Qualified identifiers opinion

Christian Maeder Christian.Maeder at dfki.de
Fri Aug 17 06:53:11 EDT 2007

Hi Isaac,

just to give you a reply at all, see below. I reply
glasgow-haskell-users at haskell.org since I'm not subscribed to
haskell-prime. And I don't want to subscribe, because I'm more
interested that Haskell becomes more stable (and standard). So here is
my opinion:

1. The lexer should recognize keywords.

2. I would not mind if Haskel98 rejected all keywords that are also
rejected by extensions, so that the lexer is extension independent.
(Starting with Haskell98, removing conflicting identifiers as soon as I
switch on valuable extensions does not make sense.)

3. I'm against qualified identifiers, with the unqualified part being a
keyword like "Foo.where". (The choice of qualification should be left to
the user, usually one is not forced to used qualified names.)

4. However, "Foo.where" should always be rejected and not changed to
"Foo.wher e"! (Longest matching, aka "maximal munch", must not consider

(see end of: http://www.haskell.org/onlinelibrary/lexemes.html#sect2.4)

I would not mind if a name "F. " is plainly rejected. It only makes
sense, when a data constructor is the first argument of the composition
operator "(.)"

Maybe "." and "$" as operators should require white spaces on both
sides, since "$(" also indicates template haskell.

Cheers Christian

Isaac Dupree wrote:
> Especially after writing a partial lexer for Haskell, I opine that this
> should be all legal:
> module Foo where
> --in case you didn't know, this is legal syntax:
> Foo.f = undefined
> Foo.mdo = undefined
> Foo.where = undefined
> x Foo.! y = undefined
> x Foo... y = undefined --remember ".." is reserved id, e.g. [2..5]
> {-# LANGUAGE RecursiveDo, BangPatterns #-} module Bar where
> import Foo
> hello !x = mdo { y <- Foo.mdo Foo... ({-Foo.-}f x y); return y }
> {- Haskell 98 -} module Baz where
> import Foo
> goodbye x = x ! 12
> (Foo.where) lexing as (Foo.wher e) or (Foo . where) does not make me
> happy.  (being a lexer error is a little less bad...)  Especially not
> when the set of keywords is flexible.  I don't see any good reason to
> forbid declaring keywords as identifiers/operators, since it is
> completely unambiguous, removes an extension-dependence from the lexer
> and simplifies it (at least the mental lexer); Also I hear that the
> Haskell98 lexing is (Foo.wher e), which I'm sure no one relies on...
> Well, that's my humble opinion on what should go into Haskell' on this
> issue.
> Isaac

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