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Optimize core.after in a simple way (#8351)
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42e1a12714
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@ -1,22 +1,28 @@
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local jobs = {}
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local time = 0.0
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local time_next = math.huge
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core.register_globalstep(function(dtime)
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time = time + dtime
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if #jobs < 1 then
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if time < time_next then
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return
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end
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time_next = math.huge
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-- Iterate backwards so that we miss any new timers added by
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-- a timer callback, and so that we don't skip the next timer
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-- in the list if we remove one.
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-- a timer callback.
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for i = #jobs, 1, -1 do
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local job = jobs[i]
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if time >= job.expire then
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core.set_last_run_mod(job.mod_origin)
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job.func(unpack(job.arg))
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table.remove(jobs, i)
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local jobs_l = #jobs
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jobs[i] = jobs[jobs_l]
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jobs[jobs_l] = nil
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elseif job.expire < time_next then
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time_next = job.expire
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end
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end
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end)
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@ -24,10 +30,12 @@ end)
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function core.after(after, func, ...)
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assert(tonumber(after) and type(func) == "function",
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"Invalid minetest.after invocation")
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local expire = time + after
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jobs[#jobs + 1] = {
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func = func,
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expire = time + after,
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expire = expire,
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arg = {...},
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mod_origin = core.get_last_run_mod()
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}
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time_next = math.min(time_next, expire)
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end
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