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Hask

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(+seq issue)
(The seq problem)
Line 13: Line 13:
 
should be equal to <hask>undefined</hask>, but can be distinguished from it using <hask>seq</hask>:
 
should be equal to <hask>undefined</hask>, but can be distinguished from it using <hask>seq</hask>:
   
ghci> (undefined :: Int -> Int) `seq` ()
+
ghci> <hask>(undefined :: Int -> Int) `seq` ()</hask>
 
* Exception: Prelude.undefined
 
* Exception: Prelude.undefined
ghci> (id . undefined :: Int -> Int) `seq` ()
+
ghci> <hask>(id . undefined :: Int -> Int) `seq` ()</hask>
 
()
 
()
   

Revision as of 01:11, 3 October 2009

Hask is the name usually given to the category having Haskell types as objects and Haskell functions between them as morphisms.

A type-constructor that is an instance of the Functor class is an endofunctor on Hask.

The seq problem

The right identity law fails in Hask if we distinguish values which can be distinguished by
seq
, since:
id . undefined = \x -> id (undefined x) = \x -> undefined x
should be equal to
undefined
, but can be distinguished from it using
seq
: ghci>
(undefined :: Int -> Int) `seq` ()
   * Exception: Prelude.undefined
ghci>
(id . undefined :: Int -> Int) `seq` ()
   ()


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