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Optimize core.after in a simple way (#8351)

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DS 2019-03-27 21:05:50 +01:00 committed by sfan5
parent 42e1a12714
commit ad75dba87b
1 changed files with 13 additions and 5 deletions
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18
builtin/common/after.lua
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@ -1,22 +1,28 @@
local jobs = {}
local time = 0.0
local time_next = math.huge
core.register_globalstep(function(dtime)
time = time + dtime
if #jobs < 1 then
if time < time_next then
return
end
time_next = math.huge
-- Iterate backwards so that we miss any new timers added by
-- a timer callback, and so that we don't skip the next timer
-- in the list if we remove one.
-- a timer callback.
for i = #jobs, 1, -1 do
local job = jobs[i]
if time >= job.expire then
core.set_last_run_mod(job.mod_origin)
job.func(unpack(job.arg))
table.remove(jobs, i)
local jobs_l = #jobs
jobs[i] = jobs[jobs_l]
jobs[jobs_l] = nil
elseif job.expire < time_next then
time_next = job.expire
end
end
end)
@ -24,10 +30,12 @@ end)
function core.after(after, func, ...)
assert(tonumber(after) and type(func) == "function",
"Invalid minetest.after invocation")
local expire = time + after
jobs[#jobs + 1] = {
func = func,
expire = time + after,
expire = expire,
arg = {...},
mod_origin = core.get_last_run_mod()
}
time_next = math.min(time_next, expire)
end