[Haskell-cafe] Proof that Haskell is RT
lennart at augustsson.net
Wed Nov 12 08:09:33 EST 2008
You can't write a straightforward dynamic semantics (in, say,
denotational style) for Haskell.
The problem is that with type classes you need to know the types
compute the values.
You could use a two step approach: first make a static semantics that
does type inference/checking and translates Haskell into a different
form that has resolved all overloading.
And, secondly, you can write a dynamic semantics for that language.
But since the semantics has to have the type inference engine inside
it, it's going to be a pain.
It's possible that there's some more direct approach that represents
types as some kind of runtime values, but nobody (to my knowledge) has
On Wed, Nov 12, 2008 at 12:39 PM, Luke Palmer <lrpalmer at gmail.com> wrote:
> On Wed, Nov 12, 2008 at 3:21 AM, Jules Bean <jules at jellybean.co.uk> wrote:
>> Andrew Birkett wrote:
>>> Is a formal proof that the Haskell language is referentially transparent?
>>> Many people state "haskell is RT" without backing up that claim. I know
>>> that, in practice, I can't write any counter-examples but that's a bit
>>> handy-wavy. Is there a formal proof that, for all possible haskell
>>> programs, we can replace coreferent expressions without changing the meaning
>>> of a program?
>> The (well, a natural approach to a) formal proof would be to give a formal
>> semantics for haskell.
> Haskell 98 does not seem that big to me (it's not teeny, but it's
> nothing like C++), yet we are continually embarrassed about not having
> any formal semantics. What are the challenges preventing its
> Haskell-Cafe mailing list
> Haskell-Cafe at haskell.org
More information about the Haskell-Cafe