IF条件错误中的Swift3选项链接?

问题描述

这段代码在 Swift 2.3 中工作得很好，我不明白为什么我必须打开 TestClass 来检查数字是否大于 4.这是链接选项的重点，以节省额外的费用称呼.

This code worked just fine in Swift 2.3 and I don't understand why I have to unwrap TestClass to check if number is bigger than 4. This is whole point of chaining optionals, to save additional call.

现在要完成这项工作，我必须检查是否 testClass != nil(或使用带有 if let 语句的隐式解包)，然后检查计数.

Now to make this work, I have to check if testClass != nil (or use implicit unwrap with if let statement) and then check count.

这真的是唯一的方法吗?

Is this really the only way?

import UIKit

class testClass
{
    var optionalInt:Int?
}

var test:testClass?

if test?.optionalInt > 4
{

}

推荐答案

从 Swift 3 中删除了可选的比较运算符.SE-0121

Optional comparison operators are removed from Swift 3.SE-0121

你需要这样写:

if test?.optionalInt ?? 0 > 4
{

}

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