具有功能的Typescript接口。子类型不能作为参数用于实现接口 [英] Typescript Interface with function. Subtype as parameter not accepted for implementing the interface

问题描述

我有一个可扩展另一个类的类，如下所示

I have a class which extends another class as shown below

abstract class FooAbstract{
    constructor(someProp:any){
        this.someProp = someProp;
    }
    someProp:any;
}

class Foo extends FooAbstract{
    constructor(prop:any){
        super(prop);
    }
    someRandomFunction(){
        console.log("Something")
    }
}

我有一个界面，其功能如下所示

I have an interface which has a function as shown below

interface ExampleInterface{
    someFunction: (foo:FooAbstract)=>any;
}

现在，我想实现该接口，但希望将子类型作为界面实现中的函数 someFunction ，如下所示

Now I want to implement the interface but want to pass subtype as parameter of the function someFunction in the interface implementation as shown below

class Example implements ExampleInterface{
    someFunction = (foo:Foo)=>{
        console.log("Hello World");
    }
}

Typescript警告 someFunction的实现不正确，并且Foo和FooAbstract类型不兼容。我想了解为什么我不能通过要求 FooAbstract

Typescript is warning that the implementation of someFunction is incorrect and the type Foo and FooAbstract are incompatible. I want to understand why can't I implement the function someFunction by requiring as a parameter a subtype of FooAbstract

推荐答案

实际上这是有道理的，因为这样做并不安全。请考虑以下情形：

Actually this makes sense because it is not safe to do this. Consider the following scenario:

class Example implements ExampleInterface{
    someFunction = (foo:Foo)=>{
        console.log("Hello World");
        foo.someRandomFunction() // we can call this since foo is of type Foo
    }
}
class Boo extends FooAbstract{
    constructor(prop:any){
        super(prop);
    }
    // no someRandomFunction method
}
var ex: ExampleInterface = new Example();
ex.someFunction(new Boo({})) // ok, Boo is derived from FooAbstract

如果编译器允许您提出问题，则上面的代码将编译但在运行时失败，因为 Boo <中不存在 someRandomFunction / code>。

If the compiler would allow the scenario in your question, the above code would compile but fail at runtime because someRandomFunction does not exist on Boo.

您可以使接口通用，因此可以指定哪种类型的派生 FooAbsrtact 您将使用：

You can make the interface generic so you can specify what type of derived FooAbsrtact you will use:

interface ExampleInterface< T extends FooAbstract >{
    someFunction: (foo:T)=>any;
}
// now ok
class Example implements ExampleInterface<Foo>{
    someFunction = (foo:Foo)=>{
        console.log("Hello World");
        foo.someRandomFunction() 
    }
}
class Boo extends FooAbstract{
    constructor(prop:any){
        super(prop);
    }
    // no someRandomFunction method
}
var ex: ExampleInterface<Foo> = new Example();
ex.someFunction(new Boo({})) // compile error as it should be

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