打字稿：如何設置方法的返回值=>子類

Question

我有兩個子類的父類：

abstract class Point { 
    public readonly x: number; 
    public readonly y: number; 

    constructor(x: number, y: number) { 
     this.x = x; 
     this.y = y; 
    } 

    diff(point: Point): Point { 
     return this.update(this.x - point.x, this.y - point.y); 
    } 
    // many methods like diff(); and then... 

    protected abstract update(x: number, y: number): Point; 
} 



class ImmutablePoint extends Point { 
    protected update(x: number, y: number): Point { 
     return new ImmutablePoint(x, y); 
    } 
} 



class MutablePoint extends Point { 
    public x: number; 
    public y: number; 

    protected update(x: number, y: number): Point { 
     this.x = x; 
     this.y = y; 
     return this; 
    } 
} 


const pointA: ImmutablePoint = new ImmutablePoint(10, 10) 
const pointB: ImmutablePoint = new ImmutablePoint(6, 2); 

但diff()方法返回一個點，而不是一個ImmutablePoint

// Error: type 'Point' is not assignable to parameter of type 'ImmutablePoint'. 
const result: ImmutablePoint = pointA.diff(pointB); 

我尋找一種方法來重新定義子類上的方法簽名而不寫一個新的實現，這有可能嗎？

我也試圖讓diff()返回值this但它不工作，因爲ImmutablePoint不返回this而是一個新的ImmutablePoint

Playground Link

Answer 1

您可以Point通用：

abstract class Point<T extends Point<any>> { 
    public readonly x: number; 
    public readonly y: number; 

    constructor(x: number, y: number) { 
     ... 
    } 

    diff(point: Point<any>): T { 
     return this.update(this.x - point.x, this.y - point.y); 
    } 

    protected abstract update(x: number, y: number): T; 
} 

class ImmutablePoint extends Point<ImmutablePoint> { 
    protected update(x: number, y: number): ImmutablePoint { 
     return new ImmutablePoint(x, y); 
    } 
} 

class MutablePoint extends Point<MutablePoint> { 
    public x: number; 
    public y: number; 

    protected update(x: number, y: number): MutablePoint { 
     ... 
     return this; 
    } 
} 


const pointA: ImmutablePoint = new ImmutablePoint(10, 10) 
const pointB: ImmutablePoint = new ImmutablePoint(6, 2); 
const result: ImmutablePoint = pointA.diff(pointB); // fine 

（code in playground）

隨着所提供的新功能Default generic type variables你應該能夠做這樣的事情：

abstract class Point<T extends Point = Point> { 
    ... 
} 

（沒有測試它尚未）