Amendment to mapTo operator type

[alaboudi]

Bug Report

This issue is not a "bug" per se but more of an amendment suggestion to the mapTo operators's type. Didn't know what best to label this as.

Current Behavior
mapTo's current type signature is mapTo<T, R>(value: R): OperatorFunction<T, R> but there is absolutely no need for the T generic for this operator. The whole point of the operator is to start a completely new stream of values with no dependency on the source stream value.

Possible Solution
I think mapTo<R>(value: R): OperatorFunction<any, R> may better represent the intention of using the mapTo operator.

[kwonoj]

Sounds ok, presume this won't be breaking?

[alaboudi]

@kwonoj If we want to not make this a breaking change, then I think we can overload the type definition of this function to the current two generic type along with the new suggested one generic type. Otherwise, this is a breaking change. Since I see @cartant referencing this suggestion in the v7 rollout, is there a point of making this a non breaking introduction when the current implementation doesn't capture the intent of the operator?

[cartant]

I think we can overload the type definition of this function to the current two generic type along with the new suggested one generic type...

Yeah. ATM, this is a breaking change: it will break anyone who explicitly specifies the T type argument. I agree with the change, as T serves no purpose. We could do the same thing as with of and provide the current signature - that includes the T - as a deprecated signature.

[alaboudi]

I like @cartant 's suggestion to add the new signature and mark the current one as a deprecated signature. It won't be a breaking change but will push people towards the correct direction till its phased. If there are no objections, I'll put up a PR soon to address this issue.

[alaboudi]

@cartant I have put up a PR 3 days ago. Do tell if any changes are required