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RSA: blind call to mpi_inv_mod() on secret value

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Signed-off-by: Manuel Pégourié-Gonnard <manuel.pegourie-gonnard@arm.com>
This commit is contained in:
Manuel Pégourié-Gonnard 2020-06-26 11:19:12 +02:00
parent b3e3d79e1a
commit 750d3c76cb
1 changed files with 22 additions and 4 deletions
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26
library/rsa.c
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@ -782,6 +782,9 @@ static int rsa_prepare_blinding( mbedtls_rsa_context *ctx,
int (*f_rng)(void *, unsigned char *, size_t), void *p_rng )
{
int ret, count = 0;
mbedtls_mpi R;
mbedtls_mpi_init( &R );
if( ctx->Vf.p != NULL )
{
@ -801,17 +804,32 @@ static int rsa_prepare_blinding( mbedtls_rsa_context *ctx,
MBEDTLS_MPI_CHK( mbedtls_mpi_fill_random( &ctx->Vf, ctx->len - 1, f_rng, p_rng ) );
ret = mbedtls_mpi_inv_mod( &ctx->Vi, &ctx->Vf, &ctx->N );
if( ret != 0 && ret != MBEDTLS_ERR_MPI_NOT_ACCEPTABLE )
/* Compute the Vf^1 as R * (R Vf)^-1 to avoid leaks from inv_mod.
* There's a negligible but non-zero probability that R is not
* invertible mod N, in that case we'd just loop one more time,
* just as if Vf itself wasn't invertible - no need to distinguish. */
MBEDTLS_MPI_CHK( mbedtls_mpi_fill_random( &R, ctx->len - 1, f_rng, p_rng ) );
MBEDTLS_MPI_CHK( mbedtls_mpi_mul_mpi( &ctx->Vi, &ctx->Vf, &R ) );
MBEDTLS_MPI_CHK( mbedtls_mpi_mod_mpi( &ctx->Vi, &ctx->Vi, &ctx->N ) );
ret = mbedtls_mpi_inv_mod( &ctx->Vi, &ctx->Vi, &ctx->N );
if( ret == MBEDTLS_ERR_MPI_NOT_ACCEPTABLE )
continue;
if( ret != 0 )
goto cleanup;
} while( ret == MBEDTLS_ERR_MPI_NOT_ACCEPTABLE )
MBEDTLS_MPI_CHK( mbedtls_mpi_mul_mpi( &ctx->Vi, &ctx->Vi, &R ) );
MBEDTLS_MPI_CHK( mbedtls_mpi_mod_mpi( &ctx->Vi, &ctx->Vi, &ctx->N ) );
} while( 0 );
/* Blinding value: Vi = Vf^(-e) mod N */
/* Blinding value: Vi = Vf^(-e) mod N
* (Vi already contains Vf^-1 at this point) */
MBEDTLS_MPI_CHK( mbedtls_mpi_exp_mod( &ctx->Vi, &ctx->Vi, &ctx->E, &ctx->N, &ctx->RN ) );
cleanup:
mbedtls_mpi_free( &R );
return( ret );
}