类型应为通用函数采用什么协议，以在Swift中将任何数字类型作为参数？ [英] What protocol should be adopted by a Type for a generic function to take any number type as an argument in Swift?

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：以下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，因此添加自己的东西通常是不必要的。在构建自己的系统之前，请先阅读标准库的数字协议。

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.

在Swift中，这实际上是不可能的。为此，您需要创建一个新协议，并使用通用函数中要使用的任何方法和运算符进行声明。该过程将为您工作，但是确切的细节将在一定程度上取决于您的通用功能。这是对要获得数字 n 并返回（n-1）^ 2 的函数的处理方式

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，包括运算符和带有 Int 的初始化程序（这样我们就可以

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，但是此时编译器不知道它们是否符合新的 NumericType 协议。您必须明确指出这一点-Apple称其为宣布扩展的协议采用。我们将对 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 { }

现在我们可以使用 NumericType 协议作为通用约束来编写实际函数。

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

这篇关于类型应为通用函数采用什么协议，以在Swift中将任何数字类型作为参数？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IT屋！