Re: Number of bits in C99 float types?

On Mar 17, 10:08 pm, Eric Sosman <Eric.Sos...@[EMAIL PROTECTED]
> wrote:

>      So, I can safely store a few additional chars in the
> extra bytes?
>
>         union {
>             long double aligner;
>             unsigned char buff[16];
>         } u;
>         memset (u.buff, 'x', sizeof u.buff);
>         *(long double*)u.buff = 42.0L;
>         assert (u.buff[10] == 'x');
>
> All happy and hunky-dory, right?

No. On some x86 systems the compiler uses sizeof (long double) = 16,
but the hardware uses only ten bytes. What this particular assignment
of 42.0L does depends on the code that the compiler generated: The
compiler could load the number 42 into an 80 bit floating point
register, and store the content of the register. This will modify 10
bytes and leave ten bytes unchanged. Or the compiler could have 16
byte lying around somewhere and do effectively the same as

  static long double tmp = 42.0L;
  memcpy (u.buff, &tmp, sizeof (tmp));

which will overwrite 16 bytes. Or the compiler could decide that
operations with 32 bit integers are much much faster than floating
point, but 16 bit integers are slower, so it could store 12 byte using
three 32-bit integer instructions. As long as the resulting 16 byte
are a valid representation for the number 42.0L, it is all correct.