I'd like to give it a try. I guess the repo is
http://darcs.haskell.org/SoC/ghc.haddock/. Glancing around, see a version
of ghc, and I don't know how to build haddock. Is ghc.haddock a new flag to
a forked version of ghc?
Other questions: is ghc.haddock known to build & run on Windows? Is the
Hi Jeremy,
Thanks for this very informative answer! This has certainly helped to
clear up a number of points.
Thanks,
Chris.
> Many arguments have been had about what it means for a language to be
> "functional", so that's probably not a productive line of discussion.
> (ICFP carefully doesn't
Many arguments have been had about what it means for a language to be
"functional", so that's probably not a productive line of discussion.
(ICFP carefully doesn't stipulate language choice for the programming
contest, for example.)
Both eager and lazy evaluation can be "pure", providing re
Chris,
I'm not sure what exactly your question is as you are mixing up several
concepts. However, you start with:
In haskell, we can transform:
g x + f x
into:
f x + g x
Here you assume that the function + is commutative. Although this is
probably true for all numeric types in all Hask
Hi,
Is lazy evaluation necessary for a functional language to remain
functional?
The reason I ask is that because it violates beta-reduction, and also
referential transparency (I think). In haskell, we can transform:
g x + f x
into:
f x + g x
as both f and g do not change the parameter x.
If
Conal Elliott wrote:
I've heard of a version Haddock that understands all of GHC's syntax
extensions. Does anyone the current status?- Conal
That would be the next version of Haddock, courtesy of David Waern's Summer
of Code project last year. It is dependent on an unreleased version of
