Re: [Haskell-beginners] small expression evaluator

23 Mar 2011

      Hello Henk-Jan,

many thanks for your answer. Yes, I could do what you propose. But I 
still wonder if it is possible or not to the "lifting" as I mentioned 
earlier.

Again thanks for answering,
pete.

On 03/22/2011 01:43 PM, Henk-Jan van Tuyl wrote:
...
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