CODEWITHSUNDEEP
javarx-javarx-java2
CJR
CJR

Reputation: 3592

RxJava .zip() raw type?

I am trying to zip two observables together.First is of type Single<SomeClass> and second is Observable<SomeOtherClass>. However, in .zip() function the return types does not get cast to correct classes, I get raw types (T1,T2). Example:

Single<SomeClass> o1 = ....
Observable<SomeOtherClass> o2 = ....

Observable.zip(o1,o2, (u,u2) -> ...) // here I get 2 raw types

And if I try this way (since I am "zipping" 2 observables only):

 o1.zipWith(o2, (someClass, u) -> ...) //here only o1 is cast to class instance

If I try Observable.zip(Observable.range(...),Observable.interval(...), (integer,long) -> ...) I get the correct casts.

I can't figure out why it won't cast to my class objects in my example above, any suggestions?

Upvotes: 0

Views: 224

Answers (1)

Vadim
Vadim

Reputation: 161

Using zip() you get a few objects in input, and transform them to something different in OUT. So you need to specify some output type:

Observable.zip(o1, o2, (u1,u2) -> new NewOtherClass(u1, u2))

In the example above the compiler know that return type for your lambda is NewOtherClass.

Note, that Single and Observable can't be used in the same chain with no additional transformation (note .toObservable() below):

 Observable.zip(o1.toObservable(), o2, (u1,u2) -> /*return do something*/)

Upvotes: 1

Related Questions