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7fe7cc57af
BN_one() uses the expand function which calls malloc which may fail. All other places that reference BN_one() check the return value. The issue is triggered by a memory allocation failure. Detected by PR #18355 Reviewed-by: Tomas Mraz <tomas@openssl.org> Reviewed-by: Paul Dale <pauli@openssl.org> (Merged from https://github.com/openssl/openssl/pull/18697)
648 lines
19 KiB
C
648 lines
19 KiB
C
/*
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* Copyright 1995-2020 The OpenSSL Project Authors. All Rights Reserved.
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*
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* Licensed under the Apache License 2.0 (the "License"). You may not use
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* this file except in compliance with the License. You can obtain a copy
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* in the file LICENSE in the source distribution or at
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* https://www.openssl.org/source/license.html
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*/
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#include "internal/cryptlib.h"
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#include "bn_local.h"
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/*
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* bn_mod_inverse_no_branch is a special version of BN_mod_inverse. It does
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* not contain branches that may leak sensitive information.
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*
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* This is a static function, we ensure all callers in this file pass valid
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* arguments: all passed pointers here are non-NULL.
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*/
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static ossl_inline
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BIGNUM *bn_mod_inverse_no_branch(BIGNUM *in,
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const BIGNUM *a, const BIGNUM *n,
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BN_CTX *ctx, int *pnoinv)
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{
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BIGNUM *A, *B, *X, *Y, *M, *D, *T, *R = NULL;
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BIGNUM *ret = NULL;
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int sign;
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bn_check_top(a);
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bn_check_top(n);
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BN_CTX_start(ctx);
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A = BN_CTX_get(ctx);
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B = BN_CTX_get(ctx);
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X = BN_CTX_get(ctx);
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D = BN_CTX_get(ctx);
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M = BN_CTX_get(ctx);
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Y = BN_CTX_get(ctx);
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T = BN_CTX_get(ctx);
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if (T == NULL)
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goto err;
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if (in == NULL)
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R = BN_new();
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else
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R = in;
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if (R == NULL)
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goto err;
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if (!BN_one(X))
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goto err;
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BN_zero(Y);
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if (BN_copy(B, a) == NULL)
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goto err;
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if (BN_copy(A, n) == NULL)
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goto err;
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A->neg = 0;
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if (B->neg || (BN_ucmp(B, A) >= 0)) {
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/*
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* Turn BN_FLG_CONSTTIME flag on, so that when BN_div is invoked,
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* BN_div_no_branch will be called eventually.
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*/
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{
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BIGNUM local_B;
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bn_init(&local_B);
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BN_with_flags(&local_B, B, BN_FLG_CONSTTIME);
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if (!BN_nnmod(B, &local_B, A, ctx))
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goto err;
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/* Ensure local_B goes out of scope before any further use of B */
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}
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}
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sign = -1;
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/*-
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* From B = a mod |n|, A = |n| it follows that
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*
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* 0 <= B < A,
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* -sign*X*a == B (mod |n|),
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* sign*Y*a == A (mod |n|).
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*/
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while (!BN_is_zero(B)) {
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BIGNUM *tmp;
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/*-
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* 0 < B < A,
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* (*) -sign*X*a == B (mod |n|),
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* sign*Y*a == A (mod |n|)
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*/
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/*
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* Turn BN_FLG_CONSTTIME flag on, so that when BN_div is invoked,
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* BN_div_no_branch will be called eventually.
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*/
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{
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BIGNUM local_A;
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bn_init(&local_A);
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BN_with_flags(&local_A, A, BN_FLG_CONSTTIME);
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/* (D, M) := (A/B, A%B) ... */
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if (!BN_div(D, M, &local_A, B, ctx))
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goto err;
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/* Ensure local_A goes out of scope before any further use of A */
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}
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/*-
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* Now
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* A = D*B + M;
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* thus we have
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* (**) sign*Y*a == D*B + M (mod |n|).
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*/
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tmp = A; /* keep the BIGNUM object, the value does not
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* matter */
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/* (A, B) := (B, A mod B) ... */
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A = B;
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B = M;
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/* ... so we have 0 <= B < A again */
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/*-
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* Since the former M is now B and the former B is now A,
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* (**) translates into
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* sign*Y*a == D*A + B (mod |n|),
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* i.e.
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* sign*Y*a - D*A == B (mod |n|).
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* Similarly, (*) translates into
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* -sign*X*a == A (mod |n|).
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*
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* Thus,
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* sign*Y*a + D*sign*X*a == B (mod |n|),
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* i.e.
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* sign*(Y + D*X)*a == B (mod |n|).
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*
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* So if we set (X, Y, sign) := (Y + D*X, X, -sign), we arrive back at
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* -sign*X*a == B (mod |n|),
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* sign*Y*a == A (mod |n|).
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* Note that X and Y stay non-negative all the time.
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*/
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if (!BN_mul(tmp, D, X, ctx))
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goto err;
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if (!BN_add(tmp, tmp, Y))
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goto err;
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M = Y; /* keep the BIGNUM object, the value does not
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* matter */
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Y = X;
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X = tmp;
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sign = -sign;
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}
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/*-
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* The while loop (Euclid's algorithm) ends when
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* A == gcd(a,n);
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* we have
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* sign*Y*a == A (mod |n|),
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* where Y is non-negative.
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*/
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if (sign < 0) {
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if (!BN_sub(Y, n, Y))
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goto err;
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}
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/* Now Y*a == A (mod |n|). */
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if (BN_is_one(A)) {
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/* Y*a == 1 (mod |n|) */
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if (!Y->neg && BN_ucmp(Y, n) < 0) {
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if (!BN_copy(R, Y))
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goto err;
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} else {
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if (!BN_nnmod(R, Y, n, ctx))
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goto err;
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}
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} else {
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*pnoinv = 1;
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/* caller sets the BN_R_NO_INVERSE error */
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goto err;
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}
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ret = R;
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*pnoinv = 0;
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err:
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if ((ret == NULL) && (in == NULL))
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BN_free(R);
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BN_CTX_end(ctx);
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bn_check_top(ret);
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return ret;
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}
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/*
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* This is an internal function, we assume all callers pass valid arguments:
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* all pointers passed here are assumed non-NULL.
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*/
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BIGNUM *int_bn_mod_inverse(BIGNUM *in,
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const BIGNUM *a, const BIGNUM *n, BN_CTX *ctx,
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int *pnoinv)
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{
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BIGNUM *A, *B, *X, *Y, *M, *D, *T, *R = NULL;
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BIGNUM *ret = NULL;
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int sign;
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/* This is invalid input so we don't worry about constant time here */
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if (BN_abs_is_word(n, 1) || BN_is_zero(n)) {
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*pnoinv = 1;
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return NULL;
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}
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*pnoinv = 0;
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if ((BN_get_flags(a, BN_FLG_CONSTTIME) != 0)
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|| (BN_get_flags(n, BN_FLG_CONSTTIME) != 0)) {
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return bn_mod_inverse_no_branch(in, a, n, ctx, pnoinv);
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}
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bn_check_top(a);
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bn_check_top(n);
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BN_CTX_start(ctx);
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A = BN_CTX_get(ctx);
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B = BN_CTX_get(ctx);
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X = BN_CTX_get(ctx);
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D = BN_CTX_get(ctx);
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M = BN_CTX_get(ctx);
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Y = BN_CTX_get(ctx);
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T = BN_CTX_get(ctx);
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if (T == NULL)
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goto err;
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if (in == NULL)
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R = BN_new();
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else
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R = in;
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if (R == NULL)
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goto err;
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if (!BN_one(X))
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goto err;
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BN_zero(Y);
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if (BN_copy(B, a) == NULL)
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goto err;
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if (BN_copy(A, n) == NULL)
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goto err;
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A->neg = 0;
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if (B->neg || (BN_ucmp(B, A) >= 0)) {
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if (!BN_nnmod(B, B, A, ctx))
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goto err;
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}
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sign = -1;
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/*-
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* From B = a mod |n|, A = |n| it follows that
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*
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* 0 <= B < A,
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* -sign*X*a == B (mod |n|),
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* sign*Y*a == A (mod |n|).
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*/
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if (BN_is_odd(n) && (BN_num_bits(n) <= 2048)) {
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/*
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* Binary inversion algorithm; requires odd modulus. This is faster
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* than the general algorithm if the modulus is sufficiently small
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* (about 400 .. 500 bits on 32-bit systems, but much more on 64-bit
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* systems)
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*/
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int shift;
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while (!BN_is_zero(B)) {
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/*-
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* 0 < B < |n|,
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* 0 < A <= |n|,
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* (1) -sign*X*a == B (mod |n|),
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* (2) sign*Y*a == A (mod |n|)
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*/
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/*
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* Now divide B by the maximum possible power of two in the
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* integers, and divide X by the same value mod |n|. When we're
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* done, (1) still holds.
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*/
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shift = 0;
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while (!BN_is_bit_set(B, shift)) { /* note that 0 < B */
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shift++;
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if (BN_is_odd(X)) {
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if (!BN_uadd(X, X, n))
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goto err;
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}
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/*
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* now X is even, so we can easily divide it by two
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*/
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if (!BN_rshift1(X, X))
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goto err;
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}
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if (shift > 0) {
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if (!BN_rshift(B, B, shift))
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goto err;
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}
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/*
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* Same for A and Y. Afterwards, (2) still holds.
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*/
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shift = 0;
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while (!BN_is_bit_set(A, shift)) { /* note that 0 < A */
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shift++;
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if (BN_is_odd(Y)) {
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if (!BN_uadd(Y, Y, n))
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goto err;
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}
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/* now Y is even */
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if (!BN_rshift1(Y, Y))
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goto err;
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}
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if (shift > 0) {
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if (!BN_rshift(A, A, shift))
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goto err;
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}
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/*-
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* We still have (1) and (2).
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* Both A and B are odd.
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* The following computations ensure that
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*
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* 0 <= B < |n|,
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* 0 < A < |n|,
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* (1) -sign*X*a == B (mod |n|),
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* (2) sign*Y*a == A (mod |n|),
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*
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* and that either A or B is even in the next iteration.
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*/
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if (BN_ucmp(B, A) >= 0) {
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/* -sign*(X + Y)*a == B - A (mod |n|) */
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if (!BN_uadd(X, X, Y))
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goto err;
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/*
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* NB: we could use BN_mod_add_quick(X, X, Y, n), but that
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* actually makes the algorithm slower
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*/
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if (!BN_usub(B, B, A))
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goto err;
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} else {
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/* sign*(X + Y)*a == A - B (mod |n|) */
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if (!BN_uadd(Y, Y, X))
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goto err;
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/*
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* as above, BN_mod_add_quick(Y, Y, X, n) would slow things down
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*/
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if (!BN_usub(A, A, B))
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goto err;
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}
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}
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} else {
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/* general inversion algorithm */
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while (!BN_is_zero(B)) {
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BIGNUM *tmp;
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/*-
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* 0 < B < A,
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* (*) -sign*X*a == B (mod |n|),
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* sign*Y*a == A (mod |n|)
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*/
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/* (D, M) := (A/B, A%B) ... */
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if (BN_num_bits(A) == BN_num_bits(B)) {
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if (!BN_one(D))
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goto err;
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if (!BN_sub(M, A, B))
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goto err;
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} else if (BN_num_bits(A) == BN_num_bits(B) + 1) {
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/* A/B is 1, 2, or 3 */
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if (!BN_lshift1(T, B))
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goto err;
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if (BN_ucmp(A, T) < 0) {
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/* A < 2*B, so D=1 */
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if (!BN_one(D))
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goto err;
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if (!BN_sub(M, A, B))
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goto err;
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} else {
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/* A >= 2*B, so D=2 or D=3 */
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if (!BN_sub(M, A, T))
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goto err;
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if (!BN_add(D, T, B))
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goto err; /* use D (:= 3*B) as temp */
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if (BN_ucmp(A, D) < 0) {
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/* A < 3*B, so D=2 */
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if (!BN_set_word(D, 2))
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goto err;
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/*
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* M (= A - 2*B) already has the correct value
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*/
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} else {
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/* only D=3 remains */
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if (!BN_set_word(D, 3))
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goto err;
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/*
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* currently M = A - 2*B, but we need M = A - 3*B
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*/
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if (!BN_sub(M, M, B))
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goto err;
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}
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}
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} else {
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if (!BN_div(D, M, A, B, ctx))
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goto err;
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}
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/*-
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* Now
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* A = D*B + M;
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* thus we have
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* (**) sign*Y*a == D*B + M (mod |n|).
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*/
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tmp = A; /* keep the BIGNUM object, the value does not matter */
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/* (A, B) := (B, A mod B) ... */
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A = B;
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B = M;
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/* ... so we have 0 <= B < A again */
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/*-
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* Since the former M is now B and the former B is now A,
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* (**) translates into
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* sign*Y*a == D*A + B (mod |n|),
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* i.e.
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* sign*Y*a - D*A == B (mod |n|).
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* Similarly, (*) translates into
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* -sign*X*a == A (mod |n|).
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*
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* Thus,
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* sign*Y*a + D*sign*X*a == B (mod |n|),
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* i.e.
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* sign*(Y + D*X)*a == B (mod |n|).
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*
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* So if we set (X, Y, sign) := (Y + D*X, X, -sign), we arrive back at
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* -sign*X*a == B (mod |n|),
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* sign*Y*a == A (mod |n|).
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* Note that X and Y stay non-negative all the time.
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*/
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/*
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* most of the time D is very small, so we can optimize tmp := D*X+Y
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*/
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if (BN_is_one(D)) {
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if (!BN_add(tmp, X, Y))
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goto err;
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} else {
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if (BN_is_word(D, 2)) {
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if (!BN_lshift1(tmp, X))
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goto err;
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} else if (BN_is_word(D, 4)) {
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if (!BN_lshift(tmp, X, 2))
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goto err;
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} else if (D->top == 1) {
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if (!BN_copy(tmp, X))
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goto err;
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if (!BN_mul_word(tmp, D->d[0]))
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goto err;
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} else {
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if (!BN_mul(tmp, D, X, ctx))
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goto err;
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}
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if (!BN_add(tmp, tmp, Y))
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goto err;
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}
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M = Y; /* keep the BIGNUM object, the value does not matter */
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Y = X;
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X = tmp;
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sign = -sign;
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}
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}
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|
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/*-
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* The while loop (Euclid's algorithm) ends when
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* A == gcd(a,n);
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* we have
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* sign*Y*a == A (mod |n|),
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* where Y is non-negative.
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*/
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if (sign < 0) {
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if (!BN_sub(Y, n, Y))
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goto err;
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}
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/* Now Y*a == A (mod |n|). */
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if (BN_is_one(A)) {
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/* Y*a == 1 (mod |n|) */
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if (!Y->neg && BN_ucmp(Y, n) < 0) {
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if (!BN_copy(R, Y))
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goto err;
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} else {
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if (!BN_nnmod(R, Y, n, ctx))
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goto err;
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}
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} else {
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*pnoinv = 1;
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goto err;
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}
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ret = R;
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err:
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if ((ret == NULL) && (in == NULL))
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BN_free(R);
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BN_CTX_end(ctx);
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bn_check_top(ret);
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return ret;
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}
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|
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/* solves ax == 1 (mod n) */
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BIGNUM *BN_mod_inverse(BIGNUM *in,
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const BIGNUM *a, const BIGNUM *n, BN_CTX *ctx)
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{
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BN_CTX *new_ctx = NULL;
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BIGNUM *rv;
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int noinv = 0;
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|
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if (ctx == NULL) {
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ctx = new_ctx = BN_CTX_new_ex(NULL);
|
|
if (ctx == NULL) {
|
|
ERR_raise(ERR_LIB_BN, ERR_R_MALLOC_FAILURE);
|
|
return NULL;
|
|
}
|
|
}
|
|
|
|
rv = int_bn_mod_inverse(in, a, n, ctx, &noinv);
|
|
if (noinv)
|
|
ERR_raise(ERR_LIB_BN, BN_R_NO_INVERSE);
|
|
BN_CTX_free(new_ctx);
|
|
return rv;
|
|
}
|
|
|
|
/*-
|
|
* This function is based on the constant-time GCD work by Bernstein and Yang:
|
|
* https://eprint.iacr.org/2019/266
|
|
* Generalized fast GCD function to allow even inputs.
|
|
* The algorithm first finds the shared powers of 2 between
|
|
* the inputs, and removes them, reducing at least one of the
|
|
* inputs to an odd value. Then it proceeds to calculate the GCD.
|
|
* Before returning the resulting GCD, we take care of adding
|
|
* back the powers of two removed at the beginning.
|
|
* Note 1: we assume the bit length of both inputs is public information,
|
|
* since access to top potentially leaks this information.
|
|
*/
|
|
int BN_gcd(BIGNUM *r, const BIGNUM *in_a, const BIGNUM *in_b, BN_CTX *ctx)
|
|
{
|
|
BIGNUM *g, *temp = NULL;
|
|
BN_ULONG mask = 0;
|
|
int i, j, top, rlen, glen, m, bit = 1, delta = 1, cond = 0, shifts = 0, ret = 0;
|
|
|
|
/* Note 2: zero input corner cases are not constant-time since they are
|
|
* handled immediately. An attacker can run an attack under this
|
|
* assumption without the need of side-channel information. */
|
|
if (BN_is_zero(in_b)) {
|
|
ret = BN_copy(r, in_a) != NULL;
|
|
r->neg = 0;
|
|
return ret;
|
|
}
|
|
if (BN_is_zero(in_a)) {
|
|
ret = BN_copy(r, in_b) != NULL;
|
|
r->neg = 0;
|
|
return ret;
|
|
}
|
|
|
|
bn_check_top(in_a);
|
|
bn_check_top(in_b);
|
|
|
|
BN_CTX_start(ctx);
|
|
temp = BN_CTX_get(ctx);
|
|
g = BN_CTX_get(ctx);
|
|
|
|
/* make r != 0, g != 0 even, so BN_rshift is not a potential nop */
|
|
if (g == NULL
|
|
|| !BN_lshift1(g, in_b)
|
|
|| !BN_lshift1(r, in_a))
|
|
goto err;
|
|
|
|
/* find shared powers of two, i.e. "shifts" >= 1 */
|
|
for (i = 0; i < r->dmax && i < g->dmax; i++) {
|
|
mask = ~(r->d[i] | g->d[i]);
|
|
for (j = 0; j < BN_BITS2; j++) {
|
|
bit &= mask;
|
|
shifts += bit;
|
|
mask >>= 1;
|
|
}
|
|
}
|
|
|
|
/* subtract shared powers of two; shifts >= 1 */
|
|
if (!BN_rshift(r, r, shifts)
|
|
|| !BN_rshift(g, g, shifts))
|
|
goto err;
|
|
|
|
/* expand to biggest nword, with room for a possible extra word */
|
|
top = 1 + ((r->top >= g->top) ? r->top : g->top);
|
|
if (bn_wexpand(r, top) == NULL
|
|
|| bn_wexpand(g, top) == NULL
|
|
|| bn_wexpand(temp, top) == NULL)
|
|
goto err;
|
|
|
|
/* re arrange inputs s.t. r is odd */
|
|
BN_consttime_swap((~r->d[0]) & 1, r, g, top);
|
|
|
|
/* compute the number of iterations */
|
|
rlen = BN_num_bits(r);
|
|
glen = BN_num_bits(g);
|
|
m = 4 + 3 * ((rlen >= glen) ? rlen : glen);
|
|
|
|
for (i = 0; i < m; i++) {
|
|
/* conditionally flip signs if delta is positive and g is odd */
|
|
cond = (-delta >> (8 * sizeof(delta) - 1)) & g->d[0] & 1
|
|
/* make sure g->top > 0 (i.e. if top == 0 then g == 0 always) */
|
|
& (~((g->top - 1) >> (sizeof(g->top) * 8 - 1)));
|
|
delta = (-cond & -delta) | ((cond - 1) & delta);
|
|
r->neg ^= cond;
|
|
/* swap */
|
|
BN_consttime_swap(cond, r, g, top);
|
|
|
|
/* elimination step */
|
|
delta++;
|
|
if (!BN_add(temp, g, r))
|
|
goto err;
|
|
BN_consttime_swap(g->d[0] & 1 /* g is odd */
|
|
/* make sure g->top > 0 (i.e. if top == 0 then g == 0 always) */
|
|
& (~((g->top - 1) >> (sizeof(g->top) * 8 - 1))),
|
|
g, temp, top);
|
|
if (!BN_rshift1(g, g))
|
|
goto err;
|
|
}
|
|
|
|
/* remove possible negative sign */
|
|
r->neg = 0;
|
|
/* add powers of 2 removed, then correct the artificial shift */
|
|
if (!BN_lshift(r, r, shifts)
|
|
|| !BN_rshift1(r, r))
|
|
goto err;
|
|
|
|
ret = 1;
|
|
|
|
err:
|
|
BN_CTX_end(ctx);
|
|
bn_check_top(r);
|
|
return ret;
|
|
}
|