From mboxrd@z Thu Jan 1 00:00:00 1970 From: Roel Kluin Subject: Re: Is kernel optimized with dead store removal? Date: Thu, 25 Feb 2010 16:14:01 +0100 Message-ID: <4B8693B9.3060102@gmail.com> References: <4B85A49E.6000803@gmail.com> <19334.22971.970220.245930@pilspetsen.it.uu.se> Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 7bit Cc: lkml , Herbert Xu , "David S. Miller" , linux-crypto@vger.kernel.org To: Mikael Pettersson Return-path: Received: from mail-ew0-f220.google.com ([209.85.219.220]:64581 "EHLO mail-ew0-f220.google.com" rhost-flags-OK-OK-OK-OK) by vger.kernel.org with ESMTP id S1759037Ab0BYPTZ (ORCPT ); Thu, 25 Feb 2010 10:19:25 -0500 In-Reply-To: <19334.22971.970220.245930@pilspetsen.it.uu.se> Sender: linux-crypto-owner@vger.kernel.org List-ID: > > Does this optimization also occur during compilation of the Linux > > kernel? > Any such dead store removal is up to the compiler and the lifetime > of the object being clobbered. For 'auto' objects the optimization > is certainly likely. > > This is only a problem if the memory (a thread stack, say) is recycled > and leaked uninitialized to user-space, but such bugs are squashed > fairly quickly upon discovery. Thanks for comments, In the sha1_update() case I don't know whether the stack is recycled and leaked - it may be dependent on the calling function, but isn't it vulnerable? I tested this with the snippet below. If compiled with -O1 or -O2 and ON_STACK defined 1, I can read "Secret" a second time. With ON_STACK defined 0 I do not. Roel --- #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; }