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# How do I preform a certain action a certain amount of times in 1 second evenly split out?

Let's say I want to preform an action 15 times in a second. I want to make sure the action was preformed exactly 15 times in exactly 1 second. How do I do that?

I've tried this:

local oldTime = os.time();
local amount = 15;
local iteration = 1;

while os.time() == oldTime do
wait(1 / amount);
iteration += 1;
end;

print(iteration)


If you don't know what it does, there is a loop that only lasts for 1 second, and it waits 1 15th of a second and adds 1 to a variable until a second has passed

But when I print the variable that was supposed to be 15 from adding, it's almost never 15.

By evenly split out, I mean I want the action to be done all at the same time, I want the actions to be done split evenly out the second.

(I'm doing a different action, not increasing a variable. I just did that for the example)

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This can be achieved with a for loop.

A for loop runs code a certain amount of times, and to wait in-between you just use wait().

local amount = 15;
local iteration = 0;

for count = 1,amount do
wait(1 / amount);
iteration += 1;
end;

print(iteration)


In this example it runs your code 15 times. Adding 1 to the "iteration" variable every 1/15 of a second. Also I set iteration to 0 because it ran 15 times and iteration would be 16 if it started as 1. Hopefully I didn't misunderstand and hopefully it helped!

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It worked, although I know what a for loop does, no need to explain it to me. I'll mark it as an answer when I figure out how. Nevertheless, you are a legend, thanks. beybladerstew1 2 — 3d
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rabbi99 692
3 days ago

Using os.time() == oldTime is very unreliable, as the oldTime could be something like 100000.5 and os.time() could be 100001. Meaning, it has only .5 seconds to take those 15 actions.

There's a simple solution to this. You just add + 1 second to oldTime and check if os.time() is lower or equal to oldTime.

As for the 15 actions, you need to check if iteration == 15.

local oldTime = os.time() + 1;
local amount = 15;
local iteration = 0;

while os.time() <= oldTime do
wait(1 / amount);
iteration += 1;
if iteration == 15 then
break
end
end;

print(iteration)


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It sometimes returns 14 and I've added some testing code and figured out it's done in 2 seconds. beybladerstew1 2 — 3d
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I don't care about the 2 seconds part, but if it doesn't do it 15 times it's unreliable for me. beybladerstew1 2 — 3d
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That's weird, unless you have some kind of code that delays the loop. Other than that, this is the approach you should take. rabbi99 692 — 2d