RFR(L): 8203197: C2: consider all paths in loop body for loop predication

[Roland Westrelin]

Hi Vladimir,

>> With this:
>> 
>> while() {
>>    if (!null_check1) unc(null_check);
>>    if (!range_check1) unc(range_check);
>>    ...
>>    if (some_condition) { // both branches profiled taken
>>      if (!null_check2) unc(null_check);
>>      if (!range_check2) unc(range_check);
>>      ...
>>    }
>>    if (!null_check3) unc(null_check);
>>    if (!range_check3) unc(range_check);
>> }
>
> In this example what will happen if *_check1 and *_check3 are moved from 
> loop first (they are regular predicates). And then you found that 
> "profiling" *_check2 can be moved. How you enforce order in such case?

Predicates *_check1 and *_check3 are executed first because they are
part of the regular predicates block. Then only *_check2 are executed
because they're part of the profiling predicates block which is
after. My understanding is that it doesn't matter whether predicates are
executed in the same order they would be in the loop body because they
have no side effect anyway.

Order of predicates matter in cases like:

while() {
  if (object == null) unc(null_check);
  value = object.field;
  if (value == null) unc(null_check);
  result += value.field;
}

Once they are out of loop, if (object == null) .. must be before if
(value == null) .. because the first check causes the object.field load
to be becomes loop independent and the second check depends on the
object.field load. So order matters, because there's a data dependence
between the 2 checks.

> No I was asking about this case when you have "profiling" case:
>
> while() {
>    if (some_condition) { // both branches profiled taken
>      if (!null_check) unc(null_check);
>      if (!range_check) unc(range_check);
>      break;
>    }
>    ...
> }
>
> Should you take special care in such case because such checks only 
> executed on exit?

Both checks are not in the loop. If we walk through the loop body from
the tail to the head we shouldn't encounter them, right?

Roland.