协议类型的 inout 变量是否被禁止? [英] Are inout variables of protocol type prohibited?

问题描述

以下代码:

protocol SomeProtocol {}
class SomeClass: SomeProtocol {}

private func doSomethingWith(inout someVar: SomeProtocol) {}

private var someGlobalVar = SomeClass() // inferring SomeClass's type

doSomethingWith(&someGlobalVar)

产生以下错误:

无法使用类型为(inout SomeClass)"的参数列表调用doSomethingWith"

Cannot invoke 'doSomethingWith' with an argument list of type '(inout SomeClass)'

将倒数第二行更改为 private var someGlobalVar: SomeProtocol = SomeClass() 可解决错误.

Changing the penultimate line to private var someGlobalVar: SomeProtocol = SomeClass() resolves the error.

主题

推荐答案

当你在声明时将一个 SomeClass 实例分配给一个变量，变量类型被推断为 SomeClass.和写作一样

When you assign a SomeClass instance to a variable while declaring, the variable type is inferred to be SomeClass. The same as writing

private var someGlobalVar: SomeClass = SomeClass()

但是，当传递给 inout 参数时，该函数可以将另一个实例分配给该变量，例如

However, when passing to an inout parameter, the function can assign another instance to that variable, e.g.

private func doSomethingWith(inout someVar: SomeProtocol) {
    someVar = OtherClass()
}

现在你有一个类型不匹配.您看到的错误是 Swift 阻止您遇到类似问题.

Now you have a type mismatch. The error you are seeing is Swift preventing you getting a similar problem.

换句话说:如果您将一个变量传递给一个函数，并且您知道该函数可以将采用 SomeProtocol 的任何实例分配给该变量，那么您必须使用一个实际上可以保存任何采用SomeProtocol的实例:

In other words: if you are passing a variable to a function and you know that the function can assign any instance adopting SomeProtocol to that variable, you have to use a variable that can actually hold any instance adopting SomeProtocol:

private var someGlobalVar: SomeProtocol

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