4 ms·
How many times will Monads be reinvented? name.trim.filter{ _.length != 0 }.toUpperCase becomes: name >>= (return . trim) >>= (guard . ((/= 0) . leng
by DanWaterworth 13y ago
How many times will Monads be reinvented?
name.trim.filter{ _.length != 0 }.toUpperCase
becomes:
name >>= (return . trim) >>= (guard . ((/= 0) . length)) >>= (return . toUpperCase)
I've purposefully used >>= and return instead of fmap to show that '.' above is almost >>=.
EDIT: I've just noticed a type error in the use of guard, but you get the idea.
- farginay 13y agoI think the big thing is to make the case for one and the case for many look exactly the same in the language. Is that nicer to look at with do-notation?
- DanWaterworth 13y agoWith do notation, it becomes: do n <- name let trimmed = trim n guard (length trimmed /= 0) return (toUpperCase trimmed)
- farginay 13y agoWouldn't it be nice if it was just: toUpperCase . guard (\x -> length x /= 0) . trim I think that is how it would be if it was part of the language.
- deleted 13y ago[deleted]
- DanWaterworth 13y agoI like my code to do what it says. You can pretty much do that in Haskell. If you allow me to add a definition first: mfilter' :: MonadPlus m => (a -> Bool) -> a -> m a mfilter' f = mfilter f . return Then it's just: toUpperCase <=< mfilter' (\x -> length x /= 0) . trim Which is pretty close. No need to change the language.