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From: watzka@stat.uni-muenchen.de (Kurt Watzka)
Subject: Re: Which language pays most? Smalltalk, not C++ nor Java.
Date: 1997/12/28
Message-ID: <685mee$5d4$1@sparcserver.lrz-muenchen.de>#1/1
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References: <199712121931.LAA25389@sirius.infonex.com>
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Organization: Institut fuer Statistik der LMU Muenchen (Germany)
Reply-To: watzka@stat.uni-muenchen.de
Newsgroups: 
 comp.lang.fortran,comp.lang.c,comp.lang.c++,comp.lang.eiffel,comp.lang.java.programmer,comp.lang.smalltalk,comp.lang.perl.misc,comp.lang.ada,comp.edu
Date: 1997-12-28T00:00:00+00:00
List-Id: <comp.lang.ada>


Guillermo Schwarz <gschwarz@netup.cl> writes:

>Kaz Kylheku wrote:
>> Although 'memcpy' is not a reserved keyword by any means,
>> it is a reserved 'external name'. 
>It is the first time I see reserved external.

Well, looks as if there still is something to learn about C for you.
The exact wording is "are reserved as identifiers with external
linkage", but "reserved external name" looks close enough to me.

>> A strictly conforming program may not define any external
>> name that begins with 'mem' followed by any combination of letters,
>> digits or underscores. 
>Ok. Define void memfoo() { } and see if it compiles.
>It does.

The behaviour of a program that declares or defines an identifier with
the same name as an identifier reserved in that context is undefined.
Undefined behaviour means: "No diagnostic required, and whatever the
implementor chooses to do is correct behaviour as far as the definition
of the C programming language is concerned".

>> Define memcpy, if you write such a static function, you must not 
>> include the <string.h> header in
>> the same translation unit, else undefined behavior results.
>Wrong. The compiler points out two different implementations.
>The executable can't be generated.

If memcpy is not defined as a function with external linkage, and 
if <string.h> is not included, an implementation that does not
sucessfully translate the program in question is at error. Did 
you actually _try_ it, and if yes, which implementation did
"point out two different implementations"?

--------8<--------8<--------8<--------8<--------8<--------8<--------
#include <stdio.h>

static void *memcpy(void *to, const void *from, size_t n)
{
   char *cto = to;
   const char *cfrom = from;

   while (n--)
      *cto++ = *cfrom++;
}

struct foo
{
   int fred;
   char wilma[32];
   double barney;
};

int main(void)
{ 
   struct foo bar = { 0 }, baz = { 42, "Hello", 3.14159 };
   memcpy(&bar, &baz, sizeof baz);
   printf("%d, %s, %f\n", bar.fred, bar.wilma, bar.barney);
   return 0;
}
--------8<--------8<--------8<--------8<--------8<--------8<--------

is a valid C program, because the header associated with the file scope
identifier memcpy() is not included, and because memcpy() is not defined
as a symbol with external linkage.

>> There is no such thing as ``overriding'' in C. Each external name must
>> have exactly one definition if it is used anywhere in the program.
>That's true. C is not object oriented. C++ is just an Smalltalk
>wannabe.

The object model and design ideas behind C++ are completely different
from Smalltalk. You could probably call C++ a Simula wannabe (If you
are prepared to call a Mustang a Modell T wannabe), but calling
it a Smalltalk wannabe indicates that you either don't know Smalltalk
or you don't know C++.

Kurt

-- 
| Kurt Watzka                             Phone : +49-89-2178-2781
| watzka@stat.uni-muenchen.de