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Fix a few typos

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Co-authored-by: Felix Conway <felix.conway@arm.com>
Signed-off-by: Manuel Pégourié-Gonnard <manuel.pegourie-gonnard@arm.com>
This commit is contained in:
Manuel Pégourié-Gonnard
2025-08-12 09:18:28 +02:00
parent 40dfc811ef
commit 7a5447ff65
1 changed files with 3 additions and 3 deletions
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6
library/bignum.c
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@@ -2003,7 +2003,7 @@ cleanup:
* This is not obvious because our constant-time modinv function only works with
* an odd modulus, and here the modulus is even. The idea is that computing a
* a^-1 mod b is really just computing the u coefficient in the Bézout relation
* a*u + b*v = 1 (assuming gcd(a,b) = 1, ie the inverse exists). But if we know
* a*u + b*v = 1 (assuming gcd(a,b) = 1, i.e. the inverse exists). But if we know
* one of u, v in this relation then the other is easy to find. So we can
* actually start by computing N^-1 mod A with gives us "the wrong half" of the
* Bézout relation, from which we'll deduce the interesting half A^-1 mod N.
@@ -2062,7 +2062,7 @@ static int mbedtls_mpi_inv_mod_even(mbedtls_mpi *X,
/* Bring A in the range [0, N). */
MBEDTLS_MPI_CHK(mbedtls_mpi_mod_mpi(&AA, A, N));
/* We know A >= 0 but the next functions wants A > 1 */
/* We know A >= 0 but the next function wants A > 1 */
int cmp = mbedtls_mpi_cmp_int(&AA, 1);
if (cmp < 0) { // AA == 0
ret = MBEDTLS_ERR_MPI_NOT_ACCEPTABLE;
@@ -2100,7 +2100,7 @@ int mbedtls_mpi_inv_mod(mbedtls_mpi *X, const mbedtls_mpi *A, const mbedtls_mpi
return mbedtls_mpi_inv_mod_even(X, A, N);
}
/* If A and N are both even, 2 divides they GCD, so no inverse. */
/* If A and N are both even, 2 divides their GCD, so no inverse. */
return MBEDTLS_ERR_MPI_NOT_ACCEPTABLE;
}