Hello,
robin wrote:
> From: "Dan Nagle" <[log in to unmask]>
> Sent: Monday, January 16, 2006 1:45 AM>
>
>>robin wrote:
>>
>>>From: "Phillip Helbig" <[log in to unmask]>
>>
>><snip>
>>
>>>>No. The standard specifies that it takes twice the storage of REAL.
>>
>>>That's right. That's what DOUBLE PRECISION has always meant.
>>>And it gives double precision.
>>
>>No, double precision gives more precision than single precision.
>>How much more is up to the processor.
>
>
> Are you having another problem with "DOUBLE PRECISION" too?
> It gives twice the storage of "REAL".
And, pray tell, how much storage is twice the storage of real?
And, pray tell, (and more to the point) how much precision is that?
> You already have a problem with DOUBLE PRECISION :-
Yes, Robin continues to post rubbish on this topic. (no smiley)
>>>>In fact, DOUBLE PRECISION is _not_ portable,
>
>
>>>>However, precision is associated with the number of bits of the
>>>>mantissa,
>
>
> In the context of DOUBLE PRECISION, it signifies the
> use of twice the amount of storage as REAL.
And, pray tell, how much is twice the storage of real?
>>>>not the total storage occupied. Most (all?) implementations
>>>>have significantly more than twice the precision for DOUBLE PRECISION
>>>>than for REAL.
>>
>>>That's generally true, but some implementations took/take
>>>extra exponent bits for double precision.
>>
>>Which is standard-conforming
Pasting the critical part Robin snipped:
>> because the standard
>> does not specify how much more precision double precision has
>> compared to single precision.
>> Please see section 4.4.2 of the Fortran 2003 standard.
> No-one said that it wasn't.
>
> And it's portable, as I said before.
It's not portable, precisely for the reason that the numeric storage
size is processor dependent. Or, a is Robin confusing syntax
with semantics again?
--
Cheers!
Dan Nagle
Purple Sage Computing Solutions, Inc.
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