Add flap operator

[chshersh]

This one:

(??) :: Functor f => f (a -> b) - a -> f b

[vrom911]

I've never seen this function before and I can't find this function anywhere, can you tell why it's needed? And why you need to make an operator? Should I implement the function first?

[chshersh]

@vrom911 These are good points. We can have both flap and ?? operator available. I learned about this operator from this document:

• https://github.com/zudov/24-days-of-syntactic-witchery#day-5--ff--a-

For me it's mostly useful when I have a list of function and want to apply them to single element, like this:

ghci> [(+2), (*3)] ?? 5
[7, 15]

Or apply to function inside Maybe:

ghci> Just (+3) ?? 5
Just 8

I can't say that this operator is as useful as other common operators like <$>, <*>, <&>, etc. But some people like it, and it's still useful sometimes.

In the document I've send in this comment, this operator is called <@> but in Haskell world it's more popular under ?? name from lens library:

• https://hackage.haskell.org/package/lens-4.17/docs/Control-Lens-Lens.html#v:-63--63-