Re: [PATCH sched-next] sched/cputime: Fix unused value issue

[Steven Rostedt]

On Tue, 19 Nov 2024 09:36:50 +0100
Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:

> > The above adds more branches than just having:
> > 
> > 	if (stime == 0)
> > 		goto update;
> > 
> > 	if (utime == 0) {
> > 		stime = rtime;
> > 		goto update;
> > 	}
> > 
> > (or's "||" are branches)
> > 
> > And the latter is much easier to read!
> > 
> > Just fix the issue. Don't try to be clever about it.  
> 
> There is nothing to fix. Yes there is an unused assignment, but the
> compiler is free to elide it (and it does).

This has nothing to do with the compiler optimizing it.

> 
> Keep the code as is, it is simple and straight-forward.

I disagree from a stability and understandability point of view. Why is
utime assigned? Here's the full context:

	if (stime == 0) {
		utime = rtime;  <<<---- Assigns utime
		goto update;    <<<---- Jumps to "update"
	}

	if (utime == 0) {
		stime = rtime;
		goto update;
	}

	stime = mul_u64_u64_div_u64(stime, rtime, stime + utime);
	/*
	 * Because mul_u64_u64_div_u64() can approximate on some
	 * achitectures; enforce the constraint that: a*b/(b+c) <= a.
	 */
	if (unlikely(stime > rtime))
		stime = rtime;

update:                           <<<---- Jumped here
	/*
	 * Make sure stime doesn't go backwards; this preserves monotonicity
	 * for utime because rtime is monotonic.
	 *
	 *  utime_i+1 = rtime_i+1 - stime_i
	 *            = rtime_i+1 - (rtime_i - utime_i)
	 *            = (rtime_i+1 - rtime_i) + utime_i
	 *            >= utime_i
	 */
	if (stime < prev->stime)
		stime = prev->stime;
	utime = rtime - stime;   <<<---- reassigns utime

So the first assignment of "utime" is meaningless, or there's a bug here
that utime incorrectly had its value overwritten. If the first assignment
is meaningless, it leaves others still wondering if there is actually a bug
here, because they are wondering "why was utime assigned?".

-- Steve