全局函数调用协议类型的变异方法。我怎样才能摆脱 var tmp 对象？

Question

这是我的工作代码片段。但是我想摆脱函数 slide.

中的 var tmp 变量

protocol Moveable {
    mutating func move(to point: CGPoint)
}

class Car : Moveable {
    var point:CGPoint = CGPoint(x: 23, y: 42)
    func move(to point: CGPoint) {
        self.point = point
        print("Method move from class Car Point: \(point)")
    }
}

struct Shape : Moveable {
    var point:CGPoint = CGPoint(x: 23, y: 42)
    mutating func move(to point: CGPoint) {
       self.point = point
       print("Method move from struct Shape Point:\(point)")
    }
}

var prius: Car = Car()
var square: Shape = Shape()

func slide(slider: Moveable) {
    var tmp = slider  // <---- I want get rid of the tmp object.(
    tmp.move(to: CGPoint(x: 100, y: 100))
}

slide(slider: prius)
slide(slider: square)

我试过类似的方法来避免 var tmp:

func slide(slider: inout Moveable) {
    slider.move(to: CGPoint(x: 100, y: 100))
}

slider(slider: prius) // <--- Compile Error
// Cannot use mutating member on immutable value: 'slider' is a 'let' constant
slider(slider: prius as (inout Moveable) // <--- Compile Error

谢谢。

Answer 1

它可能适用于这个小改动：

func slide<M: Moveable>(slider: inout M) {
    slider.move(to: CGPoint(x: 100, y: 100))
}

我们不要求 Movable，而是要求符合 Movable.

的类型

并这样称呼它：

slide(slider: &prius)
slide(slider: &square)

全局函数调用协议类型的变异方法。我怎样才能摆脱 var tmp 对象？

A global function calls a mutating method of a protocol type. How can I get rid of the var tmp object?

protocols

swift

inout