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Re: Help with bit-field semantics in C and C++
- From: Mike Stump <mrs at apple dot com>
- To: Gabriel Dos Reis <gdr at integrable-solutions dot net>
- Cc: David Carlton <david dot carlton at sun dot com>, Joe Buck <Joe dot Buck at synopsys dot COM>, Roger Sayle <roger at eyesopen dot com>, Mark Mitchell <mark at codesourcery dot com>, gcc at gcc dot gnu dot org
- Date: Tue, 24 Aug 2004 17:52:16 -0700
- Subject: Re: Help with bit-field semantics in C and C++
On Tuesday, August 24, 2004, at 04:20 PM, Gabriel Dos Reis wrote:
for which the minimum bits of representation is 1, 7 is not an
enumeration value. A fortiori, it is not a honest unspecified
enumeration value.
The honest qualifier doesn't appear in my definition of the C++
language; it only has specified and unspecified values.
Another way to think about this is to examine:
#include <stdio.h>
#include <string.h>
enum E { foo = 3 } e;
int
main() {
memset (&e, ~(unsigned char)0, sizeof (e));
printf ("%u\n", (unsigned int)e);
}
The above program violates no rules of C++. The value printed is
either specified, or unspecified. Which do you think it is, and if
specified, which value is printed?
The main point is, either it is 3 or some other value. To get the
value 3, we must always mask or perform a bit-field operation, which is
costly, in general, we want to avoid costly operations.
If you survey all C++ compilers in the world that have ever been, I
think you'll discover they produce the same category of answer,
further, I suspect this answer matches C quite closely.