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Twey • last Friday at 9:47 PM • 1 reply • view on HN

I suppose Haskell does, as `(+) <$> f1 <*> f2`.

Replies

raluk • last Saturday at 5:12 AM

In there is also ApplicativeDo that works nicely with this.

    do 
      x <- f1
      y <- f2
      return $ x + y
this is evaluated as applicative in same way.