device: optimize Peer.String even more
This reduces the allocation, branches, and amount of base64 encoding. Signed-off-by: Jason A. Donenfeld <Jason@zx2c4.com>
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parent
25ad08a591
commit
9087e444e6
@ -7,7 +7,6 @@ package device
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import (
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"container/list"
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"encoding/base64"
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"errors"
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"sync"
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"sync/atomic"
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@ -150,19 +149,22 @@ func (peer *Peer) String() string {
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// return fmt.Sprintf("peer(%s)", abbreviatedKey)
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//
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// except that it is considerably more efficient.
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const prefix = "peer("
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b := make([]byte, len(prefix)+44)
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copy(b, prefix)
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r := b[len(prefix):]
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base64.StdEncoding.Encode(r, peer.handshake.remoteStatic[:])
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r = r[4:]
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copy(r, "…")
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r = r[len("…"):]
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copy(r, b[len(prefix)+39:len(prefix)+43])
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r = r[4:]
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r[0] = ')'
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r = r[1:]
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return string(b[:len(b)-len(r)])
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src := peer.handshake.remoteStatic
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b64 := func(input byte) byte {
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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)
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}
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b := []byte("peer(____…____)")
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const first = len("peer(")
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const second = len("peer(____…")
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b[first+0] = b64((src[0] >> 2) & 63)
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b[first+1] = b64(((src[0] << 4) | (src[1] >> 4)) & 63)
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b[first+2] = b64(((src[1] << 2) | (src[2] >> 6)) & 63)
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b[first+3] = b64(src[2] & 63)
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b[second+0] = b64(src[29] & 63)
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b[second+1] = b64((src[30] >> 2) & 63)
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b[second+2] = b64(((src[30] << 4) | (src[31] >> 4)) & 63)
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b[second+3] = b64((src[31] << 2) & 63)
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return string(b)
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}
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func (peer *Peer) Start() {
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