ios - 处理具有相同签名但参数不同的多个函数的最有效方法

Question

目前我有多个具有相同签名但在调用 base 时做不同事情的函数:

func drawPath(from: JMapDestination, to: JMapDestination) {
    guard let fromWaypoint = navigationManager.getWaypointForDestination(destination: from),
        let toWaypoint = navigationManager.getWaypointForDestination(destination: to) else {
            return
    }

    drawPath(from: fromWaypoint, to: toWaypoint)
}


func drawPathFrom(_ from: CGPoint, to: JMapDestination) {
    guard let fromWaypoint = getWaypointForPoint(point: from),
        let toWaypoint = navigationManager.getWaypointForDestination(destination: to) else {
            return
    }

    drawPath(from: fromWaypoint, to: toWaypoint)
}



func drawPath(from: CGPoint, to: JMapWaypoint) {
    guard let fromWaypoint = getWaypointForPoint(point: from) else { return }
    drawPath(from: fromWaypoint, to: to)
}

我决定使用 switch 语句创建一个枚举和一个主函数来处理不同的情况:

enum pathType {
    case jMapWaypoint
    case jMapDestination
    case cgPoint
}


func drawPath(pathType: pathType, fromJMap: JMapWaypoint?, toJMap: JMapWaypoint?, fromJDestination: JMapDestination?, toJDestination: JMapDestination?, fromCGPoint: CGPoint?) {

    switch pathType {

    case .jMapWaypoint:
        guard let mapController = viewModel?.mapController else {
            return
        }

        let pathStyle = setPathStyle(style: JMapStyle.init())
        if let from = fromJMap, let to = toJMap {
            checkPathsBetweenWaypoints(mapController: mapController, pathStyle: pathStyle, from: from, to: to)
        }
        if let currentMap = mapController.currentMap {
            mapController.zoomToPath(on: currentMap, withPadding: 100, withAnimationDuration: 1)
        }

    case .jMapDestination:
        if let from = fromJDestination, let to = toJDestination {
            guard let fromWaypoint = getWaypointForDestination(destination: from),
                let toWaypoint = getWaypointForDestination(destination: to) else {
                    return
            }
            drawPath(pathType: .jMapWaypoint, fromJMap: fromWaypoint, toJMap: toWaypoint, fromJDestination: nil, toJDestination: nil, fromCGPoint: nil)
        }

    case .cgPoint:
        if let from = fromCGPoint, let to = toJMap {
            guard let fromWaypoint = getWaypointForPoint(point: from) else { return }
            drawPath(pathType: .jMapWaypoint, fromJMap: fromWaypoint, toJMap: to, fromJDestination: nil, toJDestination: nil, fromCGPoint: nil)
        }
    }
}

带有 switch 语句的函数可以正常工作，但我想知道是否有更简洁、更有效的方法来做到这一点？仅供引用，所有功能都在同一个 viewController 上，我在考虑协议(protocol)，但如果协议(protocol)功能签名相同(即 drawPath)但参数不同，将如何做到这一点？

Answer 1

更清晰的版本(恕我直言)是一个包含路径开始和结束的结构，例如:

struct MyPath {

    var start: JMapWaypoint
    var end: JMapWaypoint

    init(start: CGPoint, end: CGPoint)
    {
        //Your logic here
    }
    init(start: JMapWaypoint, end: JMapWaypoint)
    {
        self.start = start
        self.end = end
    }
    init(start: JMapDestination, end: JMapDestination)
    {
        //Your logic here
    }
} 

然后你可以简单地用你想要的任何类型初始化这个对象，得到最终想要的类型并用这个对象实例绘制你的路径。