Correct. The compiler can see that a is being changed so optimization won't
give you an infinite loop.
On Wed, Oct 24, 2012 at 7:34 AM, mind entropy <mindentr...@gmail.com> wrote:
> On Wed, Oct 24, 2012 at 3:58 PM, David Brown <da...@westcontrol.com>
> wrote:
>
> > On 24/10/2012 09:38, mind entropy wrote:
> > > On Wed, Oct 24, 2012 at 12:36 PM, Eric Decker <cire...@gmail.com>
> > > wrote:
> > >
> > >>
> > >>
> > >> On Tue, Oct 23, 2012 at 11:14 PM, mind entropy
> > >> <mindentr...@gmail.com>wrote:
> > >>
> > >>> Hi,
> > >>>
> > >>> I was reading the mspgcc manual (
> > >>>
> > >>>
> >
> http://www.eecs.harvard.edu/~konrad/projects/motetrack/mspgcc-manual-20031127.pdf
> > >>>
> > >>>
> > )
> > >>> and on Pg 39 its written as
> > >>>
> > >>
> > >> I believe there is a more up to date version but I'm not usre where
> > >> it lives. Perhaps Peter can point to it?
> > >>
> > >>
> > >>>
> > >>> "If you execute while ((long) a & 0x80000l); the program will
> > >>> hang, unless ’a’ is declared volatile. So, do it!"
> > >>>
> > >> With optimization a will be put into a register (actually two
> > >> registers since it is a long) and loaded once.
> > >>
> > >> If a is declared as a volatile, that tells the toolchain that a can
> > >> change out from underneath any assumptions made about contents.
> > >> This forces the compiler to reload a into a register each
> > >> iteration.
> > >>
> > >
> > > Its loaded once during the start of the iteration right?
> > > Theoretically if there is no change from outside is there a point in
> > > making it volatile? i.e. if that loop block is the only thing in the
> > > loop?
> > >
> >
> > If "a" is not volatile, the compiler will generate code that is like
> this:
> >
> > long localCopyOfA = a;
> > if (localCopyOfA & 0x800001) {
> > while (true) ;
> > }
> >
> > If that is what you meant to write, then it's fine to have "a"
> > non-volatile. But I /really/ doubt that this is what you want.
> >
> > The loop is completely useless unless "a" is modified from outside in
> > some way - from an interrupt function (the typical case), via hardware
> > (maybe "a" is a register, or perhaps modified by DMA), from another RTOS
> > thread, or even via a debugger (such constructs can sometimes be useful
> > in testing and debugging). In all these cases, you have to let the
> > compiler know that "a" may be modified externally - and you do that by
> > declaring it "volatile".
> >
> >
> Yes I realized it later. The code I have below would not need to have
> volatile right since
> the change is not made externally from an interrupt etc (Assuming that code
> is the only
> code in the program)
>
> unsigned long a = 0L;
> while(a < 10L) {
> a++;
> }
>
>
>
> > > If it applies to long then it should apply to double and float also
> > > right?
> >
> > It applies to /all/ variables, of all sizes.
> >
> > If the variable is bigger than the native size (16-bit in this case),
> > then "volatile" alone is not enough - you need to do extra work to make
> > the access atomic.
> >
> > >
> > >>
> > >>> I did not get it. Why should it hang?
> > >>>
> >
> >
> >
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