Bill Page <[email protected]> writes:
| > | On Thu, Apr 2, 2009 at 12:06 AM, Gabriel Dos Reis wrote:
| > | > If you're not proposing to export the Rep, then your proposal is a bit
| > | > obscure to me. Would you mind clarifying why the above does not
| > | > amount to exporting the Rep of SomeDomain?
| > | >
| > Bill Page wrote:
| > | Because it is not part of the 'with' clause and appears to the right
| > | of ==.
| >
|
| On Thu, Apr 2, 2009 at 10:38 AM, Gabriel Dos Reis wrote:
| > But with your construct, in the capsule I would be able to tell what
| > Rep(SomeDomain) is. That reveals the Rep of SomeDomain, and
| > effectively exports it.
| >
|
| Let's take a real example from 'src/algebra/tree.spad.pamphlet':
I already addressed this in my previous mail.
[...]
| Note: This code is currently the same in both FriCAS and OpenAxiom.
OpenAxiom issues this warning:
Warnings:
[1] OpenAxiom suggests removing assignment to Rep
| Since BinarySearchTree is a particular class of BinaryTree for
| consistency in BinarySearchTree I would expect to see:
|
| Implementation == BinaryTree(S) add
| Rep == List Tree S
As I said, I would not. I would not expect any definition of Rep at
all. The domain extension says BinaryTree(S) is the Rep of
BinarySearchTree(S).
| Note: The 2nd line above is not required in OpenAxiom.
And I think that is a coherent, simple notation for a simple semantics.
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