diff options
author | Matt Caswell <matt@openssl.org> | 2015-01-05 11:30:03 +0000 |
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committer | Matt Caswell <matt@openssl.org> | 2015-01-22 09:20:10 +0000 |
commit | 50e735f9e5d220cdad7db690188b82a69ddcb39e (patch) | |
tree | 48043d67891fa563074cfe4f33fe68761b5c3aba /crypto/bn/bn_sqrt.c | |
parent | 739a5eee619fc8c03736140828891b369f8690f4 (diff) | |
download | openssl-50e735f9e5d220cdad7db690188b82a69ddcb39e.tar.gz |
Re-align some comments after running the reformat script.
This should be a one off operation (subsequent invokation of the
script should not move them)
Reviewed-by: Tim Hudson <tjh@openssl.org>
Diffstat (limited to 'crypto/bn/bn_sqrt.c')
-rw-r--r-- | crypto/bn/bn_sqrt.c | 70 |
1 files changed, 35 insertions, 35 deletions
diff --git a/crypto/bn/bn_sqrt.c b/crypto/bn/bn_sqrt.c index 772c8080bb..1b259f31c6 100644 --- 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; @@ -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)) |