如何快速进行可选的关闭?

本文介绍了如何快速进行可选的关闭?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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! } }