Swift：无法为协议属性赋值？ [英] Swift: Failed to assign value to a property of protocol?

问题描述

A类提供字符串值。 B类本身有两个A类成员，并提供一个计算属性v来选择其中一个。

Class A provides a string value. Class B has two members of A type inside itself, and provide a computed property "v" to choose one of them.

class A {
    var value: String

    init(value: String) {
        self.value = value
    }
}

class B {
    var v1: A?
    var v2: A = A(value: "2")

    private var v: A {
        return v1 ?? v2
    }

    var value: String {
        get {
            return v.value
        }
        set {
            v.value = newValue
        }
    }
}

此代码为简单而且有效。由于A和B都有一个成员value，我把它作为这样的协议：

This code is simple and it works. Since both the A and B have a member "value", I make it a protocol like this:

protocol ValueProvider {
    var value: String {get set}
}

class A: ValueProvider {
    var value: String

    init(value: String) {
        self.value = value
    }
}

class B: ValueProvider {
    var v1: ValueProvider?
    var v2: ValueProvider = A(value: "2")

    private var v: ValueProvider {
        return v1 ?? v2
    }

    var value: String {
        get {
            return v.value
        }
        set {
            v.value = newValue // Error: Cannot assign to the result of the expression
        }
    }
}

如果我更改以下代码

v.value = newValue

到

var v = self.v
v.value = newValue

再次有效！

这是Swift的错误，还是协议属性的特殊内容？

Is this a bug of Swift, or something special for the property of protocols?

推荐答案

您必须将协议定义为类协议：

You have to define the protocol as a class protocol:

protocol ValueProvider : class {
    var value: String {get set}
}

然后

var value: String {
    get { return v.value }
    set { v.value = newValue }
}

编制并按预期工作（即将新值分配给 v1 引用的
对象（如果 v1！= nil ）和对象
由 v2引用否则）。

compiles and works as expected (i.e. assigns the new value to the object referenced by v1 if v1 != nil, and to the object referenced by v2 otherwise).

v 是 ValueProvider 类型的只读计算属性。
通过将协议定义为类协议，编译器知道
v 是引用类型，因此它的 v.value
属性可以修改，即使引用本身是常量。

v is a read-only computed property of the type ValueProvider. By defining the protocol as a class protocol the compiler knows that v is a reference type, and therefore its v.value property can be modified even if the reference itself is a constant.

您的初始代码示例有效，因为 v 属性的
类型为 A ，这是一种引用类型。

Your initial code example works because there the v property has the type A which is a reference type.

您的解决方法

set {
    var tmp = v1 ?? v2
    tmp.value = newValue
}

的工作原因是（读写） ）变量的属性可以在
中设置任何情况（值类型或引用类型）。

works because (read-write) properties of variables can be set in any case (value type or reference type).

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