diff --git a/library/bignum.c b/library/bignum.c index 7a29eb641..ff6fa4438 100644 --- a/library/bignum.c +++ b/library/bignum.c @@ -1265,12 +1265,23 @@ cleanup: } /* - * Helper for mbedtls_mpi subtraction: - * d -= s where d and s have the same size and d >= s. + * Helper for mbedtls_mpi subtraction. + * + * Calculate d - s where d and s have the same size. + * This function operates modulo (2^ciL)^n and returns the carry + * (1 if there was a wraparound, i.e. if `d < s`, and 0 otherwise). + * + * \param n Number of limbs of \p d and \p s. + * \param[in,out] d On input, the left operand. + * On output, the result of the subtraction: + * \param[s] The right operand. + * + * \return 1 if `d < s`. + * 0 if `d >= s`. */ -static void mpi_sub_hlp( size_t n, - mbedtls_mpi_uint *d, - const mbedtls_mpi_uint *s ) +static mbedtls_mpi_uint mpi_sub_hlp( size_t n, + mbedtls_mpi_uint *d, + const mbedtls_mpi_uint *s ) { size_t i; mbedtls_mpi_uint c, z; @@ -1281,11 +1292,7 @@ static void mpi_sub_hlp( size_t n, c = ( *d < *s ) + z; *d -= *s; } - while( c != 0 ) - { - z = ( *d < c ); *d -= c; - c = z; d++; - } + return( c ); } /* @@ -1296,6 +1303,7 @@ int mbedtls_mpi_sub_abs( mbedtls_mpi *X, const mbedtls_mpi *A, const mbedtls_mpi mbedtls_mpi TB; int ret; size_t n; + mbedtls_mpi_uint c, z; MPI_VALIDATE_RET( X != NULL ); MPI_VALIDATE_RET( A != NULL ); MPI_VALIDATE_RET( B != NULL ); @@ -1325,7 +1333,12 @@ int mbedtls_mpi_sub_abs( mbedtls_mpi *X, const mbedtls_mpi *A, const mbedtls_mpi if( B->p[n - 1] != 0 ) break; - mpi_sub_hlp( n, X->p, B->p ); + c = mpi_sub_hlp( n, X->p, B->p ); + while( c != 0 ) + { + z = ( X->p[n] < c ); X->p[n] -= c; + c = z; n++; + } cleanup: @@ -1959,7 +1972,7 @@ static void mpi_montmul( mbedtls_mpi *A, const mbedtls_mpi *B, const mbedtls_mpi * timing attacks. */ /* Set d to A + (2^biL)^n - N. */ d[n] += 1; - mpi_sub_hlp( n, d, N->p ); + d[n] -= mpi_sub_hlp( n, d, N->p ); /* Now d - (2^biL)^n = A - N so d >= (2^biL)^n iff A >= N. * So we want to copy the result of the subtraction iff d->p[n] != 0. * Note that d->p[n] is either 0 or 1 since A - N <= N <= (2^biL)^n. */