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fix(typings): set map function parameter for Observable.from as optional #1334

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fix(typings): set map function parameter for Observable.from as optional #1334

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2 changes: 1 addition & 1 deletion src/observable/FromObservable.ts
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Original file line number Diff line number Diff line change
Expand Up @@ -20,7 +20,7 @@ export class FromObservable<T> extends Observable<T> {
super(null);
}

static create<T>(ish: any, mapFnOrScheduler: Scheduler | ((x: any, y: number) => T), thisArg?: any, lastScheduler?: Scheduler): Observable<T> {
static create<T>(ish: any, mapFnOrScheduler?: Scheduler | ((x: any, y: number) => T), thisArg?: any, lastScheduler?: Scheduler): Observable<T> {
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this makes sense for signature of generic from behavior. however in this case, does array-like observable need to check existence of mapFn in line 45? so far I remember arraylikeobservable's mapfn is required parameter. @justinwoo , is my understanding correct? or can we skip mapFn?

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oh, oops, mapFn should be optional. it's supposed to just use identity when not specified.

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Cool, thanks for clarification. So it seems PR's good :)

let scheduler: Scheduler = null;
let mapFn: (x: number, y: any) => T = null;
if (isFunction(mapFnOrScheduler)) {
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