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On Mon, 03 Jul 2006 11:41:53 +0200, Bruno Desthuilliers wrote:
> Dmitry A. Kazakov wrote:
>> On Fri, 30 Jun 2006 17:37:34 GMT, Jay Dee wrote: >> >>>Well, no. A function, let's say, an operation on integers which >>>returns a rational (approximation, of course), like DIVIDE, >>>requires that the types exist before the function -- but the >>>types don't require the function at all. Granted, many OO >>>languages bundle the methods up in the class -- but that's a >>>mistake. >> >> It is a philosophical question. In case you have missed my previous posts, >> I reject data. So my answer is no. Types don't exist without operations and >> values. All operations have to be defined as well as values.
If you add semantics to the type (as invariants, preconditions of operations etc), then it probably will.
-- Regards, Dmitry A. Kazakov http://www.dmitry-kazakov.deReceived on Mon Jul 03 2006 - 07:39:17 CDT
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