On Tue, Dec 04, 2001 at 01:36:25PM -0500, Jeff Trawick wrote:
> Francis Daly <[EMAIL PROTECTED]> writes:
>
> > 8 is a magic number. How about
> >
> > printf("Pointer size: %d bits\n", CHAR_BIT*sizeof(void*)); ?
> >
> > to allow for any machine which has some other number of bits in a
> > (C-)byte.
>
> So what is more likely?
>
> a) 8 * sizeof(void *) breaks because there aren't 8 bits in a byte
Possible in C. I don't know how likely it is in current reality.
This presumes that the point of the question is "how many bits is a
void pointer", of course.
> b) CHAR_BIT * sizeof(void *) breaks because while there may be 8 bits
> in a byte a char is two bytes
If I understand it correctly, that isn't possible in C. In C, a char
is one byte, which comprises CHAR_BIT bits (and CHAR_BIT is at least
8).
In common usage in the real world, "byte" is the same as "octet",
which is "8 bits"; but in C, "byte" is defined in terms of "sizeof
(char)", which is 1.
> c) CHAR_BIT * sizeof(void *) breaks because CHAR_BIT isn't defined or
> <limits.h> doesn't exist
If <limits.h> doesn't exist, or CHAR_BIT isn't defined in it, then
we're not in C any more either.
ISO/ANSI, that is. Are there other variants of compilers that
currently compile the code?
> :)
>
> (I don't mean to say anything bad about your suggestion.)
The only plus points I can come up with, are that it's a compile-time
constant, so has no performance impact. And it's right ;-)
f
--
Francis Daly [EMAIL PROTECTED]
