali 32 bits (Debian-based). I disabled ASLR, NX and stack canaries. This is not a consecutive ret2ret exploit since I used the main return statement to jump to my buffer where the shellcode is located. Here is my code :

void foo(char *args)
{
    char buffer[512];
    strcpy(buffer, args);
}

int main (int argc, char *argv[])
{
    if(argc > 1)
        foo(argv[1]);
    else
        printf("no input args\n");
    printf("no good\n");
    return 0;
}

void exploit()
{
    printf("bravo !!!\n");
    exit(0);
}

My overflow on ebp actually happens at offset 520. It overwrites the least significant byte of ebp with 0x00.

My first test was to try to jump to exploit() so I filled out my buffer with 516 NOP and with the adress of exploit on the last 4 bytes. It worked but here is my first question : When the ret statement is executed in main, we jump somewhere in our buffer where the NOP are executed. Why is it that when I get to the address of exploit, it jumps automatically even though in my buffer I don't have the asm instruction for jump ?

My second question is : when I want this time to execute a shellcode with the exit(0) instruction : python -c 'print "\x90"*514+"\x31\xdb\xb0\x01\xcd\x80"' why do I get this instead and what is the meaning of the error:

Program received signal SIGSEGV, Segmentation fault.
_IO_new_file_write (f=0xb7ff59b0, data=0x0, n=-1209570250) at fileops.c:1286
1286    fileops.c: Aucun fichier ou dossier de ce type.

Answer 1

You are trying to put shellcode that does exit(0) in the saved return address? If you're trying to execute exit(0), you'll need the address of the 'exit' call and push something other than 0 as an argument :)