From -6394071276200918583 X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: f78e5,27b216646bf2abb1 X-Google-Attributes: gidf78e5,public X-Google-ArrivalTime: 1991-03-09 01:47:58 PST Path: gmdzi!unido!mcsun!uunet!pdn!tscs!tct!chip From: chip@tct.uucp (Chip Salzenberg) Newsgroups: comp.std.c++ Subject: Re: Conversions to/from void*, redux Keywords: Sun, C++ Message-ID: <27D8437B.F13@tct.uucp> Date: 9 Mar 91 02:07:54 GMT References: <71031@microsoft.UUCP> <27D5708A.29CF@tct.uucp> <3568@ux.acs.umn.edu> Organization: Teltronics/TCT, Sarasota, FL Lines: 50 According to dhoyt@vx.acs.umn.edu: >In article <27D5708A.29CF@tct.uucp>, chip@tct.uucp (Chip Salzenberg) writes... >>[ANSI C requires |void*| and |char*| to be identically represented] > >The reason that ANSI made this true (mostly, anyway) is because there >existed a huge body of code that depended on this equivalence. This statement is patently false, as proven by the next statement, which is true: >K&R didn't have a void type in the first place It is impossible for K&R programs to depend on |void*|, since they don't mention |void*|. Point 1: ANSI created |void*| explicitly for the purpose of holding any data address whatsoever. That's why, for example, the first two arguments to memcpy() are both of type |void*|. ANSI C++ cannot change this most basic characteristic of |void*| without breaking compatibility with many ANSI C programs and large portions of the ANSI C library. Point 2: The ARM promises that a |char*| may hold any data address whatever. This is guaranteed in the section I quoted in the referenced article, which asserts that a pointer to any object may be cast to a pointer to any object of equal or smaller type, and then back to its original type, without problem. Conclusion: Since |void*| and |char*| must both be able to hold any address, the C requirement that they have identical representations is entirely reasonable for ANSI C++. >This forces word machines, such as the Cray's, either 1) always create >slow code ... Please describe -- exactly -- how |void*|-|char*| equivalence forces machines with word addressing into generating "slow code". (Of course, on word-addressed machines, a |T*| is typically only one word, while a |void*| is larger, to specify the exact byte within the addressed word. But that would always be true, regardless of the implementation of |char*|.) >But what do I know? An excellent question. -- Chip Salzenberg at Teltronics/TCT , "Most of my code is written by myself. That is why so little gets done." -- Herman "HLLs will never fly" Rubin