Is there a simpler version of zip, that doesn’t bu...

# getting-started

[Mark]

Is there a simpler version of zip, that doesn’t build a sequence? Rather, we just do the work in the transform block?

[marstran]

What do you mean? Can you make an example?

[Mark]

seq1.zip(seq2) { (s1, s2) -> doWork(s1, s2) }

[Mark]

So I don’t need the returned sequence

[Mark]

Oh, I still need to iterate over it, of course

[Kroppeb]

I think there is a zipWith

[marstran]

You could do this:

seq1.zip(seq2).forEach { (s1, s2) ->
  doWork(s1, s2)
}

[Kroppeb]

Oh, I misread the question

[Mark]

@marstran yep, though it’s still creating a new sequence. No big deal though.

[marstran]

Yes, but it's only 1 object creation. It doesn't process anything until forEach, like it would if it was a list.

[Kroppeb]

It allocates a bunch of pairs though

[marstran]

@Kroppeb Yeah, true, it does it when consumed. I guess seq1.zip(seq2) { a, b -> doWork(s1, s2) }.forEach {} avoids it, but it looks weird 😛