By @少仲
From 360 VulpeckerTeam
0x0 漏洞信息
影响所有Nexus手机和部分Android手机的漏洞,Google于2016/03/18发布了公告修复,具体请看链接.
http://www.cvedetails.com/cve-details.php?t=1&cve_id=cve-2015-1805X
http://source.android.com/security/advisory/2016-03-18.html
0x1 漏洞描述
在linux 内核3.16版本之前的fs/pipe.c当中,由于pipe_read和pipe_write没有考虑到拷贝过程中数据没有同步的一些临界情况,造成了拷贝越界的问题,因此有可能导致系统crash以及系统权限提升.这种漏洞又称之为” I/O vector array overrun”
0x2 代码分析
//摘自fs/pipe.c:
static ssize_t
pipe_read(struct kiocb *iocb, const struct iovec *_iov,
unsigned long nr_segs, loff_t pos)
{
struct file *filp = iocb->ki_filp;
struct pipe_inode_info *pipe = filp->private_data;
int do_wakeup;
ssize_t ret;
struct iovec *iov = (struct iovec *)_iov;
size_t total_len;
total_len = iov_length(iov, nr_segs);
/* Null read succeeds. */
if (unlikely(total_len == 0))
return 0;
do_wakeup = 0;
ret = 0;
__pipe_lock(pipe);
for (;;) {
int bufs = pipe->nrbufs;
if (bufs) {
int curbuf = pipe->curbuf;
struct pipe_buffer *buf = pipe->bufs + curbuf;
const struct pipe_buf_operations *ops = buf->ops;
void *addr;
size_t chars = buf->len;
int error, atomic;
if (chars > total_len)
chars = total_len;
error = ops->confirm(pipe, buf);
if (error) {
if (!ret)
ret = error;
break;
}
//(1)
atomic = !iov_fault_in_pages_write(iov, chars);
redo:
addr = ops->map(pipe, buf, atomic);
//(2)
error = pipe_iov_copy_to_user(iov, addr + buf->offset, chars, atomic);
ops->unmap(pipe, buf, addr);
if (unlikely(error)) {
/*
* Just retry with the slow path if we failed.
*/
//(3)
if (atomic) {
atomic = 0;
goto redo;
}
if (!ret)
ret = error;
break;
}
ret += chars;
buf->offset += chars;
buf->len -= chars;
/* Was it a packet buffer? Clean up and exit */
if (buf->flags & PIPE_BUF_FLAG_PACKET) {
total_len = chars;
buf->len = 0;
}
if (!buf->len) {
buf->ops = NULL;
ops->release(pipe, buf);
curbuf = (curbuf + 1) & (pipe->buffers - 1);
pipe->curbuf = curbuf;
pipe->nrbufs = --bufs;
do_wakeup = 1;
}
(5)//在这里更新total_len
total_len -= chars;
if (!total_len)
break; /* common path: read succeeded */
}
if (bufs) /* More to do? */
continue;
if (!pipe->writers)
break;
if (!pipe->waiting_writers) {
/* syscall merging: Usually we must not sleep
* if O_NONBLOCK is set, or if we got some data.
* But if a writer sleeps in kernel space, then
* we can wait for that data without violating POSIX.
*/
if (ret)
break;
if (filp->f_flags & O_NONBLOCK) {
ret = -EAGAIN;
break;
}
}
if (signal_pending(current)) {
if (!ret)
ret = -ERESTARTSYS;
break;
}
if (do_wakeup) {
wake_up_interruptible_sync_poll(&pipe->wait, POLLOUT | POLLWRNORM);
kill_fasync(&pipe->fasync_writers, SIGIO, POLL_OUT);
}
pipe_wait(pipe);
}
__pipe_unlock(pipe);
/* Signal writers asynchronously that there is more room. */
if (do_wakeup) {
wake_up_interruptible_sync_poll(&pipe->wait, POLLOUT | POLLWRNORM);
kill_fasync(&pipe->fasync_writers, SIGIO, POLL_OUT);
}
if (ret > 0)
file_accessed(filp);
return ret;
}
(1).首先pipe_read()函数会先循环读取iovec结构,并且通过iov_fault_in_pages_write()函数判断iov->len是否大于0,且iov->base指向的地址是否可写且处于用户态,之后返回atomic.
(2)如果atomic=1,则pipe_iov_copy_to_user -> __copy_to_user_inatomic ->
__copy_to_user_nocheck;如果atomic=0,则pipe_iov_copy_to_user -> copy_to_user -> access_ok.
(3).如果atomic为1,pipe_iov_copy_to_user拷贝出现错误,会进入redo的逻辑,将再次调用pipe_iov_copy_to_user函数进行拷贝,且将atomic置为0.但是pipe_iov_copy_to_user的第三个参数chars并没有更新,还是会拷贝total_len大小的数据
static int
pipe_iov_copy_to_user(struct iovec *iov, const void *from, unsigned long len,
int atomic)
{
unsigned long copy;
while (len > 0)
{
while (!iov->iov_len)
iov++;
copy = min_t(unsigned long, len, iov->iov_len);
if (atomic)
{
if (__copy_to_user_inatomic(iov->iov_base, from, copy))
//(4)
return -EFAULT;
}
else
{
if (copy_to_user(iov->iov_base, from, copy))
//(4)
return -EFAULT;
}
from += copy;
len -= copy;
iov->iov_base += copy;
//每次对iov->iov_len进行更新
iov->iov_len -= copy;
}
return 0;
}
4. 如果copy到某种情况出错返回,已经copy成功的iov->len会被减去但总长度total_len并不会同步减去.也就是说如果total_len是0x100,第一次消耗掉了x;再次进入redo逻辑后还是0x100,然而实际已经被消耗掉了x.
0x3 具体探究
假设有一个iov结构,total_len为0x40,len为0x20.
iov[0]: iov_base = 0xdead0000 iov_len = 0x10
iov[1]: iov_base = 0xdead1000 iov_len = 0x10
iov[2]: iov_base = 0xdead2000 iov_len = 0x10
iov[3]: iov_base = 0xdead3000 iov_len = 0x10
如果iov[1].iov_base的地址被设置成不可写入.那么第一次pipe_iov_copy_to_user()会返回失败.而iov->iov_base += copy,iov->iov_len -= copy.
iov[0]: iov_base = 0xdead0010 iov_len = 0
iov[1]: iov_base = 0xdead1000 iov_len = 0x10
iov[2]: iov_base = 0xdead2000 iov_len = 0x10
iov[3]: iov_base = 0xdead3000 iov_len = 0x10
现在,redo的逻辑发生在0xdead0010,它以某种方式被设置成可写,并且len仍未0x20.那么iov[1]和iov[2]都将被用掉.
iov[0]: iov_base = 0xdead0010 iov_len = 0
iov[1]: iov_base = 0xdead1010 iov_len = 0
iov[2]: iov_base = 0xdead2010 iov_len = 0
iov[3]: iov_base = 0xdead3000 iov_len = 0x10
在注释(5)中,根据total_len -= chars;那么total_len的大小就被设置为0x20(0x40 -0x20).如果total_len变为了0x20,可我们iov[3]的大小只有0x10.这就会导致pipe_iov_copy_to_user()函数有可能读取到一个未知的iov[4].具体来查看下代码
static int iov_fault_in_pages_write(struct iovec *iov, unsigned long len)
{
//(6)
while (!iov->iov_len)
iov++;
while (len > 0) {
unsigned long this_len;
this_len = min_t(unsigned long, len, iov->iov_len);
if (fault_in_pages_writeable(iov->iov_base, this_len))
break;
len -= this_len;
iov++;
}
return len;
}
static inline int fault_in_pages_writeable(char __user *uaddr, int size)
{
int ret;
if (unlikely(size == 0))
return 0;
/*
* Writing zeroes into userspace here is OK, because we know that if
* the zero gets there, we'll be overwriting it.
*/
ret = __put_user(0, uaddr);
if (ret == 0) {
char __user *end = uaddr + size - 1;
/*
* If the page was already mapped, this will get a cache miss
* for sure, so try to avoid doing it.
*/
if (((unsigned long)uaddr & PAGE_MASK) !=
((unsigned long)end & PAGE_MASK))
ret = __put_user(0, end);
}
return ret;
}
在iov_fault_in_pages_write()函数中的注释(6),也就意味着iov[0],iov[1],iov[2]都会被跳过,iov[3]被用掉.之后len -= this_len;len被设置为0x10.iov的指针将指向一块未知的内存区域.iov[4].iov_base将被__put_user使用.
0x4 如何利用
核心的思路就是想办法触发redo的逻辑,之后精心构造一个readv()调用.把payload结构定义在已经被校验过的iov数组后,让它成为__put_user()等函数调用的目标地址.如果我们再以某种方式让构造的slab结构在iov数组后包含一个函数指针,让它指向要写的内核地址.
1.第一次循环要保证pipe_iov_copy_to_user()函数失败,这样会进入redo逻辑
2.第二次要保证pipe_iov_copy_to_user()成功,但是不能在这里overrun,否则会走向copy_to_user,要校验地址,所以还是无法写内核地址
3.当iov->len走完之后,total_len还有剩余,所以第三次循环的时候,atomic=1.可以overrun触发
4.第一次要保证失败,也就是说需要把iov_base的地址设置成不可写,第二次要成功,就要保证iov_base的地址有效.所以这里可以通过创建竞争关系的线程,调用mmap/munmap等函数来实现.
0x5 POC
我测试的Nexus 6p 6.0.1系统会crash掉.
Talk is cheap,show me the code…
Github:
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