Re: [Haskell-beginners] small expression evaluator

22 Mar 2011


      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)

Regards,
Henk-Jan van Tuyl


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