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...t; (which is not quite the same as just having one special partition). When one partition is a refinement of the other (as should be the case here), this is known as the nested set union problem, and in fact, some very recent research was done into it: http://www.cs.princeton.edu/~dhlarkin/papers/esa14-nested-set-union.pdf On Fri, Jun 12, 2015 at 1:00 PM, Andrew Trick <atrick at apple.com> wrote: > >> On Apr 27, 2015, at 3:44 PM, Sanjoy Das <sanjoy at playingwithpointers.com> wrote: >> >>> Can you explain why the alias sets are collapsed into one here? >&...

...ame as just having one special partition). > > When one partition is a refinement of the other (as should be the case > here), this is known as the nested set union problem, and in fact, > some very recent research was done into it: > > http://www.cs.princeton.edu/~dhlarkin/papers/esa14-nested-set-union.pdf > > > I’m suggesting that the tracker have a special set that is not included in > the partition and assumed to alias with everything in the partition. So we > would have a partition over the subset of memory accesses that are actually > interesting. All the s...

...gt;> >>> When one partition is a refinement of the other (as should be the case >>> here), this is known as the nested set union problem, and in fact, >>> some very recent research was done into it: >>> >>> http://www.cs.princeton.edu/~dhlarkin/papers/esa14-nested-set-union.pdf >>> >>> >>> I’m suggesting that the tracker have a special set that is not included in >>> the partition and assumed to alias with everything in the partition. So we >>> would have a partition over the subset of memory accesses that...

...gt;> >>> When one partition is a refinement of the other (as should be the case >>> here), this is known as the nested set union problem, and in fact, >>> some very recent research was done into it: >>> >>> http://www.cs.princeton.edu/~dhlarkin/papers/esa14-nested-set-union.pdf >>> >>> >>> I’m suggesting that the tracker have a special set that is not included in >>> the partition and assumed to alias with everything in the partition. So we >>> would have a partition over the subset of memory accesses that...

> Can you explain why the alias sets are collapsed into one here? If I'm reading the code correctly, it is because the readonly call aliases all of %a, %b and %c. Since two pointers can be in two different alias sets only if they do not alias, %a has to be in the same alias set as the read only call and so does %b and %c. Therefore, all of them have to be in the same alias set. > Can

...rtition is a refinement of the other (as should be the case >>>>> here), this is known as the nested set union problem, and in fact, >>>>> some very recent research was done into it: >>>>> >>>>> http://www.cs.princeton.edu/~dhlarkin/papers/esa14-nested-set-union.pdf >>>>> >>>>> >>>>> I’m suggesting that the tracker have a special set that is not included in >>>>> the partition and assumed to alias with everything in the partition. So we >>>>> would have a partition...