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- Path: wdl1.wdl.loral.com!dst17!mab
- From: mab@dst17.wdl.loral.com (Mark A Biggar)
- Newsgroups: comp.lang.ada,comp.lang.c++
- Subject: Re: on OO differnces between Ada95 and C++
- Date: 29 Feb 1996 23:58:29 GMT
- Organization: Loral Western Development Labs
- Distribution: world
- Message-ID: <4h5ej5$168@wdl1.wdl.loral.com>
- References: <4gh204$l7n@qualcomm.com> <DnDuA4.8GC@bton.ac.uk> <4gvk2b$sjq@watnews1.watson.ibm.com>
- NNTP-Posting-Host: dst17.wdl.loral.com
-
- In article <4gvk2b$sjq@watnews1.watson.ibm.com> ncohen@watson.ibm.com writes:
- >In article <DnDuA4.8GC@bton.ac.uk>, je@bton.ac.uk (John English) writes:
- >|> Nasser Abbasi (nabbasi@qualcomm.com) wrote:
- >Furthermore, for the convenience of its clients, X can contain a
- >declaration such as
- >
- > subtype T is Y.T;
- >
- >which causes X to reexport the type originally declared in Y. Then X's
- >client can refer to the type (as X.T) without a with clause for Y.
-
- How does this interact with "use type T;"?
-
- Does the following work?
-
- package A is
- type T is new integer;
- function "+"(L,R: T) return T;
- end A;
-
- with A;
- package B is
- subtype T is A.T;
- end B;
-
- with B;
- procedure C is
- X: B.T;
- use type B.T;
- begin
- X := X + X; --- what "+" operator do I get here if any?
- end C;
-
- --
- Mark Biggar
- mab@wdl.loral.com
-
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