如何让一个类在 Typescript 中实现调用签名? [英] How to make a class implement a call signature in Typescript?

问题描述

我在打字稿中定义了以下接口:

I have defined the following interface in typescript:

interface MyInterface {
    () : string;
}

这个接口简单地引入了一个不带参数并返回一个字符串的调用签名.如何在类中实现这种类型?我尝试了以下方法:

This interface simply introduces a call signature that takes no parameters and returns a string. How do I implement this type in a class? I have tried the following:

class MyType implements MyInterface {
    function () : string {
        return "Hello World.";
    }
}

编译器一直告诉我

类 'MyType' 声明了接口 'MyInterface' 但没有实现它:类型 'MyInterface' 需要调用签名，但类型 'MyType' 缺少一个

Class 'MyType' declares interface 'MyInterface' but does not implement it: Type 'MyInterface' requires a call signature, but Type 'MyType' lacks one

如何实现调用签名?

推荐答案

类无法匹配该接口.你能得到的最接近的是这个类，它将生成在功能上与接口匹配的代码(但不是根据编译器).

Classes can't match that interface. The closest you can get, I think is this class, which will generate code that functionally matches the interface (but not according to the compiler).

class MyType implements MyInterface {
  constructor {
    return "Hello";
  }
}
alert(MyType());

这将生成工作代码，但编译器会抱怨 MyType 不可调用，因为它具有签名 new() = 'string'(即使你用 new 调用它，它将返回一个对象).

This will generate working code, but the compiler will complain that MyType is not callable because it has the signature new() = 'string' (even though if you call it with new, it will return an object).

要创建与接口实际匹配的内容而编译器不会抱怨，您必须执行以下操作:

To create something that actally matches the interface without the compiler complaining, you'll have to do something like this:

var MyType = (() : MyInterface => {
  return function() { 
    return "Hello"; 
  }
})();
alert(MyType());

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