get rid of Router.NormalisePeerAddr
There was only one call site left and it always gets invoked with an ip:port, so normalisation is a no-op. Also remove mysterious comment.
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@@ -93,10 +93,8 @@ func (peer *LocalPeer) CreateConnection(peerAddr string, acceptNewPeer bool) err
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if err := peer.checkConnectionLimit(); err != nil {
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return err
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}
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// We're dialing the remote so that means connections will come from random ports
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addrStr := peer.router.NormalisePeerAddr(peerAddr)
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tcpAddr, tcpErr := net.ResolveTCPAddr("tcp4", addrStr)
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udpAddr, udpErr := net.ResolveUDPAddr("udp4", addrStr)
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tcpAddr, tcpErr := net.ResolveTCPAddr("tcp4", peerAddr)
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udpAddr, udpErr := net.ResolveUDPAddr("udp4", peerAddr)
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if tcpErr != nil || udpErr != nil {
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// they really should have the same value, but just in case...
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if tcpErr == nil {
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@@ -417,13 +417,3 @@ func (router *Router) applyTopologyUpdate(update []byte) (PeerNameSet, PeerNameS
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}
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return origUpdate, newUpdate, nil
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}
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// given an address like '1.2.3.4:567', return the address if it has a port,
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// otherwise return the address with the default port number for the router
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func (router *Router) NormalisePeerAddr(peerAddr string) string {
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_, _, err := net.SplitHostPort(peerAddr)
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if err == nil {
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return peerAddr
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}
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return fmt.Sprintf("%s:%d", peerAddr, router.Port)
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}
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