[Haskell-beginners] small expression evaluator

[Henk-Jan van Tuyl]

On Tue, 22 Mar 2011 10:14:57 +0100, Henk-Jan van Tuyl <hjgtuyl at chello.nl>  
wrote:

> On Tue, 22 Mar 2011 08:56:45 +0100, Petr Novotnik  
> <pnovotnik at googlemail.com> 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


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