问题描述

我想让一个函数在 Swift 中接受任何数字(Int、Float、Double、...)

I want to make a function accept any number (Int, Float, Double, ...) in Swift

func myFunction <T : "What to put here"> (number : T) ->  {
    //...
}

不使用 NSNumber

without using NSNumber

推荐答案

更新: 下面的答案原则上仍然适用，但是 Swift 4 完成了对数字协议的重新设计，以便添加您自己的通常是不必要的.在构建自己的系统之前，请查看标准库的数字协议.

Update: The answer below still applies in principle, but Swift 4 completed a redesign of the numeric protocols, such that adding your own is often unnecessary. Take a look at the standard library's numeric protocols before you build your own system.

这实际上在 Swift 中是不可能开箱即用的.为此，您需要创建一个新协议，使用您将在泛型函数中使用的任何方法和运算符进行声明.此过程对您有用，但具体细节将在一定程度上取决于您的通用函数的作用.以下是获取数字 n 并返回 (n - 1)^2 的函数的方法.

This actually isn't possible out of the box in Swift. To do this you'll need to create a new protocol, declared with whatever methods and operators you're going to use inside your generic function. This process will work for you, but the exact details will depend a little on what your generic function does. Here's how you'd do it for a function that gets a number n and returns (n - 1)^2.

首先，定义您的协议，使用运算符和一个采用 Int 的初始化程序(这样我们就可以减去一个).

First, define your protocol, with the operators and an initializer that takes an Int (that's so we can subtract one).

protocol NumericType {
    func +(lhs: Self, rhs: Self) -> Self
    func -(lhs: Self, rhs: Self) -> Self
    func *(lhs: Self, rhs: Self) -> Self
    func /(lhs: Self, rhs: Self) -> Self
    func %(lhs: Self, rhs: Self) -> Self
    init(_ v: Int)
}

所有数字类型已经实现了这些，但此时编译器并不知道它们是否符合新的NumericType 协议.你必须明确这一点——Apple 称之为通过扩展声明采用协议".我们将为 Double、Float 和所有整数类型执行此操作:

All of the numeric types already implement these, but at this point the compiler doesn't know that they conform to the new NumericType protocol. You have to make this explicit -- Apple calls this "declaring protocol adoption with an extension." We'll do this for Double, Float, and all the integer types:

extension Double : NumericType { }
extension Float  : NumericType { }
extension Int    : NumericType { }
extension Int8   : NumericType { }
extension Int16  : NumericType { }
extension Int32  : NumericType { }
extension Int64  : NumericType { }
extension UInt   : NumericType { }
extension UInt8  : NumericType { }
extension UInt16 : NumericType { }
extension UInt32 : NumericType { }
extension UInt64 : NumericType { }

现在我们可以编写我们的实际函数，使用 NumericType 协议作为通用约束.

Now we can write our actual function, using the NumericType protocol as a generic constraint.

func minusOneSquared<T : NumericType> (number : T) -> T {
    let minusOne = number - T(1)
    return minusOne * minusOne
}

minusOneSquared(5)              // 16
minusOneSquared(2.3)            // 1.69
minusOneSquared(2 as UInt64)    // 1

这篇关于Type 应该采用什么协议来让泛型函数将任何数字类型作为 Swift 中的参数?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跟版网！