Two loops in short succession, Logic help?
Hi all,
need help with this logical conundrum. (simplified for ease of solution).
For the purpose of the solution imagine both clients are running in the same script.
--Client 1
function TimeOut(secs)
TimeOutEnd = false
coroutine.resume(coroutine.create(function()
for i = secs, 0, -1 do
if TimeOutEnd then return end
--irrelevant
end
end))
end
--Client 2
TimeOut(20)
-- this function will most likely fire before the original TimeOut function has completed it's cycle.
button.MouseButton1Click:connect(function()
TimeOutEnd = true
TimeOut(20)
end
What I would want is for the original TimeOut sequence to terminate it's cycle, and a new cycle to begin once again. Now the example above won't work because the means the instantaneous change from true back to false will have no effect with the one second interval.
I'd prefer not to yield the existing coroutine as I don't want useless threads lying around.
Thanks for your help guysh.
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1 answer
Nice question! I've run into this too. Here's my solution for this type of problem:
version = 0
function TimeOut(secs)
version = version + 1
local curVersion = version
coroutine.resume(coroutine.create(function()
for i = secs, 0, -1 do
--Do whatever
wait(1)
if version ~= curVersion then break end
end
end))
end
Every time TimeOut is called, the version variable is changed, causing any previously running TimeOut threads to stop executing (once they return from their 'wait' command).
If you're concerned about version overflowing, you can modify the 3rd line to version = (version + 1) % 10000 or something, but it's probably not an issue because a number in lua can be precise until it hits a bit over 9x10^15 (exactly 2^53), and obviously that would take a lot of function calls.
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