On 3/1/24 15:26, Jarkko Sakkinen wrote:
On Thu Feb 29, 2024 at 4:57 PM EET, Stefan Berger wrote:
On 2/29/24 04:11, Lukas Wunner wrote:
On Fri, Feb 23, 2024 at 03:41:40PM -0500, Stefan Berger wrote:
+static inline void ecc_digits_from_bytes(const u8 *in, unsigned int nbytes,
+ u64 *out, unsigned int ndigits)
+{
+ unsigned int sz = ndigits << ECC_DIGITS_TO_BYTES_SHIFT;
+ u8 tmp[ECC_MAX_DIGITS << ECC_DIGITS_TO_BYTES_SHIFT];
+ unsigned int o = sz - nbytes;
+
+ memset(tmp, 0, o);
+ memcpy(&tmp[o], in, nbytes);
+ ecc_swap_digits(tmp, out, ndigits);
+}
Copying the whole key into tmp seems inefficient. You only need
special handling for the first few bytes of "in" (6 bytes in the
P521 case) and could use ecc_swap_digits() to convert the rest
of "in" directly to "out" without using tmp.
So it would be sufficient to allocate the first digit on the stack,
memset + memcpy, then convert that to native byte order into "in[0]"
and use ecc_swap_digits() for the rest.
And the special handling would be conditional on "!o", so is skipped
for existing curves.
Thanks. It looks like this now:
static inline void ecc_digits_from_bytes(const u8 *in, unsigned int nbytes,
u64 *out, unsigned int ndigits)
{
unsigned int o = nbytes & 7;
u64 msd = 0;
size_t i;
if (o == 0) {
ecc_swap_digits(in, out, ndigits);
} else {
for (i = 0; i < o; i++)
msd = (msd << 8) | in[i];
out[ndigits - 1] = msd;
ecc_swap_digits(&in[o], out, ndigits - 1);
This would be more stream-lined IMHO:
unsigned int o = nbytes & 7;
unsigned int n = ndigits;
u64 msd = 0;
size_t i;
if (o != 0) {
for (i = 0; i < o; i++)
msd = (msd << 8) | in[i];
out[--n] = msd;
}
ecc_swap_digits(in, out, n);
You forgot to advance 'in'.
BR, Jarkko
