Evenly-spread task placement #1576
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ssalinas
reviewed
Jul 5, 2017
| @@ -333,7 +333,8 @@ public void testFrozenSlaveCanBeDecommissioned() { | |||
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| saveAndSchedule(request.toBuilder().setSlavePlacement(Optional.of(SlavePlacement.OPTIMISTIC)).setInstances(Optional.of(2))); | |||
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feel free to change this to GREEDY or SEPARATE so that the frozen slave test isn't relying on calculations from optimistic
ssalinas
reviewed
Jul 5, 2017
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| final boolean isSlaveOk = numOnSlave < numPerSlave; | ||
| // If no tasks are active for this request yet, we can fall back to greedy. | ||
| if (currentlyActiveTasksForRequestClusterwide.size() > 0) { |
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would it make sense to make this more explicit with a == 0 check + early return rather than having to add a comment?
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Hmm, so the general pattern in this method is to return only if the SlaveMatchState is determined to be something other than OK. Otherwise, flow is allowed to trickle down to that final return SlaveMatchState.OK statement.
ssalinas
reviewed
Jul 5, 2017
| final boolean isSlaveOk = numOnSlave <= numPerSlave; | ||
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| if (!isSlaveOk) { | ||
| LOG.trace("Rejecting OPTIMISTIC task {} from slave {} ({}) due to numOnSlave {}", taskRequest.getRequest().getId(), slaveId, host, numOnSlave); |
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might want to add the other values you're using in he calculation in here for debugging purposes
…lacement strategy logic for this test.
baconmania
changed the title
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