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

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		Twey 4 months ago \| parent \| context \| favorite \| on: Asynchrony is not concurrency I suppose Haskell does, as `(+) <$> f1 <*> f2`.

raluk 4 months ago [–]

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.

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