From mboxrd@z Thu Jan 1 00:00:00 1970 X-Spam-Checker-Version: SpamAssassin 3.4.4 (2020-01-24) on polar.synack.me X-Spam-Level: X-Spam-Status: No, score=-1.3 required=5.0 tests=BAYES_00,INVALID_MSGID autolearn=no autolearn_force=no version=3.4.4 X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: 103376,4ee5611d3fbf05b7 X-Google-Attributes: gid103376,public From: bralick@seas.smu.edu (William Bralick) Subject: Re: Enumeration literal visibility and use type Date: 1998/05/27 Message-ID: <6kh5r5$6ig$1@hermes.seas.smu.edu>#1/1 X-Deja-AN: 356977641 Distribution: na References: <6kej65$dnh$1@hermes.seas.smu.edu> Organization: SMU - School of Engineering & Applied Science Newsgroups: comp.lang.ada Date: 1998-05-27T00:00:00+00:00 List-Id: In article , Matthew Heaney wrote: | |No, use type will NOT give you direct visibility to the enumeration literals. Acknowledged. |... |The issue is that "use type" only gives you direct visibility to the |predefined "operators" of a type; it does not give you direct visibility |to the predefined "operations" of a type. Enumeration literals are |operations, not operators, and so aren't made directly visible by "use |type." I think that this is the key point where my confusion arose. I recall understanding the concept of enumerals being parameterless functions and I guess that I conflated them with the concept of "operators" perhaps because I was simultaneously dealing with the definition of "primitive operations" and "primitive subprograms" which, of course, include operators ... Oh, well. |Does that explain it? Yes, thanks. You have all been most helpful ... Best regards, -- Will Bralick ............_/......._/__/..._/.......__/..._____/ ........._/....._/_/._/.._/......_/._/.._/..._/ ........................._/..._/______/._/...________/._/___/ .............__/_/......._______/.......__/..._/