swift - 如何以参数作为参数传递闭包并执行它？

Question

A 类的初始化器接受一个可选的闭包作为参数:

class A {
   var closure: ()?

   init(closure: closure()?) {
      self.closure = closure
      self.closure()
   }
}

我想传递一个带有参数的函数作为闭包:

class B {
    let a = A(closure: action(1)) // This throws the error: Cannot convert value of type '()' to expected argument type '(() -> Void)?'

    func action(_ i: Int) {
       //...
    }
}

类 A 应该执行带有参数 i 的闭包 action。

我不确定如何正确编写此代码，请参阅上面代码注释中的错误。必须更改什么？

Answer 1

请确保您的“what-you-have-now”代码没有错误。

假设你的类 A 是这样的:

class A {
    typealias ClosureType = ()->Void

    var closure: ClosureType?

    init(closure: ClosureType?) {
        self.closure = closure
        //`closure` would be used later.
    }

    //To use the closure in class A
    func someMethod() {
        //call the closure
        self.closure?()
    }
}

使用上面给出的A，您需要将类B重写为:

class B {
    private(set) var a: A!
    init() {
        //initialize all instance properties till here
        a = A(closure: {[weak self] in self?.action(1)})
    }

    func action(i: Int) {
        //...
    }
}