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swift便利init和泛型类

[英]swift convenience init and generic class

I have a problem creating a convenience init method that then calls a designated init on a class with generic type parameters. 我在创建一个方便的init方法时遇到问题,然后在具有泛型类型参数的类上调用指定的init。 Here is the swift 3.1 XCode Version 8.3.2 (8E2002) playground 这是swift 3.1 XCode版本8.3.2(8E2002)游乐场

protocol A {
    var items: [String] { get set }
    func doSomething()
}

struct Section : A {
    var items: [String] = []

    func doSomething() {
        print("doSomething")
        items.forEach { print($0) }
    }
}

class DataSource<T: A> {
    var sections: [T]

    init(sections: [T]) {
        self.sections = sections
    }

    func process() {
        sections.forEach { $0.doSomething() }
    }

    convenience init() {
        var section = Section()
        section.items.append("Goodbye")
        section.items.append("Swift")

        self.init(sections: [section])
    }
}

/*: Client */
var section = Section()
section.items.append("Hello")
section.items.append("Swift")

let ds = DataSource(sections: [section])
ds.process()

If no convenience init exists, then the code beneath the /*: Client */ section compiles and executes without issue. 如果不存在便利init,那么/ *:Client * /部分下面的代码将编译并执行而不会出现问题。 If I add in the convenience init I get the following compilation error: 如果我添加了方便init,我得到以下编译错误:

cannot convert value of type '[Section]' to expected argument type '[_]'
        self.init(sections: [section])

I wouldn't think that this would be an issue since in the convenience init I am creating a Section struct which implements the protocol A which satisfies the generic constraint on the DataSource class. 我不认为这会是一个问题,因为在方便初始化我创建了一个Section结构,它实现了满足DataSource类的通用约束的协议A. The convenience init is performing the same operations as the client code, yet it is unable to convert a [Section] into a [A]. 便利init正在执行与客户端代码相同的操作,但它无法将[部分]转换为[A]。 Is this an initialization sequencing issue? 这是初始化排序问题吗?

Generic placeholders are satisfied at the usage of the given generic type – therefore inside your convenience init , you cannot assume that T is a Section . 通用占位符对给定泛型类型的使用感到满意 - 因此在您的convenience init ,您不能假设 T是一个Section It's an arbitrary concrete type that conforms to A . 它是符合A的任意具体类型。

For example, it would be perfectly legal for the caller to define a 例如,调用者定义a是完全合法的

struct SomeOtherSection : A {...}

and then call your convenience initialiser with T being SomeOtherSection . 然后使用TSomeOtherSection调用您的便利初始化程序。

The solution in this case is simple, you can just add your convenience initialiser in an extension of DataSource , where T is constrained to being Section – therefore allowing you to 在这种情况下的解决方案很简单,您只需在DataSource的扩展中添加便利初始化器,其中T被约束为Section - 因此允许您
call init(sections:) with a [Section] : [Section]调用init(sections:)

extension DataSource where T == Section {

    convenience init() {
        var section = Section()
        section.items.append("Goodbye")
        section.items.append("Swift")

        self.init(sections: [section])
    }
}

// ...

// compiler will infer that T == Section here.
let ds = DataSource()

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