pgprndvms.c

著名的加密软件的应用于电子邮件中

/*
* pgpRndVMS.c -- Get environmental noise to seed the random number generator.
* VMS version.
*
* Copyright (C) 1995-1997 Pretty Good Privacy, Inc. All rights reserved.
*
* This hasn't been examined in a long time (December, 1995).
* Could someone who knows VMS please have a look at it and determine
* if this is still good?
*
* $Id: pgpRndVMS.c,v 1.1.2.1 1997/06/07 09:51:36 mhw Exp $
*/
#ifdef HAVE_CONFIG_H
#include "config.h"
#endif

#include "pgpRnd.h"
#include "pgpRndom.h"
#include "pgpUsuals.h"

/*
* Add as much environmentally-derived random noise as possible
* to the randPool. Typically, this involves reading the most
* accurate system clocks available.
*
* Returns the number of ticks that have passed since the last call,
* for entropy estimation purposes.
*/
word32
ranGetEntropy(struct PgpRandomContext const *rc)
	{
			word32 delta;
			word32 d1;	/* MSW of difference */
			word32 t[2];	 /* little-endian 64-bit timer */
			static word32 prevt[2];

			SYS$GETTIM(t);	/* VMS hardware clock increments by 100000 per tick */
			pgpRandomAddBytes(rc, (byte const *)t, sizeof(t));
			/* Get difference in d1 and delta, and old time in prevt */
			d1 = t[1] - prevt[1] + (t[0] < prevt[0]);
			prevt[1] = t[1];
			delta = t[0] - prevt[0];
			prevt[0] = t[0];
		
			/* Now, divide the 64-bit value by 100000 = 2^5 * 5^5 = 32 * 3125 */
			/* Divide value, MSW in d1 and LSW in delta, by 32 */
			delta >>= 5;
			delta |= d1 << (32-5);
			d1 >>= 5;
			/*
			* Divide by 3125. This fits into 16 bits, so the following
			* code is possible. 2^32 = 3125 * 1374389 + 1671.
			*
			* This code has confused people reading it, so here's a detailed
			* explanation. First, since we only want a 32-bit result,
			* reduce the input mod 3125 * 2^32 before starting. This
			* amounts to reducing the most significant word mod 3125 and
			* leaving the least-significant word alone.
			*
			* Then, using / for mathematical (real, not integer) division, we
			* want to compute floor((d1 * 2^32 + d0) / 3125), which I'll denote
			* using the old [ ] syntax for floor, so it's
			* [ (d1 * 2^32 + d0) / 3125 ]
			* = [ (d1 * (3125 * 1374389 + 1671) + d0) / 3125 ]
			* = [ d1 * 1374389 + (d1 * 1671 + d0) / 3125 ]
			* = d1 * 137438 + [ (d1 * 1671 + d0) / 3125 ]
			* = d1 * 137438 + [ d0 / 3125 ] + [ (d1 * 1671 + d0 % 3125) / 3125 ]
			*
			* The C / operator, applied to integers, performs [ a / b ], so
			* this can be implemented in C, and since d1 < 3125 (by the first
			* modulo operation), d1 * 1671 + d0 % 3125 < 3125 * 1672, which
			* is 5225000, less than 2^32, so it all fits into 32 bits.
			*/
			d1 %= 3125;	/* Ignore overflow past 32 bits */
			delta = delta/3125 + d1*1374389 + (delta%3125 + d1*1671) / 3125;

return delta;
}