Thanks, I had indeed misunderstood Ken's point.
However, if I change the code thus:
-----------------------------------
void func(void)
{
volatile int timeout;
return;
timeout = 0xFFFF;
while (timeout--);
}
-----------------------------------
the only difference is that the warning is printed twice, rather than 3
times. :)
Any other ideas?
Mary-Ann
Maarten Brock wrote:
> 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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