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use go built-in min & max functions to replace lo.Min and lo.Max (#4007)
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@ -224,7 +224,7 @@ func (mhr *MakeHoleRecords) ReportSuccess(mode int, index int) {
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}
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score.Score += 2
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score.Score = lo.Min([]int{score.Score, 10})
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score.Score = min(score.Score, 10)
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return
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}
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}
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@ -317,7 +317,7 @@ func (c *Controller) analysis(session *Session) (*msg.NatHoleResp, *msg.NatHoleR
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session.cBehavior = cBehavior
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session.vBehavior = vBehavior
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timeoutMs := lo.Max([]int{cBehavior.SendDelayMs, vBehavior.SendDelayMs}) + 5000
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timeoutMs := max(cBehavior.SendDelayMs, vBehavior.SendDelayMs) + 5000
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if cBehavior.ListenRandomPorts > 0 || vBehavior.ListenRandomPorts > 0 {
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timeoutMs += 30000
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}
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@ -384,8 +384,8 @@ func getRangePorts(addrs []string, difference, maxNumber int) []msg.PortsRange {
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return nil
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}
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ports = append(ports, msg.PortsRange{
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From: lo.Max([]int{port - difference - 5, port - maxNumber, 1}),
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To: lo.Min([]int{port + difference + 5, port + maxNumber, 65535}),
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From: max(port-difference-5, port-maxNumber, 1),
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To: min(port+difference+5, port+maxNumber, 65535),
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})
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return ports
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}
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