> On Feb. 9, 2017, 12:46 a.m., Michael Park wrote: > > What's the subsequent tie-breaker if the # of allocations is the same?
It's the client name (see `DRFComparator::operator()`). This test also covers that case (not super explicitly though). - Neil ----------------------------------------------------------- This is an automatically generated e-mail. To reply, visit: https://reviews.apache.org/r/56427/#review164819 ----------------------------------------------------------- On Feb. 8, 2017, 3:40 a.m., Neil Conway wrote: > > ----------------------------------------------------------- > This is an automatically generated e-mail. To reply, visit: > https://reviews.apache.org/r/56427/ > ----------------------------------------------------------- > > (Updated Feb. 8, 2017, 3:40 a.m.) > > > Review request for mesos and Michael Park. > > > Repository: mesos > > > Description > ------- > > When two clients have the same share, the sorter uses the total number > of allocations made to each client as a tiebreaker. > > > Diffs > ----- > > src/tests/sorter_tests.cpp a57a4fa4dbf7a9184e29b5dca4f636fe65a77773 > > Diff: https://reviews.apache.org/r/56427/diff/ > > > Testing > ------- > > `make check` > > > Thanks, > > Neil Conway > >
