[Info-vax] Volatile, was: Re: yet another sys$qiow question

[Johnny Billquist]

On 2015-08-20 21:41, John Reagan wrote:
> On Thursday, August 20, 2015 at 3:15:38 PM UTC-4, JF Mezei wrote:
>> On 15-08-20 12:12, Simon Clubley wrote:
>>
>>> Will yes, that's how all compilers work because sys$qiow doesn't return
>>> control back to the program until it's completed so it's just like a
>>> normal subroutine call.
>>
>> The compiler doesn't know that. The compiler would/should treat both QIO
>> and QIOW as possibly modifying a value during their execution. Even if
>> QIO only modifies it later, that variable should be marked as "possibly
>> modified" by the compiler for the next time you access it, whether the
>> instriuction right after a QIOW or 10 instructions after QIO. Right ?
>>
>
> Well, yes, the FIRST access after $QIO will be a memory read.  However, since the asynchronous update may occur later, the compiler won't re-read the memory.
>
> For something like
>
> while (iosb == 0) do;
>
> It is not a loop of memory fetches (well, it might be at /NOOPT).  The optimizer will quickly turn that into:
>
> tmp = iosb;
> while (tmp == 0) do;

I think an even better optimization would be:

if (iosb == 0) while (1) {};

And of course, when this ends up in assembler, it would actually be 
(excuse my MACRO-11):

	TST	IOSB
	BNE	10$
20$:	BR	20$
10$:

I'm sure you can translate that to Macro-32 easily. :-)

	Johnny