为什么我可以输入别名函数并使用它们而不用转换？ [英] Why can I type alias functions and use them without casting?

问题描述

在Go中，如果您定义一个新类型，例如：

In Go, if you define a new type e.g.:

type MyInt int

然后，您不能将 MyInt 传递给需要int的函数反之：

You can't then pass a MyInt to a function expecting an int, or vice versa:

func test(i MyInt) {
    //do something with i
}

func main() {
    anInt := 0
    test(anInt) //doesn't work, int is not of type MyInt
}

很好。但为什么它不是同样不适用于函数？例如：

Fine. But why is it then that the same does not apply to functions? e.g.:

type MyFunc func(i int)
func (m MyFunc) Run(i int) {
    m(i)
}

func run(f MyFunc, i int) {
    f.Run(i)
}

func main() {
    var newfunc func(int) //explicit declaration
    newfunc = func(i int) {
        fmt.Println(i)
    }
    run(newfunc, 10) //works just fine, even though types seem to differ
}

现在，我不抱怨，因为它节省了我不得不显式转换 newfunc 键入 MyFunc 我将在第一个例子中做;它只是似乎不一致。我相信有一个很好的理由;任何人都可以启发我？

Now, I'm not complaining because it saves me having to explicitly cast newfunc to type MyFunc, as I would have to do in the first example; it just seems inconsistent. I'm sure there is a good reason for it; can anyone enlighten me?

我问的原因主要是因为我想以这种方式缩短一些相当长的函数类型，但我想确保它预期和可接受的做到这一点：）

The reason I ask is mainly because I would like to shorten some of my rather long function types in this way, but I want to make sure it's expected and acceptable to do this :)

推荐答案

结果，这是一个误会，这可以通过阅读规范的相关部分来解决：

Turns out, this is a misunderstanding that I had about how Go dealt with types, which can be resolved by reading the relevant part of the spec:

http： //golang.org/ref/spec#Type_identity

我不知道的相关区别是命名和未命名类型。

The relevant distinction that I was unaware of was that of named and unnamed types.

命名类型是具有名称的类型，例如int，int64，float， bool。此外，使用'type'创建的任何类型都是命名类型。

Named types are types with a name, such as int, int64, float, string, bool. In addition, any type you create using 'type' is a named type.

未命名类型是[] string，map [ string] string，[4] int。它们没有名称，只是对应于它们的结构的描述。

Unnamed types are those such as []string, map[string]string, [4]int. They have no name, simply a description corresponding to how they are to be structured.

如果比较两个命名类型，则名称必须匹配才能互换。如果您比较已命名和未命名的类型，则只要基本表示符合，就可以了！

If you compare two named types, the names must match in order for them to be interchangeable. If you compare a named and an unnamed type, then as long as the underlying representation matches, you're good to go!

给定以下类型：

type MyInt int
type MyMap map[int]int
type MySlice []int
type MyFunc func(int)

以下内容无效：

var i int = 2
var i2 MyInt = 4
i = i2 //both named (int and MyInt) and names don't match, so invalid

以下是正确的：

is := make([]int)
m := make(map[int]int)
f := func(i int){}

//OK: comparing named and unnamed type, and underlying representation
//is the same:
func doSlice(input MySlice){...}
doSlice(is)

func doMap(input MyMap){...}
doMap(m)

func doFunc(input MyFunc){...}
doFunc(f)

我有点不懂我不知道更快，所以我希望澄清类型lark一点别人！并且意味着比我第一次想到的更少的投射：）

I'm a bit gutted I didn't know that sooner, so I hope that clarifies the type lark a little for someone else! And means much less casting than I at first thought :)

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