Re-align some comments after running the reformat script.

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

diff --git a/crypto/bn/bn_sqrt.c b/crypto/bn/bn_sqrt.c
index 772c8080bb5d818fb3472eb566a98837feb8d727..232af99a216d853552d8260dcaf9f02908ad6aa5 100644 (file)
--- a/crypto/bn/bn_sqrt.c
+++ b/crypto/bn/bn_sqrt.c
@@ -132,14 +132,14 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
     /* we'll set  q  later (if needed) */
 
     if (e == 1) {
-                /*-
-                 * 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
-                 *     2 * (|p|+1)/4 == 1   (mod (|p|-1)/2),
-                 * so we can use exponent  (|p|+1)/4,  i.e.  (|p|-3)/4 + 1.
-                 */
+        /*-
+         * 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
+         *     2 * (|p|+1)/4 == 1   (mod (|p|-1)/2),
+         * so we can use exponent  (|p|+1)/4,  i.e.  (|p|-3)/4 + 1.
+         */
         if (!BN_rshift(q, p, 2))
             goto end;
         q->neg = 0;
@@ -152,32 +152,32 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
     }
 
     if (e == 2) {
-                /*-
-                 * |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.
-                 * So if  a  really is a square, then  2*a  is a non-square.
-                 * Thus for
-                 *      b := (2*a)^((|p|-5)/8),
-                 *      i := (2*a)*b^2
-                 * we have
-                 *     i^2 = (2*a)^((1 + (|p|-5)/4)*2)
-                 *         = (2*a)^((p-1)/2)
-                 *         = -1;
-                 * so if we set
-                 *      x := a*b*(i-1),
-                 * then
-                 *     x^2 = a^2 * b^2 * (i^2 - 2*i + 1)
-                 *         = a^2 * b^2 * (-2*i)
-                 *         = a*(-i)*(2*a*b^2)
-                 *         = a*(-i)*i
-                 *         = a.
-                 *
-                 * (This is due to A.O.L. Atkin,
-                 * <URL: http://listserv.nodak.edu/scripts/wa.exe?A2=ind9211&L=nmbrthry&O=T&P=562>,
-                 * November 1992.)
-                 */
+        /*-
+         * |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.
+         * So if  a  really is a square, then  2*a  is a non-square.
+         * Thus for
+         *      b := (2*a)^((|p|-5)/8),
+         *      i := (2*a)*b^2
+         * we have
+         *     i^2 = (2*a)^((1 + (|p|-5)/4)*2)
+         *         = (2*a)^((p-1)/2)
+         *         = -1;
+         * so if we set
+         *      x := a*b*(i-1),
+         * then
+         *     x^2 = a^2 * b^2 * (i^2 - 2*i + 1)
+         *         = a^2 * b^2 * (-2*i)
+         *         = a*(-i)*(2*a*b^2)
+         *         = a*(-i)*i
+         *         = a.
+         *
+         * (This is due to A.O.L. Atkin,
+         * <URL: http://listserv.nodak.edu/scripts/wa.exe?A2=ind9211&L=nmbrthry&O=T&P=562>,
+         * November 1992.)
+         */
 
         /* t := 2*a */
         if (!BN_mod_lshift1_quick(t, A, p))
@@ -277,24 +277,24 @@ 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
-         * k,  0 <= k < 2^e,  such that
-         *
-         *      a^q * y^k == 1   (mod p).
-         *
-         * As  a^q  is a square and  y  is not,  k  must be even.
-         * q+1  is even, too, so there is an element
-         *
-         *     X := a^((q+1)/2) * y^(k/2),
-         *
-         * and it satisfies
-         *
-         *     X^2 = a^q * a     * y^k
-         *         = a,
-         *
-         * so it is the square root that we are looking for.
-         */
+    /*-
+     * 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).
+     *
+     * As  a^q  is a square and  y  is not,  k  must be even.
+     * q+1  is even, too, so there is an element
+     *
+     *     X := a^((q+1)/2) * y^(k/2),
+     *
+     * and it satisfies
+     *
+     *     X^2 = a^q * a     * y^k
+     *         = a,
+     *
+     * so it is the square root that we are looking for.
+     */
 
     /* t := (q-1)/2  (note that  q  is odd) */
     if (!BN_rshift1(t, q))
@@ -333,15 +333,15 @@ BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
         goto end;
 
     while (1) {
-                /*-
-                 * 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).
-                 *
-                 * We have  a*b = x^2,
-                 *    y^2^(e-1) = -1,
-                 *    b^2^(e-1) = 1.
-                 */
+        /*-
+         * 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).
+         *
+         * We have  a*b = x^2,
+         *    y^2^(e-1) = -1,
+         *    b^2^(e-1) = 1.
+         */
 
         if (BN_is_one(b)) {
             if (!BN_copy(ret, x))