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

[John Reagan]

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;