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);
BR, Jarkko
