使用“如果让..."有很多表达 [英] Using "if let..." with many expressions

问题描述

Swift 的这个习语很有道理

This idiom of Swift makes good sense

if let x = someDict[someKey] { ... }

然而，我真正想要的是

if let x = someDict[someKey], y = someDict[someOtherKey] { ... }

正如所写的那样，这并没有错，但是这个想法可能吗?

As written this is not incorrect, but is this idea possible?

推荐答案

Swift 1.2 更新

从 Swift 1.2 开始，if let 允许解包多个选项，因此您现在可以编写此代码，如您的示例所示:

Since Swift 1.2, if let allows unwrapping multiple optionals, so you can now just write this, as in your example:

if let x = someDict[someKey], y = someDict[someOtherKey] { … }

您甚至可以交错条件，例如:

You can even interleave conditions such as:

if let x = someDict[someKey] where x == "value", y = someDict[someOtherKey] { … }

---

这在 Swift 1.2 之前是有效的

如果没有丑陋的强制包装，您将如何做到这一点:

Here's how you would do it without an ugly force-upwrapping:

switch (dict["a"], dict["b"]) {
case let (.Some(a), .Some(b)):
    println("match")
default:
    println("no match")
}

实际上还是很冗长.

之所以有效，是因为表单 Type? 的可选类型实际上是 Optional 的简写，它是一个大致如下所示的枚举:

This works because an optional type of the form Type? is actually shorthand for Optional<Type>, which is an enum that looks roughly like this:

enum Optional<T> {
    case None
    case Some(T)
}

然后您可以像任何其他枚举一样使用模式匹配.

You can then use pattern matching as for any other enum.

我见过有人写过这样的辅助函数(抱歉没有注明出处，我不记得在哪里看到的):

I've seen people write helper functions like this one (sorry for the lack of attribution, I don't remember where I saw it):

func unwrap<A, B>(a: A?, b: B?) -> (A, B)? {
    switch (a, b) {
    case let (.Some(a), .Some(b)):
        return (a, b)
    default:
        return nil
    }
}

然后你可以继续使用 if let 构造，即像这样:

Then you can keep using the if let construct, namely like this:

if let (a, b) = unwrap(dict["a"], dict["b"]) {
    println("match: \(a), \(b)")
} else {
    println("no match")
}

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