Upgrade openssl from 1.1.0e to 1.1.1b, with source code. 4.0.78

trunk/3rdparty/openssl-1.1-fit/crypto/bn/bn_recp.c

@@ -0,0 +1,194 @@
+/*
+ * Copyright 1995-2018 The OpenSSL Project Authors. All Rights Reserved.
+ *
+ * Licensed under the OpenSSL license (the "License").  You may not use
+ * this file except in compliance with the License.  You can obtain a copy
+ * in the file LICENSE in the source distribution or at
+ * https://www.openssl.org/source/license.html
+ */
+
+#include "internal/cryptlib.h"
+#include "bn_lcl.h"
+
+void BN_RECP_CTX_init(BN_RECP_CTX *recp)
+{
+    memset(recp, 0, sizeof(*recp));
+    bn_init(&(recp->N));
+    bn_init(&(recp->Nr));
+}
+
+BN_RECP_CTX *BN_RECP_CTX_new(void)
+{
+    BN_RECP_CTX *ret;
+
+    if ((ret = OPENSSL_zalloc(sizeof(*ret))) == NULL) {
+        BNerr(BN_F_BN_RECP_CTX_NEW, ERR_R_MALLOC_FAILURE);
+        return NULL;
+    }
+
+    bn_init(&(ret->N));
+    bn_init(&(ret->Nr));
+    ret->flags = BN_FLG_MALLOCED;
+    return ret;
+}
+
+void BN_RECP_CTX_free(BN_RECP_CTX *recp)
+{
+    if (recp == NULL)
+        return;
+    BN_free(&recp->N);
+    BN_free(&recp->Nr);
+    if (recp->flags & BN_FLG_MALLOCED)
+        OPENSSL_free(recp);
+}
+
+int BN_RECP_CTX_set(BN_RECP_CTX *recp, const BIGNUM *d, BN_CTX *ctx)
+{
+    if (!BN_copy(&(recp->N), d))
+        return 0;
+    BN_zero(&(recp->Nr));
+    recp->num_bits = BN_num_bits(d);
+    recp->shift = 0;
+    return 1;
+}
+
+int BN_mod_mul_reciprocal(BIGNUM *r, const BIGNUM *x, const BIGNUM *y,
+                          BN_RECP_CTX *recp, BN_CTX *ctx)
+{
+    int ret = 0;
+    BIGNUM *a;
+    const BIGNUM *ca;
+
+    BN_CTX_start(ctx);
+    if ((a = BN_CTX_get(ctx)) == NULL)
+        goto err;
+    if (y != NULL) {
+        if (x == y) {
+            if (!BN_sqr(a, x, ctx))
+                goto err;
+        } else {
+            if (!BN_mul(a, x, y, ctx))
+                goto err;
+        }
+        ca = a;
+    } else
+        ca = x;                 /* Just do the mod */
+
+    ret = BN_div_recp(NULL, r, ca, recp, ctx);
+ err:
+    BN_CTX_end(ctx);
+    bn_check_top(r);
+    return ret;
+}
+
+int BN_div_recp(BIGNUM *dv, BIGNUM *rem, const BIGNUM *m,
+                BN_RECP_CTX *recp, BN_CTX *ctx)
+{
+    int i, j, ret = 0;
+    BIGNUM *a, *b, *d, *r;
+
+    BN_CTX_start(ctx);
+    d = (dv != NULL) ? dv : BN_CTX_get(ctx);
+    r = (rem != NULL) ? rem : BN_CTX_get(ctx);
+    a = BN_CTX_get(ctx);
+    b = BN_CTX_get(ctx);
+    if (b == NULL)
+        goto err;
+
+    if (BN_ucmp(m, &(recp->N)) < 0) {
+        BN_zero(d);
+        if (!BN_copy(r, m)) {
+            BN_CTX_end(ctx);
+            return 0;
+        }
+        BN_CTX_end(ctx);
+        return 1;
+    }
+
+    /*
+     * We want the remainder Given input of ABCDEF / ab we need multiply
+     * ABCDEF by 3 digests of the reciprocal of ab
+     */
+
+    /* i := max(BN_num_bits(m), 2*BN_num_bits(N)) */
+    i = BN_num_bits(m);
+    j = recp->num_bits << 1;
+    if (j > i)
+        i = j;
+
+    /* Nr := round(2^i / N) */
+    if (i != recp->shift)
+        recp->shift = BN_reciprocal(&(recp->Nr), &(recp->N), i, ctx);
+    /* BN_reciprocal could have returned -1 for an error */
+    if (recp->shift == -1)
+        goto err;
+
+    /*-
+     * d := |round(round(m / 2^BN_num_bits(N)) * recp->Nr / 2^(i - BN_num_bits(N)))|
+     *    = |round(round(m / 2^BN_num_bits(N)) * round(2^i / N) / 2^(i - BN_num_bits(N)))|
+     *   <= |(m / 2^BN_num_bits(N)) * (2^i / N) * (2^BN_num_bits(N) / 2^i)|
+     *    = |m/N|
+     */
+    if (!BN_rshift(a, m, recp->num_bits))
+        goto err;
+    if (!BN_mul(b, a, &(recp->Nr), ctx))
+        goto err;
+    if (!BN_rshift(d, b, i - recp->num_bits))
+        goto err;
+    d->neg = 0;
+
+    if (!BN_mul(b, &(recp->N), d, ctx))
+        goto err;
+    if (!BN_usub(r, m, b))
+        goto err;
+    r->neg = 0;
+
+    j = 0;
+    while (BN_ucmp(r, &(recp->N)) >= 0) {
+        if (j++ > 2) {
+            BNerr(BN_F_BN_DIV_RECP, BN_R_BAD_RECIPROCAL);
+            goto err;
+        }
+        if (!BN_usub(r, r, &(recp->N)))
+            goto err;
+        if (!BN_add_word(d, 1))
+            goto err;
+    }
+
+    r->neg = BN_is_zero(r) ? 0 : m->neg;
+    d->neg = m->neg ^ recp->N.neg;
+    ret = 1;
+ err:
+    BN_CTX_end(ctx);
+    bn_check_top(dv);
+    bn_check_top(rem);
+    return ret;
+}
+
+/*
+ * len is the expected size of the result We actually calculate with an extra
+ * word of precision, so we can do faster division if the remainder is not
+ * required.
+ */
+/* r := 2^len / m */
+int BN_reciprocal(BIGNUM *r, const BIGNUM *m, int len, BN_CTX *ctx)
+{
+    int ret = -1;
+    BIGNUM *t;
+
+    BN_CTX_start(ctx);
+    if ((t = BN_CTX_get(ctx)) == NULL)
+        goto err;
+
+    if (!BN_set_bit(t, len))
+        goto err;
+
+    if (!BN_div(r, NULL, t, m, ctx))
+        goto err;
+
+    ret = len;
+ err:
+    bn_check_top(r);
+    BN_CTX_end(ctx);
+    return ret;
+}