1、 suppose *len = 35, sizeof(struct ipt_get_entries) = 36
2、 set get.size = 0xffffffff from user space
3、 sizeof(struct ipt_get_entries) + get.size = 36 + 0xffffffff = 35;
4、 if (*len != sizeof(struct ipt_get_entries) + get.size) was bypassed.
you can test with c code:
#include <stdio.h>
int main(void)
{
unsigned int arg = 0xffffffff;
printf("%u\n", arg + 36);
if (35 != arg + 36) {
printf("not over flow.\n");
return -1;
}
printf("arg over flow.\n");
return 0;
}
On Tue, Mar 23, 2010 at 11:04 AM, Jan Engelhardt <jengelh@xxxxxxxxxx> wrote:
>
> On Tuesday 2010-03-23 03:37, wzt wzt wrote:
>>> And, for the addition overflow, can it be caught by
>>>
>>> "if (*len != sizeof(struct ipt_get_entries) + get.size)" ???
>>
>>sizeof(struct ipt_get_entries) + get.size can be overflow as *len,
>>get.size is control by user space with copy_from_user().
>
> The != should catch it.
>
> For 64-bit environments:
> * + invoked with size_t, unsigned int
> => right side promoted to size_t, result type is size_t
> * != invoked with int and size_t
> => left-side promoted to ssize_t (probably; but something as large as size_t)
> * get.size is 32-bit bounded, as is *len,
> so no overflow to worry about at all unless you make
> sizeof(X) hilariously big close to 2^64 which is rather unlikely.
>
> For 32-bit environments:
> * Let *len be a number of choice (e.g. 36)
> * Find a sizeof(X)+get.size that equals 36 mod 2^32.
> * Since sizeof(X) is const, get.size must be 0 mod 2^32.
> * So get.size must be a multiple of 2^32 to fool the system.
> * Since get.size itself is only a 32-bit quantity, you cannot
> represent any value larger than 4294967295.
>
>
> What Was What Was Wanted.
>
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