swift - 我必须沮丧还是我错过了什么？

Question

所以我有一个我想在调用站点动态化的类型:

    struct DynamicView<Content: View>: View {
    let content: Content
    var body: some View {
              content
           }
    }

使用某种类型的函数将其转换为具体的东西

    static func contentCreator<T: View>(id: Int) -> T {
    switch id {
    case 0:
        return FirstView() //<-- Xcode is demanding that I state as! T here
    }
}

FirstView 在哪里

struct FirstView: View {
    var body: some View {
        Circle()
    }
}

我是在接近这个错误还是什么？只要它们都符合通用协议(protocol)，它们应该是可以互换的，对吧？

Answer 1

这是因为类型不是通过检查返回来推断的。
如果你试试:

let v = contentCreator(id: 0)

你会得到一个 Generic parameter T could not be inferred错误，因为 v 类型未知。
如果设置了类型，它将编译:

let v: View = contentCreator(id: 0)

所以回到你的提问，因为 View 是一个协议(protocol)，当 Xcode 构建源代码时，它无法知道类 'T'，也没有办法在你使用它时强制你使用正确的类型。

struct FirstView: View {
    var body: some View {
        Circle()
    }
}

struct SecondView: View {
    var body: some View {
        Square()
    }
}

// This will work, because once executed, we know v1 class. 
// The cast will work as long as v1 is declared as conforming to 'View' protocol. 
// Type is not inferred in this case, but known only when it returns.

let v1: View = contentCreator(0)

/// This won't work, because the inferred type is 'SecondView'. 
/// The cast 'return FirstView() as! T' will crash

let v2: SecondView = contentCreator(0)