通过Swift中的函数(或Init)传播可选内容 [英] Propagate an optional through a function (or Init) in Swift

问题描述

有人有(更好)的方法吗?

does anyone have a (better) way to do this?

假设我有一个可选的Float

Lets say I have a optional Float

let f: Float? = 2

现在我想将其转换为Double

Now I want to cast it to a Double

let d = Double(f) //fail

这显然会失败，但是有没有办法像通过计算变量那样通过函数链接可选项?我现在正在做的是这样:

This will obviously fail but is there a way to chain the optional through the function like you can with calculated variables? What I am doing now is this:

extension Float {
    var double: Double { return Double(self) }
}
let d: Double? = f?.double

但是我真的不喜欢将强制转换作为计算变量.

But I really do not like putting a cast as a calculated variable.

我考虑使用的另一个选项是:

Another option I have considered using is this:

public func optionalize<A,B>(_ λ : @escaping (A) -> B) -> (A?) -> B? {
    return { (a) in
        guard let a = a else { return nil }
        return λ(a)
    }
}
let d: Double? = optionalize(Double.init)(f)

我意识到我可以保护'f'的值来解开它.但是，在许多情况下，可选值将是返回可选值的函数的参数.这导致防护中的中间值.如本例所示:

I realize I can guard the value of 'f' to unwrap it. However in many cases the optional value will be the parameter for a function that returns an optional. This leads to intermediate values in the guard. As seen in this example:

func foo(_ a: String?) throws -> Float {
    guard 
        let a = a,
        let intermediate = Float(a)
    else { throw.something }
    return intermediate
}

在这里，从String到Float的转换也可能会失败. 至少使用计算出的变量，该foo函数会更干净

Here it is possible for the cast from String to Float to fail also. At least with a calculated variable this foo function is a bit cleaner

extension String {
    var float: Float? { return Float(self) }
}

func foo(_ a: String?) throws -> Float {
    guard 
        let a = a?.float
    else { throw.something }
    return a
}

我不想重写频繁初始化的可选版本.

I do not want to rewrite optional versions of frequent inits.

任何想法都将不胜感激.谢谢！

Any ideas will be much appreciated. Thanks!

推荐答案

您可以简单地使用Optional的

You can simply use Optional's map(_:) method, which will return the wrapped value with a given transform applied if it's non-nil, else it will return nil.

let f : Float? = 2

// If f is non-nil, return the result from the wrapped value passed to Double(_:),
// else return nil.
let d = f.map { Double($0) }

正如您在下面的评论中指出的那样，也可以这样表示:

Which, as you point out in the comments below, can also be said as:

let d = f.map(Double.init)

这是因为在这种情况下，map(_:)需要类型为(Float) -> Double的转换函数，并且

This is because map(_:) expects a transformation function of type (Float) -> Double in this case, and Double's float initialiser is such a function.

如果转换还返回了可选内容(例如，将String转换为Int时)，则可以使用

If the transform also returns an optional (such as when converting an String to a Int), you can use flatMap(_:), which simply propagates a nil transform result back to the caller:

let s : String? = "3"

// If s is non-nil, return the result from the wrapped value being passed to the Int(_:)
// initialiser. If s is nil, or Int($0) returns nil, return nil.
let i = s.flatMap { Int($0) }

这篇关于通过Swift中的函数(或Init)传播可选内容的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！