From: roel kluin Subject: Re: [PATCH] sha: prevent removal of memset as dead store in sha1_update() Date: Thu, 25 Feb 2010 16:31:36 +0100 Message-ID: <25e057c01002250731q68bb9e28ld009163f2a009f48@mail.gmail.com> References: <4B8692E3.9030509@gmail.com> <20100225.071759.98314060.davem@davemloft.net> Mime-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Cc: herbert@gondor.apana.org.au, mikpe@it.uu.se, linux-crypto@vger.kernel.org, akpm@linux-foundation.org, linux-kernel@vger.kernel.org To: David Miller Return-path: Received: from mail-bw0-f209.google.com ([209.85.218.209]:64567 "EHLO mail-bw0-f209.google.com" rhost-flags-OK-OK-OK-OK) by vger.kernel.org with ESMTP id S1759013Ab0BYPb7 (ORCPT ); Thu, 25 Feb 2010 10:31:59 -0500 In-Reply-To: <20100225.071759.98314060.davem@davemloft.net> Sender: linux-crypto-owner@vger.kernel.org List-ID: On Thu, Feb 25, 2010 at 4:17 PM, David Miller wrote: > From: Roel Kluin > 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 > > 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 #include #include #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; }