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

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In article <mr753o$1jg$3 at Iltempo.Update.UU.SE>, Johnny Billquist <bqt at softjar.se> writes:
>On 2015-08-20 22:57, "VAXman- "@SendSpamHere.ORG wrote:
>> In article <mr5au9$d3s$1 at Iltempo.Update.UU.SE>, Johnny Billquist <bqt at softjar.se> writes:
>>> On 2015-08-20 21:41, John Reagan wrote:
>>>> On Thursday, August 20, 2015 at 3:15:38 PM UTC-4, JF Mezei wrote:
>>>>> On 15-08-20 12:12, Simon Clubley wrote:
>>>>>
>>>>>> Will yes, that's how all compilers work because sys$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. :-)
>>
>> How did you define IOSB in the C?
>
>The assumption was that the IOSB was not declared as volatile. With the 
>above examples, iosb would have to be something int-like.
>
>	Johnny
>

int-like... then,...

	TSTL	IOSB
	BNEQ	10$
20$:	BRB	20$
10$:

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