Question

我想创建一个接受结构类型的函数，并且在该函数内部，我想返回创建的Struct的实例。

例如：

struct Person {
  var name: String
  func greet() {
    print("Hi person \(self.name)")
  }
}

struct Animal {
  var name: String
  func greet() {
    print("Hi animal \(self.name)")
  }
}

// T is the stuct type, so I can pass either Person or Animal.
// name is the name string.
func greet<T>(_ a: T, name: String) {
  let thingToGreet: a = a(name: name)
  thingToGreet.greet()
}

// Pass the struct type and a string.
greet(Person, name: "Johny")

这甚至可能吗？在应用程序中，我想创建一个接受URL的函数，结构类型然后在完成时我想返回基于数据任务请求创建的结构。

Answer 1

您需要使用协议向编译器解释A）这些类型具有.name，B）它们具有.greet()函数，并且最后，C）它们可以被初始化只有name。在greet()全局函数中，您可以参考协议。最后的皱纹是您传入的类型，并且您明确地调用init(name:) ...

protocol HasName {
    var name: String { get set }
    func greet()
    init(name: String)
}

struct Person: HasName {
    var name: String
    func greet() {
        print("Hi person \(self.name)")
    }
}

struct Animal: HasName {
    var name: String
    func greet() {
        print("Hi animal \(self.name)")
    }
}

// ** We demand T follows the protocol, 
// ** & declare A is a type that follows the protocol, not an instance
func greet<T: HasName>(_ A: T.Type, name: String) { 
    let thingToGreet = A.init(name: name) // ** A(name: ) doesn't work
    thingToGreet.greet()
}

// Pass the struct type and a string.
greet(Person.self, name: "Johny") // ** .self returns the type

将Struct传递给函数并在函数内创建该struct的实例

1 个答案: