Re: [Haskell-cafe] $ do?

16 Dec 2010

      Am 15.12.2010 08:36, schrieb Roman Cheplyaka:
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* Jonathan Geddes  [2010-12-14 19:59:14-0700]
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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

Re: [Haskell-cafe] $ do?

Christian Maeder