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device: optimize Peer.String even more

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This reduces the allocation, branches, and amount of base64 encoding.

Signed-off-by: Jason A. Donenfeld <Jason@zx2c4.com>
This commit is contained in:
Jason A. Donenfeld 2021-05-14 01:07:55 +02:00
parent 25ad08a591
commit 9087e444e6
1 changed files with 16 additions and 14 deletions
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30
device/peer.go
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@ -7,7 +7,6 @@ package device
import (
"container/list"
"encoding/base64"
"errors"
"sync"
"sync/atomic"
@ -150,19 +149,22 @@ func (peer *Peer) String() string {
// return fmt.Sprintf("peer(%s)", abbreviatedKey)
//
// except that it is considerably more efficient.
const prefix = "peer("
b := make([]byte, len(prefix)+44)
copy(b, prefix)
r := b[len(prefix):]
base64.StdEncoding.Encode(r, peer.handshake.remoteStatic[:])
r = r[4:]
copy(r, "…")
r = r[len("…"):]
copy(r, b[len(prefix)+39:len(prefix)+43])
r = r[4:]
r[0] = ')'
r = r[1:]
return string(b[:len(b)-len(r)])
src := peer.handshake.remoteStatic
b64 := func(input byte) byte {
return input + 'A' + byte(((25-int(input))>>8)&6) - byte(((51-int(input))>>8)&75) - byte(((61-int(input))>>8)&15) + byte(((62-int(input))>>8)&3)
}
b := []byte("peer(____…____)")
const first = len("peer(")
const second = len("peer(____…")
b[first+0] = b64((src[0] >> 2) & 63)
b[first+1] = b64(((src[0] << 4) | (src[1] >> 4)) & 63)
b[first+2] = b64(((src[1] << 2) | (src[2] >> 6)) & 63)
b[first+3] = b64(src[2] & 63)
b[second+0] = b64(src[29] & 63)
b[second+1] = b64((src[30] >> 2) & 63)
b[second+2] = b64(((src[30] << 4) | (src[31] >> 4)) & 63)
b[second+3] = b64((src[31] << 2) & 63)
return string(b)
}
func (peer *Peer) Start() {