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

更新2(2020年9月28日):这种语言在不断发展,所以如果你可以使用Partial(在v2.1中引入),那么这是我现在最喜欢的实现这一目标的方式。

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

   public constructor(b: Partial<Box> = {}) {
      Object.assign(this, b);
   }
}

// Example use
const a = new Box();
const b = new Box({x: 10, height: 99});
const c = new Box({foo: 10});          // Will fail to compile

更新(2017年6月8日):guyarad和snolflake在他们的评论中对我的回答提出了有效的观点。我建议读者看看Benson, Joe和snolflake的答案,他们的答案比我的更好

原答案(2014年1月27日)

另一个如何实现构造函数重载的例子:

class DateHour {

  private date: Date;
  private relativeHour: number;

  constructor(year: number, month: number, day: number, relativeHour: number);
  constructor(date: Date, relativeHour: number);
  constructor(dateOrYear: any, monthOrRelativeHour: number, day?: number, relativeHour?: number) {
    if (typeof dateOrYear === "number") {
      this.date = new Date(dateOrYear, monthOrRelativeHour, day);
      this.relativeHour = relativeHour;
    } else {
      var date = <Date> dateOrYear;
      this.date = new Date(date.getFullYear(), date.getMonth(), date.getDate());
      this.relativeHour = monthOrRelativeHour;
    }
  }
}

来源:http://mimosite.com/blog/post/2013/04/08/Overloading-in-TypeScript

其他回答

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 { ... }
}
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) {
        const { x, y, height, width } = { x: 0, y: 0, height: 0, width: 0, ...obj }
        this.x = x;
        this.y = y;
        this.height = height;
        this.width = width;
    }
}

下面是一个工作示例,您必须考虑每个具有更多字段的构造函数都应该将额外的字段标记为可选。

class LocalError {
  message?: string;
  status?: string;
  details?: Map<string, string>;

  constructor(message: string);
  constructor(message?: string, status?: string);
  constructor(message?: string, status?: string, details?: Map<string, string>) {
    this.message = message;
    this.status = status;
    this.details = details;
  }
}

正如chuckj所说,简单的答案是一个可选形参,但是如果我们想用多个形参重载构造函数,或者我们想改变形参顺序,该怎么办呢?

事实证明,构造函数可以像函数一样重载:

class FooBar {
  public foo?: number;
  public bar?: string;

  // Constructor A
  constructor(foo: number, bar?: string);
  // Constructor B
  constructor(bar: string, foo?: number);
  // Constructor C
  constructor(bar: string);
  // Constructor D
  constructor(foo: number);
  // Constructor E
  constructor();

  constructor(...args: any[]) {
    switch (args.length) {
      case 2:
        if (typeof args[0] === "number") {
          this.foo = args[0];
          this.bar = args[1];
        } else {
          this.bar = args[0];
          this.foo = args[1];
        }
        break;
      case 1:
        if (typeof args[0] === "number") {
          this.foo = args[0];
        } else {
          this.bar = args[0];
        }
    }

    console.log(this.foo, this.bar);
  }
}

const fooBarConstructorA = new FooBar("150", 25);
const fooBarConstructorB = new FooBar(25, "150");
const fooBarConstructorC = new FooBar("150");
const fooBarConstructorD = new FooBar("150");
const fooBarConstructorE = new FooBar();