[U-Boot] [RFC] ARM timing code refactoring

[Albert ARIBAUD <albert.aribaud@free.fr>]

Le 22/01/2011 22:26, Wolfgang Denk a ?crit :

>> Hmm... My idea with providing time() with an argument was that precisely
>> since we are interested only in elapsed time, not absolute time, our
>> basic time function should be able to tell us relative times.
>
> The disadvantage of this approach is that such calls cannot be nested,
> i. e. you must always make sure that the code run within the timeout
> loop does not attempt to set up a timeout on it's own.  From my point
> of view, this is a killing point.
>
>> Functionally this is the same as what I suggested, minus the absolute
>> get_timer() vs relative time(x) call. Either way works, I'm not going to
>> try and make a point that we use "my" time(x).
>
> See above for the fundamental difference - not in the implementation
> for a single timeout, but from a system-wide point of view.
>
>>>> 	then = time(0);
>>>> 	do {...} while ( time(then)<   ms_to_ticks(100) );
>>>
>>> We should NOT do this.  It is bound to break as soon as the code
>>> in the loop (the "..." part) needs to implement a timeout, too.
>>
>> I'm not sure to understand why. Can you develop how you think this would
>> break with an inner timeout?
>
> Sure:
>
> 	/* implement something which needs a 100 ms timeout */
> 	then = time(0);
>
> 	do {
> 		int then_nested;
> 		... do something...
> 		... do more...
> 		/* now do something which needs a 5 ms timeout */
> 		then_nested = time(0);
> 		do {
> 			...
> 		} while (time(then_nested)<  ms_to_ticks(5));
>
> 	} while (time(then)<  ms_to_ticks(100));
>
> You see the problem?

Actually no, I don't. As a reminder, I am considering the following 
definitions:

	#define time(n) (ticks-n)
	#define ms_to_ticks(ms) (ms * fast_tick_rate) / CONFIG_SYS_HZ

Neither of these has any side effect, so I am at loss as to why that 
would break when used in nested loops; each loop has its own reference 
start time by assigning time(0) to its own variable (then and 
then_nested), and each has its own elapsed time computation by passing 
its own variable to time() and comparing with its own constant timeout 
value. Can you point the exact instruction that would break in the code 
above, and why it would?

> Best regards,
>
> Wolfgang Denk

Amicalement,
-- 
Albert.