Re: [Haskell-beginners] small expression evaluator

22 Mar 2011

      On Tue, 22 Mar 2011 10:14:57 +0100, Henk-Jan van Tuyl   
wrote:
...
On Tue, 22 Mar 2011 08:56:45 +0100, Petr Novotnik  
 wrote:
...
data Person = Person {
       personName :: String
     , personAge :: Int
     }
     deriving (Show)
exampleExpr :: Bool
exampleExpr = (VConst 99) .==. (VFunc personAge) $ Person "pete" 99
I was wondering, whether it'd be possible to enable defining expression  
without the Value data constructors, i.e.
99 .==. personAge $ Person "pete" 99
You can write:
   99 == personAge (Person "pete" 99)
Or you could write:
   c .==. f = \x -> c == f x

   test = 99 .==. personAge $ Person "pete" 99

The .==. operator is not symmetrical in this case, of course

Regards,
Henk-Jan van Tuyl

-- 
http://Van.Tuyl.eu/
http://members.chello.nl/hjgtuyl/tourdemonad.html
--