TypeScript function overloading

overloadingtypescript

Section 6.3 of the TypeScript language spec talks about function overloading and gives concrete examples on how to implement this. However if I try something like this:

export class LayerFactory { 

    constructor (public styleFactory: Symbology.StyleFactory) { }

    createFeatureLayer (userContext : Model.UserContext, mapWrapperObj : MapWrapperBase) : any {           
         throw "not implemented";
    }                 

    createFeatureLayer(layerName : string, style : any) : any {
        throw "not implemented";
     }        

}

I get a compiler error indicating duplicate identifier even though function parameters are of different types. Even if I add an additional parameter to the second createFeatureLayer function, I still get a compiler error. Ideas, please.

Best Answer

When you overload in TypeScript, you only have one implementation with multiple signatures.

class Foo {
    myMethod(a: string);
    myMethod(a: number);
    myMethod(a: number, b: string);
    myMethod(a: any, b?: string) {
        alert(a.toString());
    }
}

Only the three overloads are recognized by TypeScript as possible signatures for a method call, not the actual implementation.

In your case, I would personally use two methods with different names as there isn't enough commonality in the parameters, which makes it likely the method body will need to have lots of "ifs" to decide what to do.

TypeScript 1.4

As of TypeScript 1.4, you can typically remove the need for an overload using a union type. The above example can be better expressed using:

myMethod(a: string | number, b?: string) {
    alert(a.toString());
}

The type of a is "either string or number".

Related Solutions

Javascript – Function overloading in Javascript – Best practices

The best way to do function overloading with parameters is not to check the argument length or the types; checking the types will just make your code slow and you have the fun of Arrays, nulls, Objects, etc.

What most developers do is tack on an object as the last argument to their methods. This object can hold anything.

function foo(a, b, opts) {
  // ...
  if (opts['test']) { } //if test param exists, do something.. 
}


foo(1, 2, {"method":"add"});
foo(3, 4, {"test":"equals", "bar":"tree"});

Then you can handle it anyway you want in your method. [Switch, if-else, etc.]

Typescript – Constructor overload in TypeScript

TypeScript allows you to declare overloads but you can only have one implementation and that implementation must have a signature that is compatible with all overloads. In your example, this can easily be done with an optional parameter as in,

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 && obj.x || 0
        this.y = obj && obj.y || 0
        this.height = obj && obj.height || 0
        this.width = obj && obj.width || 0;
    }   
}

or two overloads with a more general constructor as in,

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?: any) {    
        this.x = obj && obj.x || 0
        this.y = obj && obj.y || 0
        this.height = obj && obj.height || 0
        this.width = obj && obj.width || 0;
    }   
}

Best Answer

TypeScript 1.4

Related Solutions

Javascript – Function overloading in Javascript – Best practices

Typescript – Constructor overload in TypeScript

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