Re: [Haskell-cafe] Monad laws

On 2 March 2010 15:44, Luke Palmer wrote:

On Tue, Mar 2, 2010 at 1:17 PM, David Sabel wrote:

...

Hi, when checking the first monad law (left unit) for the IO-monad (and also for the ST monad):

return a >>= f ≡ f a

I figured out that there is the "distinguishing" context (seq [] True) which falsifies the law for a and f defined below

...

let a = True let f = \x -> (undefined::IO Bool) seq (return a >>= f) True True seq (f a) True *** Exception: Prelude.undefined

Is there a side-condition of the law I missed?

No, IO just doesn't obey the laws. However, I believe it does in the seq-free variant of Haskell, which is nicer for reasoning. In fact, this difference is precisely what you have observed: the distinguishing characteristic of seq-free Haskell is that (\x -> undefined) == undefined, whereas in Haskell + seq, (\x -> undefined) is defined.

In GHC 6.12.1 both expressions reduce to True, but it doesn't happen in GHC 6.10.4. Any ideas why the behaviour is different? -- Andrés