Mary-Ann,
I think you missed the point Ken Jackson was trying to
make. Even if you do NOT insert the return-statement the
code can still be optimized out by the compiler because
it has no so-called side-effects. Only if you make the
timeout variable 'volatile' you disallow the compiler to
make any assumptions about its use and thus the
generated code must actually do all the loops you
intended.
Greets,
Maarten
> Yes, I want the compiler to optimise it away, but I'd like it to do so
> without complaining about the "timeout" variable. The following code
> has similar problems:
>
> void func(void)
> {
> return;
>
> {
> int timeout;
> timeout = 0xFFFF;
> while (timeout--);
> }
> }
>
> Mary-Ann
>
>
> Ken Jackson wrote:
> > I don't know the answer, though I offer an observation.
> >
> > Most compilers will optimize away the delay that you are
> > trying to implement. Therefore you should declare timeout this way:
> >
> > volatile int timeout;
> >
> > Also, when I do it that way, I don't get the warning.
> >
> > -Ken Jackson
> >
> > Mary-Ann Johnson writes:
> > > If I compile the following code (SDCC 2.7.0):
> > >
> > > ------------------------------------------------
> > > void func(void)
> > > {
> > > int timeout;
> > >
> > > return;
> > >
> > > timeout = 0xFFFF;
> > > while (timeout--);
> > > }
> > > ------------------------------------------------
> > >
> > > SDCC produces the following warnings:
> > > "src/test.c:8: warning 84: 'auto' variable 'timeout' may
> > > be used before initialization
> > > src/test.c:8: warning 84: 'auto' variable 'timeout' may be
> > > be used before initialization
> > > src/test.c:8: warning 84: 'auto' variable 'timeout' may be
> > > used before initialization"
> > >
> > > Any idea why?
> > >
> > > And before someone says "Don't call 'return' there!" -
> > > this scenario happens when a macro is set to "return" to
> > > eliminate the 2nd half of a function.
> > >
> > > Thanks,
> > > Mary-Ann
>
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