Type tree traversals [Re: Modeling multiple inheritance]

2003-10-04 Thread oleg
This message illustrates how to get the typechecker to traverse non-flat, non-linear trees of types in search of a specific type. We have thus implemented a depth-first tree lookup at the typechecking time, in the language of classes and instances. The following test is the best illustration:

Re: Haskellsupport in KDevelop

2003-10-04 Thread Wolfgang Jeltsch
Am Samstag, 4. Oktober 2003, 19:15 schrieb Peter Robinson: Hello, I've begun to write a plugin that provides basic support for Haskell in KDevelop 3.0 alpha. (http://www.kdevelop.org). Great! I will probably use it since I like Haskell and KDE very much. By the way, wasn't KDevelop only for

Re: Haskellsupport in KDevelop

2003-10-04 Thread Peter Robinson
On Saturday 04 October 2003 20:20, Wolfgang Jeltsch wrote: Great! I will probably use it since I like Haskell and KDE very much. By the way, wasn't KDevelop only for developing in C and C++? The current stable Release 2.1.* is a C/C++ only IDE but the upcoming 3.0 will probably support: Ada,

Re: Why does this work - haskell mysteries?

2003-10-04 Thread Petter Egesund
Thanks, the Scheme-version made it even clearer! Cheers, Petter On Saturday 04 October 2003 13:53, you wrote: On Sun, 5 Oct 2003 11:02:37 + Petter Egesund [EMAIL PROTECTED] wrote: Hi thanks for answering. I think I got it - the chaning of the functions lies in the last part of

Re: Haskellsupport in KDevelop

2003-10-04 Thread Ross Paterson
On Sat, Oct 04, 2003 at 07:15:32PM +0200, Peter Robinson wrote: What's really missing is a (primitive) background parser written that reports syntax errors. It can be written in yacc, antlr, etc., anything that produces C/C++ code. The only parsers for Haskell I could find are written

Re: Newbie qustion about monads

2003-10-04 Thread Juanma Barranquero
On Thu, 02 Oct 2003 14:57:22 +0200, Juanma Barranquero [EMAIL PROTECTED] wrote: data Accum s a = Ac [s] a instance Monad (Accum s) where return x = Ac [] x Ac s1 x = f = let Ac s2 y = f x in Ac (s1++s2) y output :: a - Accum a () output x = Ac [x] () After trying this one,