ec/ecp_nistz256.c: harmonize with latest indent script.

[oweals/openssl.git] / crypto / bn / bn_sqrt.c

diff --git a/crypto/bn/bn_sqrt.c b/crypto/bn/bn_sqrt.c
index 51902703e1ed1e8b4ba3e8d621894079feeeacd7..04cf4a0bf83e37f84b602fd5d8a30280315614d2 100644 (file)
--- a/crypto/bn/bn_sqrt.c
+++ b/crypto/bn/bn_sqrt.c
@@ -1,4 +1,4 @@
-/* crypto/bn/bn_mod.c */
+/* crypto/bn/bn_sqrt.c */
 /* Written by Lenka Fibikova <fibikova@exp-math.uni-essen.de>
  * and Bodo Moeller for the OpenSSL project. */
 /* ====================================================================
@@ -83,7 +83,8 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
                                goto end;
                        if (!BN_set_word(ret, BN_is_bit_set(a, 0)))
                                {
-                               BN_free(ret);
+                               if (ret != in)
+                                       BN_free(ret);
                                return NULL;
                                }
                        bn_check_top(ret);
@@ -102,7 +103,8 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
                        goto end;
                if (!BN_set_word(ret, BN_is_one(a)))
                        {
-                       BN_free(ret);
+                       if (ret != in)
+                               BN_free(ret);
                        return NULL;
                        }
                bn_check_top(ret);
@@ -133,7 +135,8 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
 
        if (e == 1)
                {
-               /* The easy case:  (|p|-1)/2  is odd, so 2 has an inverse
+               /*-
+                * The easy case:  (|p|-1)/2  is odd, so 2 has an inverse
                 * modulo  (|p|-1)/2,  and square roots can be computed
                 * directly by modular exponentiation.
                 * We have
@@ -150,7 +153,8 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
        
        if (e == 2)
                {
-               /* |p| == 5  (mod 8)
+               /*-
+                * |p| == 5  (mod 8)
                 *
                 * In this case  2  is always a non-square since
                 * Legendre(2,p) = (-1)^((p^2-1)/8)  for any odd prime.
@@ -260,7 +264,8 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
                goto end;
                }
 
-       /* Now we know that (if  p  is indeed prime) there is an integer
+       /*-
+        * Now we know that (if  p  is indeed prime) there is an integer
         * k,  0 <= k < 2^e,  such that
         *
         *      a^q * y^k == 1   (mod p).
@@ -288,7 +293,7 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
                if (BN_is_zero(t))
                        {
                        /* special case: a == 0  (mod p) */
-                       if (!BN_zero(ret)) goto end;
+                       BN_zero(ret);
                        err = 0;
                        goto end;
                        }
@@ -301,7 +306,7 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
                if (BN_is_zero(x))
                        {
                        /* special case: a == 0  (mod p) */
-                       if (!BN_zero(ret)) goto end;
+                       BN_zero(ret);
                        err = 0;
                        goto end;
                        }
@@ -316,7 +321,8 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
 
        while (1)
                {
-               /* Now  b  is  a^q * y^k  for some even  k  (0 <= k < 2^E
+               /*- 
+                * Now  b  is  a^q * y^k  for some even  k  (0 <= k < 2^E
                 * where  E  refers to the original value of  e,  which we
                 * don't keep in a variable),  and  x  is  a^((q+1)/2) * y^(k/2).
                 *