[Haskell-cafe] Re: (flawed?) benchmark : sort
Denis Bueno
dbueno at gmail.com
Mon Mar 10 12:54:30 EDT 2008
On Mon, Mar 10, 2008 at 12:19 PM, Adrian Hey <ahey at iee.org> wrote:
> Denis Bueno wrote:
> > On Mon, Mar 10, 2008 at 10:10 AM, Adrian Hey <ahey at iee.org> wrote:
> >> >> The Eq instance you've given violates the law that (x == y) = True
> >> >> implies x = y. Of course the Haskell standard doesn't specify this law,
> >> >> but it should.
> >
> > Unless I'm missing something obvious, the example Neil gave earlier
> > should make it clear how impossible this requirement is:
> >
> > What if I had made the definition of Foo:
> >
> > data Foo = Foo Int (Int -> Int)
> >
> > There is no way in general to decide the observational equivalence of
> > two values of this data type (by reduction to the halting problem).
> > Therefore it is impossible to write any function implementing such an
> > equality test.
>
> Did you read my original response to this example?
>
> http://www.haskell.org/pipermail/haskell-cafe/2008-March/040356.html
Yes. You would argue that one should not export the data constructor
Foo. That is a decision that requires more details about the code
providing Foo, although it certainly seems a reasonable approach in
many cases. Supposing I wanted to export Foo, though, the condition
you'd like to put on == breaks down. Even if I don't export Foo, how
do I ensure that any standard library functions called from the Foo
library don't depend on the condition you'd like to put on ==? Do I
have to examine them individually? Wouldn't it be easier to reason
about code if we constrain the semantics of == as *little* as possible
(as an equivalence relation)?
--
Denis
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