3 ms·
I suppose Haskell does, as `(+) <$> f1 <*> f2`.
by Twey 1y ago
I suppose Haskell does, as `(+) <$> f1 <*> f2`.
- raluk 1y agoIn 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.