无法在非接口值上键入开关 [英] Cannot type switch on non-interface value

问题描述

我正在使用以下虚拟代码进行类型断言，但出现错误:

I am playing with type assertion using the following dummy code, and I got the error:

不能在非接口值上键入开关

cannot type switch on non-interface value

有人知道这是什么意思吗?

Does anyone know what does that mean?

package main

import "fmt"
import "strconv"

type Stringer interface {
    String() string
}

type Number struct {
    v int
}

func (number *Number) String() string {
    return strconv.Itoa(number.v)
}

func main() {
    n := &Number{1}
    switch v := n.(type) {
    case Stringer:
        fmt.Println("Stringer:", v)
    default:
        fmt.Println("Unknown")
    }
}

http://play.golang.org/p/Ti4FG0m1mc

推荐答案

类型开关需要一个接口来进行自省.如果您将已知类型的值传递给它，它就会爆炸.如果您创建一个接受接口作为参数的函数，它将起作用:

Type switches require an interface to introspect. If you are passing a value of known type to it it bombs out. If you make a function that accepts an interface as a parameter, it will work:

func typeSwitch(tst interface{}) {
    switch v := tst.(type) {
        case Stringer:
           fmt.Println("Stringer:", v)
        default:
           fmt.Println("Unknown")
    }
}

在此处查看完整代码 http://play.golang.org/p/QNyf0eG71_ 和 golang 文档接口 http://golang.org/doc/effective_go.html#interfaces.

See the full code here http://play.golang.org/p/QNyf0eG71_ and the golang documentation on interfaces http://golang.org/doc/effective_go.html#interfaces.

这篇关于无法在非接口值上键入开关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！