diff options
Diffstat (limited to 'crypto/bn/bn_sqrt.c')
| -rw-r--r-- | crypto/bn/bn_sqrt.c | 12 |
1 files changed, 8 insertions, 4 deletions
diff --git a/crypto/bn/bn_sqrt.c b/crypto/bn/bn_sqrt.c index 6beaf9e5e5..04cf4a0bf8 100644 --- a/crypto/bn/bn_sqrt.c +++ b/crypto/bn/bn_sqrt.c @@ -135,7 +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 @@ -152,7 +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. @@ -262,7 +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). @@ -318,7 +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). * |