如何在打字稿中使用通用返回值注释函数 [英] How to annotate a function with a generic return value in typescript

问题描述

给出一个返回工厂的函数，如何注释该函数/工厂，使其包含正确的类型定义?

Given a function that returns a factory, how can I annotate the function/factory so that it contains the correct type definitions?

这是我的例子:

class item<T> {
  constructor(a: T) {
    this.a = a;
  }
  a: T
}

function generate(c) {
  return function a(a) {
    return new c(a);
  }
}

const factory = generate(item); // I want this to always be annotated as <T> (a: T) => item<T>

const instance = factory('string'); // instance should now be of type item<string>

这可能是打字稿中的问题吗，还是应该将其作为一项新功能推荐?

Is this possible in typescript or should I suggest it as a new feature?

推荐答案

对于非通用类，我们可以在3.0中使用

For non generic classes we can in 3.0 use Tuples in rest parameters and spread expressions and InstanceType to map the constructor to a similar function.

不幸的是，对于泛型类，在映射时无法保留类型参数.唯一的解决方案是在类中添加一个字段，该字段将告诉generate结果类型应该是什么.可以使用接口类合并来完成此操作，因此原始类不了解generate.

For generic classes unfortunately there is no way to preserve the type argument when mapping. The only solution is to add a field to the class that will tell generate what the result type should be. This can be done using interface-class merging so the original class does not know about generate.

使用这两种方法的解决方案(在可能的情况下为自动，在必要的情况下为手动)可能看起来像这样:

A possible solution using both approaches (automatic where possible, manual where necessary) could look something like this:

class item<T> {
    constructor(a: T) {
        this.a = a;
    }
    a: T
}

const generateResult = Symbol.for("generate")
interface item<T> {
    [generateResult] : <T>(a: T) => item<T>    
}

type d = item<any>[typeof generateResult]

type SupportsGeneration<R> =  { [generateResult] : R }
type ConstructorArguments<T> = T extends new (...a:infer A ) => any ? A : [];


function generate<T extends { new (...a:any[]) : SupportsGeneration<any> }>(c:T) :  InstanceType<T> extends SupportsGeneration<infer R> ? R: never
function generate<T extends new (...a:any[]) => any>(c:T) :  (a: ConstructorArguments<T>) => InstanceType<T>
function generate(c: new (...a: any[]) => any) : any {
    return function a(...a: any[]) {
        return new c(...a);
    }
}

const factory = generate(item); 

const instance = factory('string');

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