RFR: 8196340: (coll) Examine overriding inherited methods in ArrayList and ArrayList.SubList

[Martin Buchholz]

Claes,

Looks good.

I would move the size check up to the beginning of the method.

583         int expectedModCount = modCount;
 584         int otherModCount = other.modCount;
 585         int s = size;
 586         if (s != other.size) {
 587             return false;
 588         }

I would prefer having only one comodification check for a bulk operation,
but I understand that checking at each step is more compatible with the
default implementation.

 594         for (int i = 0; i < s; i++) {
 595             if (!Objects.equals(es[i], otherEs[i])) {
 596                 other.checkForComodification(otherModCount);
 597                 checkForComodification(expectedModCount);
 598                 return false;
 599             }
 600         }



On Mon, May 14, 2018 at 6:37 AM, Claes Redestad <claes.redestad at oracle.com>
wrote:

> Hi Ivan,
>
> right, checkForComodification() alone should be sufficient here.
>
> Updated in-place: http://cr.openjdk.java.net/~redestad/8196340/open.01/
>
> Thanks!
>
> /Claes
>
>
> On 2018-05-12 03:38, Ivan Gerasimov wrote:
>
>> Hi Claes!
>>
>> One thing I can't figure out is why both these two checks are necessary:
>>
>> 1303             checkForComodification();
>> 1304             root.checkForComodification(expectedModCount);
>>
>> The former compares the current root.modCount with the one at the time
>> this subList was created.
>>
>> The later one compares the current root.modCount with the one at the time
>> the method was called.
>>
>> If the later fails, wouldn't it imply the former should also have failed?
>>
>>
>> With kind regards,
>>
>> Ivan
>>
>