无法在Swift中创建符合协议类型的数组

问题描述

我有以下协议和符合该协议的类:

I have the following protocol and a class that conforms to it:

protocol Foo{
    typealias BazType

    func bar(x:BazType) ->BazType
}


class Thing: Foo {
    func bar(x: Int) -> Int {
        return x.successor()
    }
}

当我尝试创建foos数组时，出现一个奇怪的错误:

When I try to create an Array of foos, I get an odd error:

var foos: Array<Foo> = [Thing()]

好，因此只有在具有关联的类型要求(确实如此)的情况下才可以使用它，但是由于某种原因，这是一个错误? WTF?！

OK, so it can only be used if it has an associated type requirement (which it does), but for some reason this is an error?? WTF?!

我不确定我是否完全理解编译器要告诉我的内容...

I'm not sure I fully understand what the compiler is trying to tell me...

推荐答案

比方说，如果我们可以将Thing的实例放入数组foos，会发生什么?

Let's say, if we could put an instance of Thing into array foos, what will happen?

protocol Foo {
    typealias BazType

    func bar(x:BazType) -> BazType
}

class Thing: Foo {
    func bar(x: Int) -> Int {
        return x.successor()
    }
}

class AnotherThing: Foo {
    func bar(x: String) -> String {
        return x
    }
}

var foos: [Foo] = [Thing()]

因为AnotherThing也符合Foo，所以我们也可以将其放入foos.

Because AnotherThing conforms to Foo too, so we can put it into foos also.

foos.append(AnotherThing())

现在，我们从foos中随机抓取一个foo.

Now we grab a foo from foos randomly.

let foo = foos[Int(arc4random_uniform(UInt32(foos.count - 1)))]

然后我要调用方法bar，你能告诉我应该向bar发送字符串还是整数?

and I'm going to call method bar, can you tell me that I should send a string or an integer to bar?

foo.bar("foo")或foo.bar(1)

雨燕不能.

因此它只能用作一般约束.

So it can only be used as a generic constraint.

什么情况下需要这样的协议?

What scenario requires a protocol like this?

示例:

class MyClass<T: Foo> {
        let fooThing: T?

        init(fooThing: T? = nil) {
                self.fooThing = fooThing
        }

        func myMethod() {
                let thing = fooThing as? Thing // ok
                thing?.bar(1) // fine

                let anotherThing = fooThing as? AnotherThing // no problem
                anotherThing?.bar("foo") // you can do it

                // but you can't downcast it to types which doesn't conform to Foo
                let string = fooThing as? String // this is an error
        }
}

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