Question

给出

Class X{
}

我希望将类作为参数传递给：

someMethod(theType: ?X?){
   theType.staticMethod //or
   new theType();
}

构造签名（接口）是我唯一的选择吗？

我可能会使用构造签名，但它感觉不自然，构造签名似乎并没有真正传达出新的特定类型。我想我可能会遗漏一些细节。这是一个更详细的示例，但泛型不是运行时类型。

abstract class GreeterBase<T extends GreeterBase<T>>{
    message:string;

    constructor(message:string){
        this.message=message;
    }

    public static create<T>(message:string):T{
        //This is the only part that doesn't work.
        return new T(message);  //This doesn't work.
    }

    public abstract greet():string;

    public logMessage(){
        console.log(this.message);
    }

}

class EnglishGreeter extends GreeterBase<EnglishGreeter>{
    constructor(message: string) {
        super(message);
    }

    greet() {
        return "Hello, " + this.message;
    }
}

class FriendlyMessager<T extends GreeterBase<T>>{

    private _myGreeter:T;

    constructor(andSoForth:string){
        this._myGreeter = GreeterBase.create<T>(andSoForth);
        this._myGreeter.logMessage();
    }

    public sendGreeting(){
        this._myGreeter.greet();
    }
}

let messager = new FriendlyMessager<EnglishGreeter>(', how are you');
messager.sendGreeting();

TS Playground Link

Answer 1

您需要将构造函数作为参数传递，因为<T>中的public static create<T>(message:string)只是帮助打字稿来推断类型信息，这些信息将在js中丢失。

public static create<T>(message:string, cTor: {new(string): T;}):T{
    //This is the only part that doesn't work.
    return new cTor(message);  //This doesn't work.
}

我认为你应该重新设计课程。也许like this

Answer 2

使用typeof类型运算符获取特定值的类型（在您的情况下为X）：

someMethod(myCtor: typeof X) {
   let f: X = new myCtor(); // OK
}

如何在Typescipt中声明参数是一个类型而不是类型实例？

2 个答案: