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lukevalenta
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Jul 15, 2026
lukevalenta
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LGTM. Some suggestions and one fix to the comments.
| c.Rsh(c, 1) | ||
| c.Sub(c, &bi) | ||
| si := s[i%m.p.D] | ||
| kbit := uint64(k[i>>3]>>(i&7)) & 1 |
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Possibly worth extracting this into a helper function for readability (used twice in this function and once in signs):
Suggested change
| kbit := uint64(k[i>>3]>>(i&7)) & 1 | |
| kbit := uint64(bit(k, i)) |
The new function:
// bit returns the i-th bit of buf in little-endian order, as 0 or 1.
func bit(buf []byte, i uint) byte {
return byte(buf[i>>3]>>(i&7)) & 1
}
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prefer to keep it inline for easier reading
Co-authored-by: Luke Valenta <lvalenta@cloudflare.com>
Co-authored-by: Luke Valenta <lvalenta@cloudflare.com>
Co-authored-by: Luke Valenta <lvalenta@cloudflare.com>
Co-authored-by: Luke Valenta <lvalenta@cloudflare.com>
Co-authored-by: Luke Valenta <lvalenta@cloudflare.com>
Co-authored-by: Luke Valenta <lvalenta@cloudflare.com>
Co-authored-by: Luke Valenta <lvalenta@cloudflare.com>
lukevalenta
reviewed
Jul 15, 2026
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| // Original algorithm starts with c := k >> D, then computes | ||
| // | ||
| // b_(i-D) = s_(j%D) * lsb(c) |
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Suggested change
| // b_(i-D) = s_(j%D) * lsb(c) | |
| // b_(i-D) = s_(i%D) * lsb(c) |
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