From: tb <tb@openbsd.org>
Date: Mon, 20 Jun 2022 19:38:25 +0000 (+0000)
Subject: Flip roles of lowercase and uppercase A and B.
X-Git-Url: http://artulab.com/gitweb/?a=commitdiff_plain;h=ec07a3cad870cfc90b624b386e4200c1264fe56a;p=openbsd

Flip roles of lowercase and uppercase A and B.

This matches Cohen's text better and makes the entire thing easier to
read.

suggested by jsing
---

diff --git a/lib/libcrypto/bn/bn_kron.c b/lib/libcrypto/bn/bn_kron.c
index c7bc53535e8..774e9cef30d 100644
--- a/lib/libcrypto/bn/bn_kron.c
+++ b/lib/libcrypto/bn/bn_kron.c
@@ -1,4 +1,4 @@
-/* $OpenBSD: bn_kron.c,v 1.7 2022/06/20 19:32:35 tb Exp $ */
+/* $OpenBSD: bn_kron.c,v 1.8 2022/06/20 19:38:25 tb Exp $ */
 /* ====================================================================
  * Copyright (c) 1998-2000 The OpenSSL Project.  All rights reserved.
  *
@@ -66,36 +66,36 @@
  */
 
 int
-BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)
+BN_kronecker(const BIGNUM *A, const BIGNUM *B, BN_CTX *ctx)
 {
 	/* tab[BN_lsw(n) & 7] = (-1)^((n^2 - 1)) / 8) for odd values of n. */
 	static const int tab[8] = {0, 1, 0, -1, 0, -1, 0, 1};
-	BIGNUM *A, *B, *tmp;
+	BIGNUM *a, *b, *tmp;
 	int k, v;
 	int ret = -2;
 
-	bn_check_top(a);
-	bn_check_top(b);
+	bn_check_top(A);
+	bn_check_top(B);
 
 	BN_CTX_start(ctx);
 
-	if ((A = BN_CTX_get(ctx)) == NULL)
+	if ((a = BN_CTX_get(ctx)) == NULL)
 		goto end;
-	if ((B = BN_CTX_get(ctx)) == NULL)
+	if ((b = BN_CTX_get(ctx)) == NULL)
 		goto end;
 
-	if (BN_copy(A, a) == NULL)
+	if (BN_copy(a, A) == NULL)
 		goto end;
-	if (BN_copy(B, b) == NULL)
+	if (BN_copy(b, B) == NULL)
 		goto end;
 
 	/*
 	 * Cohen's step 1:
 	 */
 
-	/* If B is zero, output 1 if |A| is 1, otherwise output 0. */
-	if (BN_is_zero(B)) {
-		ret = BN_abs_is_word(A, 1);
+	/* If b is zero, output 1 if |a| is 1, otherwise output 0. */
+	if (BN_is_zero(b)) {
+		ret = BN_abs_is_word(a, 1);
 		goto end;
 	}
 
@@ -104,36 +104,36 @@ BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)
 	 */
 
 	/* If both are even, they have a factor in common, so output 0. */
-	if (!BN_is_odd(A) && !BN_is_odd(B)) {
+	if (!BN_is_odd(a) && !BN_is_odd(b)) {
 		ret = 0;
 		goto end;
 	}
 
-	/* Factorize B = 2^v * u with odd u and replace B with u. */
+	/* Factorize b = 2^v * u with odd u and replace b with u. */
 	v = 0;
-	while (!BN_is_bit_set(B, v))
+	while (!BN_is_bit_set(b, v))
 		v++;
-	if (!BN_rshift(B, B, v))
+	if (!BN_rshift(b, b, v))
 		goto end;
 
-	/* If v is even set k = 1, otherwise set it to (-1)^((A^2 - 1) / 8). */
+	/* If v is even set k = 1, otherwise set it to (-1)^((a^2 - 1) / 8). */
 	k = 1;
 	if (v % 2 != 0)
-		k = tab[BN_lsw(A) & 7];
+		k = tab[BN_lsw(a) & 7];
 
 	/*
-	 * If B is negative, replace it with -B and if A is also negative
+	 * If b is negative, replace it with -b and if a is also negative
 	 * replace k with -k.
 	 */
-	if (BN_is_negative(B)) {
-		BN_set_negative(B, 0);
+	if (BN_is_negative(b)) {
+		BN_set_negative(b, 0);
 
-		if (BN_is_negative(A))
+		if (BN_is_negative(a))
 			k = -k;
 	}
 
 	/*
-	 * Now B is positive and odd, so compute the Jacobi symbol (A/B)
+	 * Now b is positive and odd, so compute the Jacobi symbol (a/b)
 	 * and multiply it by k.
 	 */
 
@@ -142,55 +142,55 @@ BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)
 		 * Cohen's step 3:
 		 */
 
-		/* B is positive and odd. */
+		/* b is positive and odd. */
 
-		/* If A is zero output k if B is one, otherwise output 0. */
-		if (BN_is_zero(A)) {
-			ret = BN_is_one(B) ? k : 0;
+		/* If a is zero output k if b is one, otherwise output 0. */
+		if (BN_is_zero(a)) {
+			ret = BN_is_one(b) ? k : 0;
 			goto end;
 		}
 
-		/* Factorize A = 2^v * u with odd u and replace A with u. */
+		/* Factorize a = 2^v * u with odd u and replace a with u. */
 		v = 0;
-		while (!BN_is_bit_set(A, v))
+		while (!BN_is_bit_set(a, v))
 			v++;
-		if (!BN_rshift(A, A, v))
+		if (!BN_rshift(a, a, v))
 			goto end;
 
-		/* If v is odd, multiply k with (-1)^((B^2 - 1) / 8). */
+		/* If v is odd, multiply k with (-1)^((b^2 - 1) / 8). */
 		if (v % 2 != 0)
-			k *= tab[BN_lsw(B) & 7];
+			k *= tab[BN_lsw(b) & 7];
 
 		/*
 		 * Cohen's step 4:
 		 */
 
 		/*
-		 * Apply the reciprocity law: multiply k by (-1)^((A-1)(B-1)/4).
+		 * Apply the reciprocity law: multiply k by (-1)^((a-1)(b-1)/4).
 		 *
-		 * This expression is -1 if and only if A and B are 3 (mod 4).
+		 * This expression is -1 if and only if a and b are 3 (mod 4).
 		 * In turn, this is the case if and only if their two's
 		 * complement representations have the second bit set.
-		 * A could be negative in the first iteration, B is positive.
+		 * a could be negative in the first iteration, b is positive.
 		 */
-		if ((BN_is_negative(A) ? ~BN_lsw(A) : BN_lsw(A)) & BN_lsw(B) & 2)
+		if ((BN_is_negative(a) ? ~BN_lsw(a) : BN_lsw(a)) & BN_lsw(b) & 2)
 			k = -k;
 
 		/*
-		 * (A, B) := (B mod |A|, |A|)
+		 * (a, b) := (b mod |a|, |a|)
 		 *
-		 * Once this is done, we know that 0 < A < B at the start of the
-		 * loop. Since B is strictly decreasing, the loop terminates.
+		 * Once this is done, we know that 0 < a < b at the start of the
+		 * loop. Since b is strictly decreasing, the loop terminates.
 		 */
 
-		if (!BN_nnmod(B, B, A, ctx))
+		if (!BN_nnmod(b, b, a, ctx))
 			goto end;
 
-		tmp = A;
-		A = B;
-		B = tmp;
+		tmp = a;
+		a = b;
+		b = tmp;
 
-		BN_set_negative(B, 0);
+		BN_set_negative(b, 0);
 	}
 
  end: