Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

TypeScript function overloading

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.

like image 838
Klaus Nji Avatar asked Nov 03 '12 19:11

Klaus Nji


1 Answers

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".

like image 71
Fenton Avatar answered Oct 15 '22 01:10

Fenton