通过打字稿中的此类型从派生类型调用构造函数 [英] Call constructor from derived type via this in typescript

问题描述

在我的打字稿中，我试图通过基类中的方法创建/克隆子对象.这是我的(简化)设置.

In my typescript I'm trying to create/clone an child-object via a method in the base-class. This is my (simplified) setup.

abstract class BaseClass<TCompositionProps> {
    protected props: TCompositionProps;

    protected cloneProps(): TCompositionProps { return $.extend(true, {}, this.props); } // can be overwriten by childs

    constructor(props: TCompositionProps){
        this.props = props;
    }

    clone(){
        const props = this.cloneProps();
        return this.constructor(props);
    }   
}

interface IProps {
    someValues: string[];
}

class Child extends BaseClass<IProps>{
    constructor(props: IProps){
        super(props);
    }
}

现在，我要创建一个新对象

Now, I'm going to create a new object

const o1 = new Child({someValues: ["This","is","a","test"]};

// get the clone
const clone = o1.clone();

构造函数被命中(但这只是对函数的调用)，这意味着没有创建新对象.当使用返回Child.prototype.constructor(props)时，我得到了新对象.

The constructor is hit (but it's just the call to the function), meaning there is no new object created. When using return Child.prototype.constructor(props) instead I get my new object.

那么如何在其基类中调用 Child 的构造函数?

So how can I call the constructor of Child in it's base-class?

还尝试了此

推荐答案

您可以使用new运算符调用构造函数，这似乎可行.另外，我将使用 this 作为返回类型，以便clone方法将返回派生类型而不是基本类型

You can invoke the constructor with the new operator, that seems to work. Also I would use this for the return type so that the clone method will return the derived type not the base type

abstract class BaseClass<TCompositionProps> {
    protected props: TCompositionProps;

    protected cloneProps(): TCompositionProps { return $.extend(true, {}, this.props); } 

    constructor(props: TCompositionProps){
        this.props = props;
    }

    clone() : this{
        const props = this.cloneProps();
        return new (<any>this.constructor)(props);
    }   
}

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