[Haskell-cafe] Re: Laziness question
nicolas.pouillard at gmail.com
Wed Aug 4 11:01:48 EDT 2010
On Wed, 04 Aug 2010 15:41:54 +0200, Janis Voigtländer <jv at informatik.uni-bonn.de> wrote:
> Nicolas Pouillard schrieb:
> >>> Actually I think we can keep the old generic seq, but cutting its full
> >>> polymorphism:
> >>> seq :: Typeable a => a -> b -> b
> >> I guess I don't know enough about Typeable to appreciate that.
> > Basically the Typeable constraints tells that we dynamically know the identity
> > of the type being passed in. So this may be a bit challenging to cleanly
> > explain how this safely disable the parametricity but in the mean time
> > this is the net result the type is dynamically known at run time.
> > ...
> > However I would like to here more comments about this seq variant, anyone?
> On reflection, isn't Typeable actually much too strong a constraint?
It is indeed "too strong", or "not precise enough" we could say.
However at least this simple change make it "correct" (i.e. restore
the parametricity results).
I would call this function genericSeq :: Typeable a => a -> b -> b
> Given that it provides runtime type inspection, probably one cannot
> derive any parametricity results at all for a type variable constrained
> by Typeable.
Exactly. We could say that we no longer car derive wrong parametricity
results about it.
> In contrast, for a type variable constrained via a
> hypothetical (and tailored to seq) Eval-constraint, one still gets
> something which looks like a standard free theorem, just with some side
> conditions relating to _|_ (strictness, totality, ...).
Indeed, that's why I want both!
In particular for the instance on functions which could be defined using
> >>> OK, I better understand now where we disagree. You want to see in the type
> >>> whether or not the free theorem apply,
> >> Oh, YES. That's the point of a free theorem, isn't it: that I only need
> >> to look at the type of the function to derive some property about it.
> >>> I want them to always apply when
> >>> no call to unsafe function is made.
> >> Well, the question is what you mean by "no call to unsafe function is
> >> made". Where? In the function under consideration, from whose type the
> >> free theorem is derived? Are you sure that this is enough? Maybe that
> >> function f does not contain a call to unsafeSeq, but it has an argument
> >> which is itself a function. Maybe in some function application,
> >> unsafeSeq is passed to f in that argument position, directly or
> >> indirectly. Maybe f does internally apply that function argument to
> >> something. Can you be sure that this will not lead to a failure of the
> >> free theorem you derived from f's type (counting on the fact that f does
> >> not call an unsafe function)?
> >> Of course, preventing the *whole program* from calling unsafeSeq is
> >> enough to guarantee validity of the free theorems thus derived. But
> >> that's equivalent to excluding seq from Haskell altogether.
> > It depends on the unsafe function that is used. Using unsafeCoerce
> > or unsafePerformIO (from which we can derive unsafeCoerce) badely
> > anywhere suffice to break anything. So while seq is less invasive
> > I find it too much invasive in its raw form.
> Hmm, from this answer I still do not see what you meant when you said
> you want free theorems to always apply when no call to seq is made. You
> say that seq is less invasive, so do you indeed assume that as soon as
> you are sure a function f does not itself (syntactically) contain a call
> to seq you are safe to use the standard free theorem derived from f's
> type unconstrained? Do you have any justification for that? Otherwise,
> we are back to banning seq completely from the whole program/language,
> in which case it is trivial that no seq-related side conditions will be
Actually given genericSeq, I no longer advocate the need for a polymorphic
seq function. So both genericSeq and the seq from the type class would
However the rule is still the same when using an unsafe function you are on
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