Dear all,
On 10/27/14, Kim-Ee Yeoh
Consider
let u = return () :: IO () in expr >> u
where expr ranges over (return undefined), (undefined), and (error "urk").
Ok, so this is what we get: (1) return undefined >> return () :: IO () ==> evaluates to 'IO ()' (2)
undefined >> return () *** Exception: Prelude.undefined
(3)
error "urk" >> return () *** Exception: ur
Are their executions surprising?
In my understanding, in a strict state monad, the '>>' operator will evaluate the (value,state) pair of the left expression up to the (,) constructor. Assuming this is correct, in (1) 'return undefined' gives a '(undefined,state)', which is not evaluated further, hence the computation continues. In (2), there is no 'return' to create a (value,state) pair, so when '>>' tries to evaluate the left expression, it evaluates directly to undefined. The same holds for error, which (I guess) stops the execution immediately upon being evaluated. Is my reasoning correct?
Compare to
let u = return () :: State Int () in evalState (expr >> u) 0
for both strict and lazy state.
Ok, here we go:
Strict.evalState (return undefined >> return () :: Strict.State Int ()) 0 ()
Strict.evalState (undefined >> return () :: Strict.State Int ()) 0 *** Exception: Prelude.undefined
Strict.evalState (error "urk" >> return () :: Strict.State Int ()) 0 *** Exception: urk
Lazy.evalState (return undefined >> return () :: Lazy.State Int ()) 0 () Lazy.evalState (undefined >> return () :: Lazy.State Int ()) 0 () Lazy.evalState (error "urk" >> return () :: Lazy.State Int ()) 0 ()
So, strict-state monad matches the behavior of 'IO', which is what I expected because I know 'IO' is a strict-state monad. Furthermore, lazy-state monad never inspects the left argument of '>>', so all three cases compute.
IO above matches one of them. Suppose IO's behavior matches the other, what happens?
I guess the IO monad would behave in an undesirable way because we would not be able to raise an exception nor catch undefined computations. Is that what you meant?
Now with the definition of mapM in mind, consider the difference between your original:
Lazy.evalStateT (head `fmap` mapM return (1:undefined)) 0
and
Lazy.evalStateT (head `fmap` mapM return [1,undefined]) 0
Did you mean the latter?
No, that's not what I meant. But I assume that in the latter case we won't get a bottom because we don't evaluate beyond the spine of the list:
Lazy.evalState (head `fmap` mapM return [1,undefined]) 0 1 Lazy.evalStateT (head `fmap` mapM return [1,undefined]) 0 1
Is my interpretation correct? But what confuses me why we get different behavior here:
Lazy.evalState (head `fmap` mapM return (1:undefined)) 0 1 Lazy.evalStateT (head `fmap` mapM return (1:undefined)) 0 *** Exception: Prelude.undefined
From this I conclude that the second monad is strict, whereas the first is not. Is that correct? Assuming that is correct, does this mean that every monad stacked on top of an IO monad becomes strict?? Here's another example that illustrates this point: http://pastebin.com/0fAFmufA Why is (4) not computed in constant space, whereas (3) is? Finally, what ultimately bothers me is the following: is it true that if my state monad is built on top of an IO monad I cannot lazily consume a result of a mapM computation? (Here I am assuming, perhaps wrongly, that if '>>=' is non-strict in its left argument, then 'sequence' is tail recursive modulo cons. But I'm actually not convinced why this should be the case.) Many thanks for any input! Cheers, Jan