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Johannes Berg authored
Jouni found a bug in the remain-on-channel logic: when a short item is queued, a long item is combined with it extending the original one, and then the long item is deleted, the timeout doesn't go back to the short one, and the short item ends up taking a long time. In this case, this showed as blocking scan when running two test cases back to back - the scan from the second was delayed even though all the remain-on-channel items should long have been gone. Fixing this with the current data structures turns out to be a bit complicated, we just remove the long item from the dependents list right now and don't recalculate the timeouts. There's a somewhat similar bug where we delete the short item and all the dependents go with it; to fix this we'd have to move them from the dependents to the real list. Instead of trying to do that, rewrite the code to not have all this complexity in the data structures: use a single list and allow more than one entry in it being marked as started. This makes the code a bit more complex, the worker needs to understand that it might need to just remove one of the started items, while keeping the device off-channel, but that's not more complicated than the nested data structures. This then fixes both issues described, and makes it easier to also limit the overall off-channel time when combining. TODO: as before, with hardware remain-on-channel, deleting an item after combining results in cancelling them all - we can keep track of the time elapsed and only cancel after that to fix this. Signed-off-by: Johannes Berg <johannes.berg@intel.com>
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