> First of all, we have got the following warning on 64-bits model

   First of all:

      uname -a

Then, which compiler are you using, and with which options?  As usual,
showing actual commands with their actual output can be more helpful
than vague descriptions or interpretations.

> warning: shift count is too large

   Knowing nothing about your environment, I'd guess that your
(unspecified) compiler on your (unspecified) system thinks that "0x1FF"
is a 32-bit integer, while "mask" is a larger (perhaps 64-bit) unsigned
integer, so it thinks that "0x1FF << 55" would shift all the bits off
the end of the 32-bit "0x1FF" value.

deb64# cat shift.c
#include <stdio.h>

int main( int argc, char **argv)
{
    unsigned long mask;

#ifdef GOOD
    mask = 0x1FFUL << ((8 * (sizeof(long) - 1)) - 1);
#else
    mask = 0x1FF << ((8 * (sizeof(long) - 1)) - 1);
#endif

    printf( " sizeof long = %lu, mask = %016lx .\n",
     sizeof(long), mask);

    return 0;
}

deb64# cc -o shift_gcc shift.c
shift.c: In function 'main':
shift.c:10: warning: left shift count >= width of type

deb64# ./shift_gcc
 sizeof long = 8, mask = 0000000000000000 .

deb64# cc -o shift_gcc_G -DGOOD shift.c

deb64# ./shift_gcc_G
 sizeof long = 8, mask = ff80000000000000 .


> Second. What about little-endian machine?
>         Should mask be equal 0xFF800000?
>         Should mask be equal 0?

   What would _zero_ do in an experssion like "integer & mask"?

> [...] the most significant end [...]

> /* mask is 0xFF800000 on a big-endian machine */
> /* mask is 0xFF000000 on a big-endian machine */

   These comments make no sense to me, because the "most significant
end" of a number is the most significant end, regardless of the
endianity of the system.

> Copyright (c) 1999

   Find newer code to steal?