[Haskell-cafe] $ do?
Christian.Maeder at dfki.de
Thu Dec 16 13:45:48 CET 2010
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.
Of course, you can supply more arguments, because "do", "let" and "case"
will be terminated by layout. This is not true for "\" (lambda) and
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