有人在TypeScript中做过构造函数重载吗?在语言规范(v 0.8)的第64页,有描述构造函数重载的语句,但没有给出任何示例代码。

我现在正在尝试一个非常基本的类声明;它是这样的,

interface IBox {    
    x : number;
    y : number;
    height : number;
    width : number;
}

class Box {
    public x: number;
    public y: number;
    public height: number;
    public width: number;

    constructor(obj: IBox) {    
        this.x = obj.x;
        this.y = obj.y;
        this.height = obj.height;
        this.width = obj.width;
    }   

    constructor() {
        this.x = 0;
        this.y = 0;
        this.width = 0;
        this.height = 0;
    }
}

当运行tsc BoxSample。Ts,它抛出一个重复的构造函数定义——这是显而易见的。任何帮助都是感激的。


当前回答

注意,你也可以通过TypeScript中的默认参数来解决在实现级别上缺乏重载的问题,例如:

interface IBox {    
    x : number;
    y : number;
    height : number;
    width : number;
}

class Box {
    public x: number;
    public y: number;
    public height: number;
    public width: number;

    constructor(obj : IBox = {x:0,y:0, height:0, width:0}) {    
        this.x = obj.x;
        this.y = obj.y;
        this.height = obj.height;
        this.width = obj.width;
    }   
}

编辑: 截至2016年12月5日,请参阅本森的回答,以获得更灵活的更详细的解决方案。

其他回答

TypeScript允许你声明重载,但是你只能有一个实现,而且这个实现必须有一个与所有重载兼容的签名。在你的例子中,这可以很容易地用一个可选参数来完成,

interface IBox {    
    x : number;
    y : number;
    height : number;
    width : number;
}
    
class Box {
    public x: number;
    public y: number;
    public height: number;
    public width: number;

    constructor(obj?: IBox) {    
        this.x = obj?.x ?? 0
        this.y = obj?.y ?? 0
        this.height = obj?.height ?? 0
        this.width = obj?.width ?? 0;
    }   
}

或者使用更通用的构造函数进行两次重载,

interface IBox {    
    x : number;
    y : number;
    height : number;
        width : number;
}
    
class Box {
    public x: number;
    public y: number;
    public height: number;
    public width: number;

    constructor();
    constructor(obj: IBox); 
    constructor(obj?: IBox) {    
        this.x = obj?.x ?? 0
        this.y = obj?.y ?? 0
        this.height = obj?.height ?? 0
        this.width = obj?.width ?? 0;
    }   
}

参见游乐场

As commented in @Benson answer, I used this example in my code and I found it very useful. However I found with the Object is possibly 'undefined'.ts(2532) error when I tried to make calculations with my class variable types, as the question mark leads them to be of type AssignedType | undefined. Even if undefined case is handled in later execution or with the compiler type enforce <AssignedType> I could not get rid of the error, so could not make the args optional.I solved creating a separated type for the arguments with the question mark params and the class variables without the question marks. Verbose, but worked.

下面是原始代码,给出了类方法()中的错误,如下所示:

/** @class */

class Box {
  public x?: number;
  public y?: number;
  public height?: number;
  public width?: number;

  // The Box class can work double-duty as the interface here since they are identical
  // If you choose to add methods or modify this class, you will need to
  // define and reference a new interface for the incoming parameters object 
  // e.g.:  `constructor(params: BoxObjI = {} as BoxObjI)` 
  constructor(params: Box = {} as Box) {
    // Define the properties of the incoming `params` object here. 
    // Setting a default value with the `= 0` syntax is optional for each parameter
    const {
      x = 0,
      y = 0,
      height = 1,
      width = 1,
    } = params;

    //  If needed, make the parameters publicly accessible
    //  on the class ex.: 'this.var = var'.
    /**  Use jsdoc comments here for inline ide auto-documentation */
    this.x = x;
    this.y = y;
    this.height = height;
    this.width = width;
  }

  method(): void {
    const total = this.x + 1; // ERROR. Object is possibly 'undefined'.ts(2532)
  }
}

const box1 = new Box();
const box2 = new Box({});
const box3 = new Box({ x: 0 });
const box4 = new Box({ x: 0, height: 10 });
const box5 = new Box({ x: 0, y: 87, width: 4, height: 0 });

所以变量不能在类方法中使用。 如果像这样纠正,例如:

method(): void {
    const total = <number> this.x + 1;
}

现在出现这个错误:

Argument of type '{ x: number; y: number; width: number; height: number; }' is not 
assignable to parameter of type 'Box'.
Property 'method' is missing in type '{ x: number; y: number; width: number; height: 
number; }' but required in type 'Box'.ts(2345)

好像整个arguments bundle不再是可选的了。

因此,如果创建了带有可选参数的类型,并且从可选参数中删除了类变量,我就实现了我想要的,参数是可选的,并且能够在类方法中使用它们。下面是解决方案代码:

type BoxParams = {
  x?: number;
  y?: number;
  height?: number;
  width?: number;
}

/** @class */
class Box {
  public x: number;
  public y: number;
  public height: number;
  public width: number;

  // The Box class can work double-duty as the interface here since they are identical
  // If you choose to add methods or modify this class, you will need to
  // define and reference a new interface for the incoming parameters object 
  // e.g.:  `constructor(params: BoxObjI = {} as BoxObjI)` 
  constructor(params: BoxParams = {} as BoxParams) {
    // Define the properties of the incoming `params` object here. 
    // Setting a default value with the `= 0` syntax is optional for each parameter
    const {
      x = 0,
      y = 0,
      height = 1,
      width = 1,
    } = params;

    //  If needed, make the parameters publicly accessible
    //  on the class ex.: 'this.var = var'.
    /**  Use jsdoc comments here for inline ide auto-documentation */
    this.x = x;
    this.y = y;
    this.height = height;
    this.width = width;
  }

  method(): void {
    const total = this.x + 1;
  }
}

const box1 = new Box();
const box2 = new Box({});
const box3 = new Box({ x: 0 });
const box4 = new Box({ x: 0, height: 10 });
const box5 = new Box({ x: 0, y: 87, width: 4, height: 0 });

感谢任何花时间阅读并试图理解我想要表达的观点的人的评论。

提前谢谢你。

一般来说,对于N个重载,最好使用:

constructor(obj?: {fromType1: IType1} | {fromType2: IType2}) {    
    if(obj){
      if(obj.fromType1){
        //must be of form IType1
      } else if(obj.fromType2){
        //must have used a IType2
      } else {
        throw "Invalid argument 1"
      }
    } else {
      //obj not given
    }
}   

至少现在我们可以确定走哪条路,并采取相应的行动

Box类试图定义多个构造函数实现。

只有最后一个构造函数重载签名被用作类构造函数实现。

在下面的示例中,请注意构造函数实现的定义与前面的任何重载签名都不冲突。

interface IBox = {
    x: number;
    y: number;
    width: number;
    height: number;
}

class Box {
    public x: number;
    public y: number;
    public width: number;
    public height: number;

    constructor() /* Overload Signature */
    constructor(obj: IBox) /* Overload Signature */
    constructor(obj?: IBox) /* Implementation Constructor */ {
        if (obj) {
            this.x = obj.x;
            this.y = obj.y;
            this.width = obj.width;
            this.height = obj.height;
        } else {
            this.x = 0;
            this.y = 0;
            this.width = 0;
            this.height = 0
        }
    }

    get frame(): string {
        console.log(this.x, this.y, this.width, this.height);
    }
}

new Box().frame; // 0 0 0 0
new Box({ x:10, y:10, width: 70, height: 120 }).frame; // 10 10 70 120



// You could also write the Box class like so;
class Box {
    public x: number = 0;
    public y: number = 0;
    public width: number = 0;
    public height: number = 0;

    constructor() /* Overload Signature */
    constructor(obj: IBox) /* Overload Signature */
    constructor(obj?: IBox) /* Implementation Constructor */ {
        if (obj) {
            this.x = obj.x;
            this.y = obj.y;
            this.width = obj.width;
            this.height = obj.height;
        }
    }

    get frame(): string { ... }
}

我使用以下替代方法来获得默认/可选参数和“kind-of-overloaded”的构造函数,参数数量可变:

private x?: number;
private y?: number;

constructor({x = 10, y}: {x?: number, y?: number}) {
 this.x = x;
 this.y = y;
}

我知道这不是最漂亮的代码,但人们会习惯它。不需要额外的接口,它允许私有成员,这在使用接口时是不可能的。