如何快速进行可选的关闭? [英] How does one make an optional closure in swift?

问题描述

我正在尝试在Swift中声明一个参数，该参数需要一个可选的闭包.我声明的函数看起来像这样:

I'm trying to declare an argument in Swift that takes an optional closure. The function I have declared looks like this:

class Promise {

 func then(onFulfilled: ()->(), onReject: ()->()?){       
    if let callableRjector = onReject {
      // do stuff! 
    }
 }

}

但是Swift抱怨说，"if let"声明时，条件中的绑定值必须是Optional类型".

But Swift complains that "Bound value in a conditional must be an Optional type" where the "if let" is declared.

推荐答案

您应该将可选的闭包括在括号中.这样可以正确确定?运算符的作用域.

You should enclose the optional closure in parentheses. This will properly scope the ? operator.

func then(onFulfilled: ()->(), onReject: (()->())?){       
    if let callableRjector = onReject {
      // do stuff! 
    }
 }

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