* Jonathan Geddes
Quick question:
Why do I need the $ in the following bits of code?
main = withSocketsDo $ do --do something with sockets
foo = fromMaybe 0 $ do --do something in the maybe monad
I don't see (after admittedly only a minute or so thinking about it) where any grammar ambiguities would be if 'do' had an implicit $ in front of it:
foo = fromMaybe 0 do --do something in maybe
Though now that I've written it down, that is hard for me to visually parse at a glance. I'm still curious though.
Hi Jonathan, it's not clear whether you ask how this implies from the standard or what the rationale of such syntax is. Regarding the former, there are two distinct syntactic categories in Haskell: lexp and aexp. aexp produces all the constructs that can be used as function arguments and does not include do expressions. lexp is more broad and produces do expressions as well. Next look at these productions to see the difference between function application and operator application (sections 3.3 and 3.4 of 2010 report): fexp → [fexp] aexp infixexp → lexp qop infixexp Regarding the rationale, I'm not so sure and I'd like to hear an explanation from someone competent. But I assume it has something to do with the fact that if you supply a 'do' argument, you cannot supply any more arguments (because 'do' extends to the right as far as possible). Not that I'm convinced that it is a valid reason to prohibit such construct. -- Roman I. Cheplyaka :: http://ro-che.info/ Don't worry what people think, they don't do it very often.