[Haskell-cafe] $ do?

[Christian Maeder]

Am 15.12.2010 08:36, schrieb Roman Cheplyaka:
> * Jonathan Geddes <geddes.jonathan at gmail.com> [2010-12-14 19:59:14-0700]
>> 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.

I made such a proposal 3 years ago, without much feedback.
http://www.haskell.org/pipermail/haskell-cafe/2007-July/028203.html

Of course, you can supply more arguments, because "do", "let" and "case"
will be terminated by layout. This is not true for "\" (lambda) and
if-then-else, though.

HTH Christian