Re: Another question about fixed point types.

["Peter C. Chapin" <chapinp@acm.org>]

On 2010-08-29 02:32, Yannick Duchêne (Hibou57) wrote:

> If it is symmetric around zero, that should explain your observations,
> as I would be very surprised if there would be provision for a hole
> somewhere. If it is symmetric around zero, and there is no two disjoint
> range, then this is include zero.
> 
> A tiny test program which would write first and last of the base type
> may tell more.

You're right. Thanks for pointing that out. Here is the range of
Julian_Day'Base that I obtained using a test program:

-9007199254740992.000 .. 9007199254740991.999

So my understanding was that for integral types the compiler is allowed
to allocate only the number of bits necessary to hold the range. For
example:

type Example_Type is range 1000 .. 1001;

The compiler is permitted (not required) to represent objects of type
Example_Type using only a single bit. But I guess this is a separate
issue. The base type, in which computations are done, might be wide even
if the compiler chooses to store the objects in a much small space.

So in my case as Phil explained computations on Julian Day are being
done, most likely, with a 64 bit type. Yet despite that, storage could
(perhaps) be using a smaller space since 64 bits is not needed to
represent the full range of Julian_Day at my specified resolution.

I understand that these details are compiler specific. However, it does
seem like my code, at least, should be portable in its effect.

Peter