Hash and xor each entry instead, then write final xor'ed result.
name old time/op new time/op delta
Hash-4 33.6µs ± 4% 34.6µs ± 3% +3.03% (p=0.013 n=10+9)
name old alloc/op new alloc/op delta
Hash-4 1.86kB ± 0% 1.77kB ± 0% -5.10% (p=0.000 n=10+9)
name old allocs/op new allocs/op delta
Hash-4 51.0 ± 0% 49.0 ± 0% -3.92% (p=0.000 n=10+10)
Signed-off-by: Brad Fitzpatrick <bradfitz@tailscale.com>
Yes, it printed, but that was an implementation detail for hashing.
And coming optimization will make it print even less.
Signed-off-by: Brad Fitzpatrick <bradfitz@tailscale.com>