[Haskell-cafe] Smarter do notation

[Thomas Schilling]

On 5 September 2011 13:41, Sebastian Fischer <fischer at nii.ac.jp> wrote:

> Hi again,
>
> I think the following rules capture what Max's program does if applied
> after the usual desugaring of do-notation:
>
> a >>= \p -> return b
>  -->
> (\p -> b) <$> a
>
> a >>= \p -> f <$> b -- 'free p' and 'free b' disjoint
>  -->
> ((\p -> f) <$> a) <*> b
>

Will there also be an optimisation for some sort of simple patterns?  I.e.,
where we could rewrite this to:

  liftA2 (\pa pb -> f ...) a b

I think I remember someone saying that the one-at-a-time application of <*>
inhibits certain optimisations.


>
> a >>= \p -> f <$> b -- 'free p' and 'free f' disjoint
>  -->
> f <$> (a >>= \p -> b)
>
> a >>= \p -> b <*> c -- 'free p' and 'free c' disjoint
>  -->
> (a >>= \p -> b) <*> c
>
> a >>= \p -> b >>= \q -> c -- 'free p' and 'free b' disjoint
>  -->
> liftA2 (,) a b >>= \(p,q) -> c
>
> a >>= \p -> b >> c -- 'free p' and 'free b' disjoint
>  -->
> (a << b) >>= \p -> c
>

I find (a << b) confusing.  The intended semantics seem to be "effect a",
then "effect b", return result of "a".  That doesn't seem intuitive to me
because it contradicts with the effect ordering of (=<<) which reverses the
effect ordering of (>>=).  We already have (<*) and (*>) for left-to-right
effect ordering and pointed result selection.  I understand that (>>) = (*>)
apart from the Monad constraint, but I would prefer not to have (<<) =
(<*).



>
> The second and third rule overlap and should be applied in this order.
> 'free' gives all free variables of a pattern 'p' or an expression
> 'a','b','c', or 'f'.
>
> If return, >>, and << are defined in Applicative, I think the rules also
> achieve the minimal necessary class constraint for Thomas's examples that do
> not involve aliasing of return.
>
> Sebastian
>
> On Mon, Sep 5, 2011 at 5:37 PM, Sebastian Fischer <fischer at nii.ac.jp>wrote:
>
>> Hi Max,
>>
>> thanks for you proposal!
>>
>> Using the Applicative methods to optimise "do" desugaring is still
>>> possible, it's just not that easy to have that weaken the generated
>>> constraint from Monad to Applicative since only degenerate programs
>>> like this one won't use a Monad method:
>>>
>>
>> Is this still true, once Monad is a subclass of Applicative which defines
>> return?
>>
>> I'd still somewhat prefer if return get's merged with the preceding
>> statement so sometimes only a Functor constraint is generated but I think, I
>> should adjust your desugaring then..
>>
>> Sebastian
>>
>>
>


-- 
Push the envelope. Watch it bend.
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