[Haskell-cafe] Monad laws

[Andrés Sicard-Ramírez]

On 2 March 2010 15:44, Luke Palmer <lrpalmer at gmail.com> wrote:

> On Tue, Mar 2, 2010 at 1:17 PM, David Sabel
> <sabel at ki.informatik.uni-frankfurt.de> 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
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