On Thu, Feb 25, 2010 at 4:17 PM, David Miller <davem@xxxxxxxxxxxxx> wrote:
> From: Roel Kluin <roel.kluin@xxxxxxxxx>
> Date: Thu, 25 Feb 2010 16:10:27 +0100
>
>> Due to optimization A call to memset() may be removed as a dead store when
>> the buffer is not used after its value is overwritten.
>>
>> Signed-off-by: Roel Kluin <roel.kluin@xxxxxxxxx>
>
> Solution is wrong and overkill in my mind.
>
> It's overkill because the whole reason it's using a stack buffer is to
> avoid the overhead of a kmalloc() call.
>
> And it's wrong because the reason the memset() is there seems to be
> to clear out key information that might exist kernel stack so that
> it's more difficult for rogue code to get at things.
If the memset is optimized away then the clear out does not occur. Do you
know a different way to fix this? I observed this with:
$ gcc -O2 test.c;./a.out
and It shows (on my box) "...S.e.c.r.e.t..."
$ cat test.c
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define ON_STACK 1
void foo()
{
char password[] = "secret";
password[0]='S';
printf ("Don't show again: %s\n", password);
memset(password, 0, sizeof(password));
}
void foo2()
{
char* password = malloc(7);
strncpy (password, "secret" , 7);
password[6] = '\0';
password[0] = 'S';
printf ("Don't show again: %s\n", password);
//memset(password, 0, 7);
free(password);
}
int main(int argc, char* argv[])
{
#if ON_STACK == 1
foo();
#else
foo2();
#endif
int i;
char foo3[] = "hoi";
printf ("foo1:%s\n", foo3);
char* bar = &foo3[0];
for (i = -50; i < 50; i++)
printf ("%c.", bar[i]);
printf("\n");
return 0;
}
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