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

Desour · sfan5 · commit ad75dba87bb3 · 2019-03-27T21:05:50.000+01:00
diff --git a/builtin/common/after.lua b/builtin/common/after.lua
@@ -1,33 +1,41 @@
 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)
 
 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
