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[Caml-list] Why are arithmetic functions not polymorph?
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| Date: | -- (:) |
| From: | Brian Hurt <brian.hurt@q...> |
| Subject: | Re: [Caml-list] Why are arithmetic functions not polymorph? |
On Fri, 23 May 2003 brogoff@speakeasy.net wrote:
> I understand the difference between operator redefinition, which OCaml and
> SML have, and user defined overloading, which neither has, but which can be
> found in Haskell and Clean (sort of, through type classes) and C++ and Ada.
>
> I actually think overloading can be *really* *really* good. The problem, or
> rather, one of the many problems, with C++ IMO is that it has overloading and
> implicit conversions of types. That's a bad combination.
>
> One nice thing about GCaml is that it shouldn't bother people like you who
> dislike overloading. The overloading is fairly explicit and closed world.
> It gracefully handles the most important, very simple cases, and sneaks
> in the ability to type a much wider range of functions than can be typed now.
> You should at least take a look at the README in the prototype to get an idea of
> what I mean here.
OK. I'm reading the readme. First off, congratulations, you're dodging a
lot of the bullets that C++ didn't.
Here's a question. Consider the following code:
generic one = | int => 1 | float => 1.0
generic two = | int => 2 | float => 2.0
generic plus = | int -> int -> int = (+)
| float -> float -> float = (+.)
plus one two;;
What's the result? Is it the int 3, or the float 3.0?
Another problem already arises with the generic (aka overloaded)
comparisons- you oftentimes need to arbitrarily specify types in order to
replace the expensive call to the generic compare with a much cheaper
inlined type-specific compare. If you start having to specify types a lot
more often, that reduces the advantage of having type inference.
> New operators are not sufficient, and SML is more powerful in it's ability to
> define new operators than OCaml (minus CamlP4) is.
Yeah- I'd like to be able to define accessor operators somehow. Say being
able to define $[ ] as hashtbl lookups, so that h$[3] ==> Hashtbl.find h
3.
Brian
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