No, that doesn't work; it's close, but not quite.  liftM doesn't have the right type signature.

 

liftM :: Monad m => (a -> r) -> (m a1 -> m r)

 

What would work is if you could define a function

liftLast :: Monad m => (a0 -> a1 -> ... -> aN -> r) -> (a0 -> a1 -> ... -> aN -> m r)

 

then

 

nary' f = runIdentity . nary (liftLast f)

 

  -- ryan

 

On 12/5/07, Dan Weston <westondan@imageworks.com> wrote:

Wouldn't any isomorphism do (like the Identity monad)? How about

nary' f = runIdentity . nary (liftM f) . return


Brandon S. Allbery KF8NH wrote:
>
> On Dec 5, 2007, at 16:00 , Philipp N. wrote:
>
>> the odd thing is. you can get this to work, if you have a terminating
>> type
>> as result type (for example (IO x)). then you can work with all types (IO
>> x), (a -> IO x), (a -> b -> IO x), ...
>>
>> but i don't want this delimiter IO! any ideas?
>
> Use ST instead?  (just tossing ideas in the wind...)
>


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