Re: Did you *really* zeroize that key?

[Patrick Chkoreff]

From: "Trei, Peter" <[EMAIL PROTECTED]>

[Moderator's note: FYI: no "pragma" is needed. This is what C's
"volatile" keyword is for. Unfortunately, not everyone writing in C
knows the language. --Perry]

Thanks for the reminder about "volatile." It is an ancient and valuable feature of C and I suppose it's implemented correctly under gcc and some of the Windoze compilers even with high optimization options like -O2.

>From RISKS:
http://catless.ncl.ac.uk/Risks/22.35.html#subj6

Those of us who write code need to be reminded of this
now and then.

Everybody probably also knows about the gnupg trick, where they define a recursive routine called "burn_stack":

static void
burn_stack (int bytes)
{
char buf[64];

memset (buf, 0, sizeof buf);
bytes -= sizeof buf;
if (bytes > 0)
burn_stack (bytes);
}

Then there's the vararg technique discussed in Michael Welschenbach's book "Cryptography in C and C++":

static void purgevars_l (int noofvars, ...)
{
va_list ap;
size_t size;
va_start (ap, noofvars);
for (; noofvars > 0; --noofvars)
{
switch (size = va_arg (ap, size_t))
{
case 1: *va_arg (ap, char *) = 0;
break;
case 2: *va_arg (ap, short *) = 0;
break;
case 4: *va_arg (ap, long *) = 0;
break;
default:
memset (va_arg(ap, char *), 0, size);
}
}
va_end (ap);
}

Here's an example of how you might call the routine:

purgevars_l(2, sizeof (la), &la,
sizeof (lb), &lb);


But hey, if "volatile" keyword works then so much the better. I would recommend examining the assembly language output of your compiler to verify that it honours "volatile."

-- Patrick
http://fexl.com