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RSA: use CT gcd-modinv in prepare_blinding()
While at it, draw the blinding value uniformly in the permissible range. Signed-off-by: Manuel Pégourié-Gonnard <manuel.pegourie-gonnard@arm.com>
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@@ -1304,33 +1304,16 @@ static int rsa_prepare_blinding(mbedtls_rsa_context *ctx,
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}
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}
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/* Unblinding value: Vf = random number, invertible mod N */
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/* Unblinding value: Vf = random number, invertible mod N */
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mbedtls_mpi_lset(&R, 0);
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do {
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do {
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if (count++ > 10) {
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if (count++ > 10) {
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ret = MBEDTLS_ERR_RSA_RNG_FAILED;
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ret = MBEDTLS_ERR_RSA_RNG_FAILED;
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goto cleanup;
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goto cleanup;
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}
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}
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MBEDTLS_MPI_CHK(mbedtls_mpi_fill_random(&ctx->Vf, ctx->len - 1, f_rng, p_rng));
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MBEDTLS_MPI_CHK(mbedtls_mpi_random(&ctx->Vf, 1, &ctx->N, f_rng, p_rng));
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MBEDTLS_MPI_CHK(mbedtls_mpi_gcd_modinv_odd(&R, &ctx->Vi, &ctx->Vf, &ctx->N));
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/* Compute Vf^-1 as R * (R Vf)^-1 to avoid leaks from inv_mod. */
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} while (mbedtls_mpi_cmp_int(&R, 1) != 0);
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MBEDTLS_MPI_CHK(mbedtls_mpi_fill_random(&R, ctx->len - 1, f_rng, p_rng));
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MBEDTLS_MPI_CHK(mbedtls_mpi_mul_mpi(&ctx->Vi, &ctx->Vf, &R));
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MBEDTLS_MPI_CHK(mbedtls_mpi_mod_mpi(&ctx->Vi, &ctx->Vi, &ctx->N));
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/* At this point, Vi is invertible mod N if and only if both Vf and R
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* are invertible mod N. If one of them isn't, we don't need to know
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* which one, we just loop and choose new values for both of them.
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* (Each iteration succeeds with overwhelming probability.) */
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ret = mbedtls_mpi_inv_mod(&ctx->Vi, &ctx->Vi, &ctx->N);
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if (ret != 0 && ret != MBEDTLS_ERR_MPI_NOT_ACCEPTABLE) {
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goto cleanup;
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}
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} while (ret == MBEDTLS_ERR_MPI_NOT_ACCEPTABLE);
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/* Finish the computation of Vf^-1 = R * (R Vf)^-1 */
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MBEDTLS_MPI_CHK(mbedtls_mpi_mul_mpi(&ctx->Vi, &ctx->Vi, &R));
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MBEDTLS_MPI_CHK(mbedtls_mpi_mod_mpi(&ctx->Vi, &ctx->Vi, &ctx->N));
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/* Blinding value: Vi = Vf^(-e) mod N
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/* Blinding value: Vi = Vf^(-e) mod N
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* (Vi already contains Vf^-1 at this point) */
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* (Vi already contains Vf^-1 at this point) */
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