为什么print()将我的String打印为可选? [英] Why print() is printing my String as an optional?

问题描述

我有一本字典，我想将其某些值用作另一本字典的键:

I have a dictionary and I want to use some of its values as a key for another dictionary:

let key: String = String(dictionary["anotherKey"])

在这里，dictionary["anotherKey"]是42，但是当我在调试器中打印key时，会看到以下内容:

here, dictionary["anotherKey"] is 42 but when I print key in the debugger I see the following:

(lldb) expression print(key)
Optional(42)

那怎么可能?据我了解，String()构造函数不会不返回一个可选内容(当我编写let key: String而不是let key: String?时，编译器不会抱怨).有人可以解释这是怎么回事吗?

How is that possible? To my understanding, the String() constructor does not return an optional (and the compiler does not complain when I write let key: String instead of let key: String?). Can someone explain what's going on here?

作为旁注，我目前正在使用description()方法解决此问题.

As a sidenote, I am currently solving this using the description() method.

推荐答案

这是设计使然-这是Swift的Dictionary实现方式:

This is by design - it is how Swift's Dictionary is implemented:

Swift的Dictionary类型将其键值下标实现为采用并返回可选类型的下标. [...] Dictionary类型使用可选的下标类型来对并非每个键都具有值的事实进行建模，并通过为该键分配nil值来提供一种删除键值的方法. (链接到文档)

Swift’s Dictionary type implements its key-value subscripting as a subscript that takes and returns an optional type. [...] The Dictionary type uses an optional subscript type to model the fact that not every key will have a value, and to give a way to delete a value for a key by assigning a nil value for that key. (link to documentation)

您可以将结果包装到if let构造中，以消除可选的内容，如下所示:

You can unwrap the result in an if let construct to get rid of optional, like this:

if let val = dictionary["anotherKey"] {
    ... // Here, val is not optional
}

例如，如果您确定该值在那里，因为您将其放在字典中的前几步，则也可以使用!运算符强制展开:

If you are certain that the value is there, for example, because you put it into the dictionary a few steps before, you could force unwrapping with the ! operator as well:

let key: String = String(dictionary["anotherKey"]!)

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