Swift函数返回两种不同的类型 [英] Swift function returning two different types

问题描述

我需要一个可以根据输入的参数返回String或Int的函数，例如:

I need a function that can return either a String or an Int depending on the parameters entered eg:

func getValue (type: String) -> (String || Int) {  //this line is obviously wrong
    if type == "type1" {
        return "exampleString"
    }
    else if type == "type2"
        return 56
    }
}

推荐答案

使用枚举

您可以使用带有关联值的枚举来实现所需的行为.它们很像是C联合的更好版本.

Use an Enumeration

You can use an enumeration with associated values to achieve the behaviour you're looking for. They're much like a nicer version of C's unions.

enum Foo { //TODO: Give me an appropriate name.
    case type1(String)
    case type2(Int)
    
    static func getValue(type: String) -> Foo {
        switch (type) {
            case "type1": return type1("exampleString")
            case "type2": return type2(56)
            default: fatalError("Invalid \"type\"");
        }
    }
}

let x = Foo.getValue(type: "type1")

x必须通过打开其类型并做出相应的响应来有条件地使用:

x must be consumed conditionally, by switching on its type and responding accordingly:

switch x {
    case .type1(let string): funcThatExpectsString(string)
    case .type2(let int): funcThatExpectsInt(int)
}

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