在Go中创建具有约束的自定义类型 [英] Create a custom type with constraint in Go

问题描述

如何在Go中创建只能接受有效值的自定义类型?例如，我要创建一个名为名称"的类型，其基础类型为字符串.但是，它只能接受值"John"，"Rob"或"Paul".其他任何值都将返回错误.我已经以一种非常简单的方式创建了以下程序，以表示我想要实现的目标.

How do I create a custom type in Go where only valid values can be accepted? For example, I want to create a type called "Names" where its underlying type is a string. However it can only accept the values "John", "Rob" or "Paul". Any other value will return an error. I've create the following program in a very simplistic way just to represent what I would like to achieve.

http://play.golang.org/p/jzZwALsiXz

编写此代码的最佳方法是什么?

What would be the best way to write this code?

推荐答案

您可以执行以下操作(http://play.golang.org/p/JaIr_0a5_- ):

You could do something like this (http://play.golang.org/p/JaIr_0a5_-):

type Name struct {
    string
}

func (n *Name) String() string {
    return n.string
}

func NewName(name string) (*Name, error) {
    switch name {
    case "John":
    case "Paul":
    case "Rob":
    default:
        return nil, fmt.Errorf("Wrong value")
    }
    return &Name{string: name}, nil
}

Golang不提供运算符重载，因此您无法在强制转换或影响值时进行检查.

Golang does not provide operator overload so you can't make the check while casting or affecting value.

根据您要尝试执行的操作，您可能需要执行以下操作( http://play.golang.org/p/uXtnHKNRxk ):

Depending on what you are trying to do, you might want to do something like this (http://play.golang.org/p/uXtnHKNRxk):

type Name string

func (n Name) String() string {
    switch n {
    case "John":
    case "Paul":
    case "Rob":
    default:
        return "Error: Wrong value"
    }
    return string(n)
}

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