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Avoid reallocating during subtraction

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mbedtls_mpi_sub_abs systematically allocated a new mpi when the result
was aliased with the right operand (i.e. X = A - X). This aliasing
very commonly happens during ECP operations. Rewrite the function to
allocate only if the result might not fit otherwise.

This costs a few bytes of code size in bignum.o, and might make
mbedtls_mpi_sub_abs very very slightly slower when no reallocation is
done. However, there is a substantial performance gain in ECP
operations with Montgomery curves (10-20% on my PC).

test_suite_ecp drops from 1422794 to 1271506 calls to calloc().

This commit also fixes a bug whereby mbedtls_mpi_sub_abs would leak
memory when X == B (so TB was in use) and the result was negative.

Signed-off-by: Gilles Peskine <Gilles.Peskine@arm.com>
This commit is contained in:
Gilles Peskine
2020-07-23 01:03:22 +02:00
parent d10e8fae9e
commit 1acf7cb76c
1 changed files with 31 additions and 37 deletions
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68
library/bignum.c
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@@ -1339,29 +1339,32 @@ cleanup:
/**
* Helper for mbedtls_mpi subtraction.
*
* Calculate d - s where d and s have the same size.
* Calculate l - r where l and r 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).
* (1 if there was a wraparound, i.e. if `l < r`, 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[in] s The right operand.
* d may be aliased to l or r.
*
* \return 1 if `d < s`.
* 0 if `d >= s`.
* \param n Number of limbs of \p d, \p l and \p r.
* \param[out] d The result of the subtraction.
* \param[in] l The left operand.
* \param[in] r The right operand.
*
* \return 1 if `l < r`.
* 0 if `l >= r`.
*/
static mbedtls_mpi_uint mpi_sub_hlp( size_t n,
mbedtls_mpi_uint *d,
const mbedtls_mpi_uint *s )
const mbedtls_mpi_uint *l,
const mbedtls_mpi_uint *r )
{
size_t i;
mbedtls_mpi_uint c, z;
mbedtls_mpi_uint c = 0, t, z;
for( i = c = 0; i < n; i++, s++, d++ )
for( i = 0; i < n; i++ )
{
z = ( *d < c ); *d -= c;
c = ( *d < *s ) + z; *d -= *s;
z = ( l[i] < c ); t = l[i] - c;
c = ( t < r[i] ) + z; d[i] = t - r[i];
}
return( c );
@@ -1372,7 +1375,6 @@ static mbedtls_mpi_uint mpi_sub_hlp( size_t n,
*/
int mbedtls_mpi_sub_abs( mbedtls_mpi *X, const mbedtls_mpi *A, const mbedtls_mpi *B )
{
mbedtls_mpi TB;
int ret = MBEDTLS_ERR_ERROR_CORRUPTION_DETECTED;
size_t n;
mbedtls_mpi_uint carry;
@@ -1380,29 +1382,21 @@ int mbedtls_mpi_sub_abs( mbedtls_mpi *X, const mbedtls_mpi *A, const mbedtls_mpi
MPI_VALIDATE_RET( A != NULL );
MPI_VALIDATE_RET( B != NULL );
mbedtls_mpi_init( &TB );
if( X == B )
{
MBEDTLS_MPI_CHK( mbedtls_mpi_copy( &TB, B ) );
B = &TB;
}
if( X != A )
MBEDTLS_MPI_CHK( mbedtls_mpi_copy( X, A ) );
/*
* X should always be positive as a result of unsigned subtractions.
*/
X->s = 1;
ret = 0;
for( n = B->n; n > 0; n-- )
if( B->p[n - 1] != 0 )
break;
carry = mpi_sub_hlp( n, X->p, B->p );
MBEDTLS_MPI_CHK( mbedtls_mpi_grow( X, A->n ) );
/* Set the high limbs of X to match A. Don't touch the lower limbs
* because X might be aliased to B, and we must not overwrite the
* significant digits of B. */
if( A->n > n )
memcpy( X->p + n, A->p + n, ( A->n - n ) * ciL );
if( X->n > A->n )
memset( X->p + A->n, 0, ( X->n - A->n ) * ciL );
carry = mpi_sub_hlp( n, X->p, A->p, B->p );
if( carry != 0 )
{
/* Propagate the carry to the first nonzero limb of X. */
@@ -1418,10 +1412,10 @@ int mbedtls_mpi_sub_abs( mbedtls_mpi *X, const mbedtls_mpi *A, const mbedtls_mpi
--X->p[n];
}
/* X should always be positive as a result of unsigned subtractions. */
X->s = 1;
cleanup:
mbedtls_mpi_free( &TB );
return( ret );
}
@@ -2065,7 +2059,7 @@ static void mpi_montmul( mbedtls_mpi *A, const mbedtls_mpi *B, const mbedtls_mpi
* do the calculation without using conditional tests. */
/* Set d to d0 + (2^biL)^n - N where d0 is the current value of d. */
d[n] += 1;
d[n] -= mpi_sub_hlp( n, d, N->p );
d[n] -= mpi_sub_hlp( n, d, d, N->p );
/* If d0 < N then d < (2^biL)^n
* so d[n] == 0 and we want to keep A as it is.
* If d0 >= N then d >= (2^biL)^n, and d <= (2^biL)^n + N < 2 * (2^biL)^n