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打字稿泛型返回类型

[英]Typescript generics return type

I am new to typescript and I am trying to wrap my head around how generics work.我是打字稿的新手,我正试图围绕泛型的工作原理进行思考。 I am wondering why the following code doesn't fly:我想知道为什么下面的代码不飞:

function identity<T>(arg: T): T {
    let num: number = 2;
    return num;
}

let output = identity<number>(1);

I get the error: Type 'number' is not assignable to type 'T'.我收到错误消息:“数字”类型不可分配给“T”类型。 If the input to the function is a number, does this not mean that the return type of number should work as well, since we are saying that T is of type number?如果函数的输入是数字,这是否意味着 number 的返回类型也应该起作用,因为我们说 T 是数字类型?

You have to treat T like any type in existence.你必须像对待现有的任何类型一样对待T Your function head says: I want an argument of type T and return something of type T.你的函数负责人说:我想要一个 T 类型的参数并返回 T 类型的东西。

Your code however always returns a number , which does not have to be T .但是,您的代码始终返回一个number ,该number不必是T In your example call it is, but you could call it with identity<string>("test") and expect a string to come back, but it would still be a number .在您的示例中调用它,但您可以使用identity<string>("test")调用它并期望返回一个字符串,但它仍然是一个number Thus the confict.冲突由此而来。

Let's have a look at the type signature of your identity function.让我们看看你的身份函数的类型签名。

function identity<T>(arg: T)

What we are saying here is that this function acceps an argument of type T (which we are not specifying in advance, it could be anything), and it return a value of type T .我们在这里说的是这个函数接受一个T类型的参数(我们没有提前指定,它可以是任何东西),并且它返回一个T类型的值。

Given that there is no information whatsoever about T , the only the function can return a value of such type is to reuse the provided argument.鉴于没有关于T任何信息,唯一可以返回这种类型值的函数是重用提供的参数。 In other words, there is exactly one and only one possible implementation of the identity function - the one that returns its argument.换句话说,恒等函数只有一种且只有一种可能的实现——返回其参数的实现。

function identity<T>(arg: T) { return arg }

The error message is telling you exactly this: Type 'number' is not assignable to type 'T' , which is true because T is, well, generic :)错误消息准确地告诉您: Type 'number' is not assignable to type 'T' ,这是正确的,因为T是通用的 :)

Short story , you should correct your function like this:短篇小说,你应该像这样更正你的功能:

function identity<T>(arg: T): T {
    let num: T = arg;
    return num;
}

let output = identity<number>(1)


console.log(output);

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