I am running Node v0.8.14 with --nouse_idle_notification --expose_gc
--max_old_space_size=19000 --max_new_space_size=19000000.
I have a large object used as part of a BitCask style store, keeping a few
million entries.
Calling gc() manually takes a 3 seconds which is fine as I call it every 2
minutes.
The machine has 32GB of RAM and all of this is available to the process,
there is nothing else running.
The process sits at around 1.9GB of RAM.
I have found an interesting test case where async reading a 1mb file in
Node takes longer and longer depending on how many entries are in the large
object discussed above:
Node.fs.readFile('test', 'binary', End.timer())
347745 ms: Scavenge 1617.4 (1660.4) -> 1611.1 (1660.4) MB, 0 ms
[allocation failure].
350900 ms: Mark-sweep 1611.5 (1660.4) -> 1512.2 (1633.4) MB, 3153 ms
[last resort gc].
354072 ms: Mark-sweep 1512.2 (1633.4) -> 1512.0 (1592.4) MB, 3171 ms
[last resort gc].
357247 ms: Mark-sweep 1512.0 (1592.4) -> 1512.0 (1568.4) MB, 3175 ms
[last resort gc].
360426 ms: Mark-sweep 1512.0 (1568.4) -> 1512.0 (1567.4) MB, 3178 ms
[last resort gc].
363620 ms: Mark-sweep 1512.0 (1567.4) -> 1512.0 (1567.4) MB, 3193 ms
[last resort gc].
366802 ms: Mark-sweep 1512.0 (1567.4) -> 1511.6 (1567.4) MB, 3182 ms
[last resort gc].
369967 ms: Mark-sweep 1511.6 (1567.4) -> 1511.6 (1567.4) MB, 3164 ms
[last resort gc].
2012-11-04T14:59:30.700Z INFO 22230ms
Reading the 1mb file before the large object is created is fast, the bigger
the object becomes the slower the file is to read.
Why is last resort gc being called if gc is exposed and if the machine has
more than enough RAM?
What was interesting was that this behabiour does not happen for V8
3.6.6.25 and earlier.
The reason I can't use 3.6.6.25 however is that the heap is limited to
1.9GB and I need more head room than that.
Is there anyway I can disable the last resort GC?
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