如何使用两个描述“get”函数的typescript 0.9.5接口？

Question

I made this example of what seems to be valid typescript (playground link and inline as well): http://www.typescriptlang.org/Playground/#src=interface%20I1%20%7B%0A%09%09get(name%3A%20%22templatePath%22)%3A%20string%3B%0A%09%09get(name%3A%20string)%3A%20any%3B%09%0A%7D%0A%0Ainterface%20I2%20%7B%0A%09%09get(name%3A%20%22baseClass%22)%3A%20string%3B%0A%09%09get(name%3A%20string)%3A%20any%3B%0A%7D%0A%0Ainterface%20I3%20extends%20I1%2C%20I2%20%7B%0A%09%0A%7D%0A%0Aclass%20C3%20implements%20I3%20%7B%0A%09%09get(name%3A%20string)%20%7B%0A%09%09%09return%20name%3B%0A%09%09%7D%09%0A%7D

interface I1 {
        get(name: "templatePath"): string;
        get(name: string): any; 
}

interface I2 {
        get(name: "baseClass"): string;
        get(name: string): any;
}

interface I3 extends I1, I2 {

}

class C3 implements I3 {
        get(name: string) {
            return name;
        }   
}

And I get an error indication types of property 'get' are not identical in these two interfaces. (1) I don't see how that matters and (2) I don't know how to get around this.

Specifically my interface looks like this:

interface IGotoPane extends dijit._WidgetBase, dijit._TemplatedMixin, dijit._WidgetsInTemplateMixin {
    inputCoordinates: dijit._Widget;
}

I've found that I can eliminate the compiler errors by re-declaring several methods:

interface I3 extends I1, I2 {
        get(name: string): any;     
}

Or in the case of my specific issue,

interface IGotoPane extends dijit._WidgetBase, dijit._TemplatedMixin, dijit._WidgetsInTemplateMixin {
    inputCoordinates: dijit._Widget;
    get(name: string): any;
    set(name: string, value: any, raiseChangeEvent?: boolean): void;
    set(values: Dojo.PropertiesMap): void;
    watch(prop: string, callback: Dojo.WatchCallback<any>): Dojo.WatchHandle;
}

Why is this necessary?

Answer 1

The error about the properties needing to be identical is sort of common-sense -- what is the type of I3.get if left unspecified as in your first example? There isn't any notion of "merging" function types in TypeScript, so it isn't clear what exactly this would mean.

You can reason about simple cases, but it quickly becomes very confusing for nontrivial cases, for example:

interface I1 {
    foo(a: string, b: "y"): number;
    foo(a: string, b: string): any;
}

interface I2 {
    foo(a: "x", b: string): boolean;
    foo(a: string, b: string): any;
}

interface I3 extends I1, I2 {
}

var x: I3;
var y = x.foo('x', 'y'); // y: number? boolean? any?

The fix of explicitly declaring get in I3 is the correct one. You can copy down the constant-specialized signatures if they apply in that case.