Re: How do you bitwise operations in Ada '83 and '95

["Dmitry A. Kazakov" <mailbox@dmitry-kazakov.de>]

On Fri, 21 Jul 2006 20:03:09 +0100, Simon Wright wrote:

> "Dmitry A. Kazakov" <mailbox@dmitry-kazakov.de> writes:
> 
>> It depends on endianness. Provided that storage elements are octets
>> on both machines, which assumption could be wrong, 128 shifted left
>> might yield 0, 256 or even 64 (what is "left"?).
> 
> You might as well argue that the C operator<< could produce 64 or 256.

I can't tell for C. Does ANSI C standard require (x<<1) == (x*2)?

As for Ada 95, ARM certainly does not. This is why Shift_* is not portable.
Then Interfaces, where Shift_* is declared, does not refer to C, anyway.

>> When considering *any* thinkable architecture, it might be even
>> impossible to implement Shift_* at all. Let some crazy hardware
>> store numbers in only odd bits, reserving even ones for something
>> else (like a check-sum). On such imaginary hardware Shift_Left (*,
>> 1) wouldn't even yield a valid number! Actually "they operate on a
>> bit-by-bit basis" is quite hazy. What is bit? Which bits are
>> adjacent? How are they ordered? What does "left" mean? Which bits
>> are of the number? etc.
> 
> One would think the market for hardware like that would be pretty
> thin; not really a "thinkable" architecture. Certainly not
> interesting.

Ah, but the standard is silent about interesting vs. uninteresting bit
representations. Anyway, middle-endian machines still exist and produced.

-- 
Regards,
Dmitry A. Kazakov
http://www.dmitry-kazakov.de