属性 getter 和 setter [英] Property getters and setters

问题描述

通过这个简单的类，我得到了编译器警告

With this simple class I am getting the compiler warning

试图在自己的 setter/getter 中修改/访问 x

Attempting to modify/access x within its own setter/getter

当我像这样使用它时:

var p: point = Point()
p.x = 12

我收到了 EXC_BAD_ACCESS.如果没有明确的支持 ivars，我该如何做到这一点?

I get an EXC_BAD_ACCESS. How can I do this without explicit backing ivars?

class Point {

    var x: Int {
        set {
            x = newValue * 2 //Error
        }
        get {
            return x / 2 //Error
        }
    }
    // ...
}

推荐答案

Setter 和 Getter 适用于 计算属性;此类属性在实例中没有存储 - 来自 getter 的值旨在从其他实例属性计算.在您的情况下，没有要分配的 x.

Setters and Getters apply to computed properties; such properties do not have storage in the instance - the value from the getter is meant to be computed from other instance properties. In your case, there is no x to be assigned.

明确表示:如果没有明确的支持 ivars，我怎么能做到这一点".你不能 - 你需要东西来备份计算属性.试试这个:

Explicitly: "How can I do this without explicit backing ivars". You can't - you'll need something to backup the computed property. Try this:

class Point {
  private var _x: Int = 0             // _x -> backingX
  var x: Int {
    set { _x = 2 * newValue }
    get { return _x / 2 }
  }
}

特别是在 Swift REPL 中:

Specifically, in the Swift REPL:

 15> var pt = Point()
pt: Point = {
  _x = 0
}
 16> pt.x = 10
 17> pt
$R3: Point = {
  _x = 20
}
 18> pt.x
$R4: Int = 10

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