Hello,
I'm wondering whether there is an exploitable TOCTTOU race condition in the way user pointers are handled in the kernel. Consider the following code:
1: struct st { int *u; };
2: void syscall(struct st * stp) {
3: if (!access_ok(VERIFY_READ,stp,sizeof(struct st)))
4: return;
5: if (!access_ok(VERIFY_WRITE,stp->u,sizeof(int)))
6: return;
7: foo(); //user app writes a kernel address to stp->u
8: *(stp->u) = 0;
9:}
Suppose syscall is some system call and, thus, stp and stp->u are user pointers. The function checks the stp and stp->u pointers using the access_ok macro on lines 3 and 5. Also suppose that the call to foo on line 7 takes a non-trivial amount of time to execute. During the time it takes foo to execute, the user application writes a kernel address to stp->u. Note that this write occurs after the check on line 5. Then, on line 8, the kernel writes to stp->u which contains a kernel address. So, the user application could force the kernel to overwrite itself. Is it possible to exploit this race condition? If so, does Sparse check for this?
-SKB
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On Wed, May 16, 2007 at 10:56:22PM -0600, sk b wrote:
>
> Hello,
>
> I'm wondering whether there is an exploitable TOCTTOU race condition in the way user pointers are handled in the kernel. Consider the following code:
>
> 1: struct st { int *u; };
> 2: void syscall(struct st * stp) {
> 3: if (!access_ok(VERIFY_READ,stp,sizeof(struct st)))
> 4: return;
> 5: if (!access_ok(VERIFY_WRITE,stp->u,sizeof(int)))
... and there's your bug - direct access to userland data. The normal
variant is to use accessors (get_user() or copy_from_user()) to fetch
the value of stp->u. At which point races of the kind you mentioned
take an obviously dumb code (explicitly copying the same struct from
userland _twice_).
From: sk b <[email protected]>
Date: Wed, 16 May 2007 22:56:22 -0600
> 3: if (!access_ok(VERIFY_READ,stp,sizeof(struct st)))
> 4: return;
> 5: if (!access_ok(VERIFY_WRITE,stp->u,sizeof(int)))
> 6: return;
This code would not exist in the kernel, the kernel cannot dereference
stp->u. The stp->u dereference would silently work on x86 and x86_64
but it would generate an exception on sparc64 and other platforms.
User space accesses must go through the proper copy_from_user(),
copy_to_user, get_user(), and put_user() interfaces.
It must first copy stp into a local kernel space copy, then it may
inspect the value of stp->u.
And yes sparse would catch this problem in your code, because the
"__user" annotations would catch the illegal "stp->u" dereference.
On Wed, 2007-05-16 at 22:56 -0600, sk b wrote:
> Hello,
>
> I'm wondering whether there is an exploitable TOCTTOU race condition in the way user pointers are handled in the kernel. Consider the following code:
>
> 1: struct st { int *u; };
> 2: void syscall(struct st * stp) {
> 3: if (!access_ok(VERIFY_READ,stp,sizeof(struct st)))
> 4: return;
> 5: if (!access_ok(VERIFY_WRITE,stp->u,sizeof(int)))
> 6: return;
this line is invalid; you are not allowed to dereference userspace
memory directly. You need to call copy_from_user() on stp first, and
then use the copy.
> 7: foo(); //user app writes a kernel address to stp->u
> 8: *(stp->u) = 0;
> 9:}
>
> Suppose syscall is some system call and, thus, stp and stp->u are user pointers. The function checks the stp and stp->u pointers using the access_ok macro on lines 3 and 5. Also suppose that the call to foo on line 7 takes a non-trivial amount of time to execute. During the time it takes foo to execute, the user application writes a kernel address to stp->u. Note that this write occurs after the check on line 5. Then, on line 8, the kernel writes to stp->u which contains a kernel address. So, the user application could force the kernel to overwrite itself. Is it possible to exploit this race condition? If so, does Sparse check for this?
>
> -SKB
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