How to delay something?

Help and Feedback Scripting Support
#1

Hey so i’m looking for a way to delay something
I tried delay however due to the variable changing it did not work.

Here’s an example of what I tried with delay:

local x = 5 
delay(2, function() 
print(x * 2)
end)

x = 12
| output: 24

What I’m looking to do is have it print 10, the original without pausing the entire script

Anyone have any ideas? Thank you!

#2

You could use Wait(amount of time goes here) but I don’t think that will solve your problem.

#3

You should try to find a tutorial/develop forum for it if you did already please check back on it so you can auto correct yourself

#4

Hmm what do you mean by that? Wouldn’t this be seen as a develop forum?

#5

Well yes I am saying where did you find this script from, I am telling you to go back where you got the script…

#6

It’s a really simple answer, most beginners would already know this.

You should use the wait function.

#7

I made the script lmao as an example here, so it’s understood.

#8

The script can’t pause, it needs to continue on.

#9

How would you even know how to create a script without looking an develop forum or tutorial unless you just put random script

#10

Then you could run it on a separate thread via spawn or a coroutine which will allow it to not interfere with the current thread and pause the script.

#11

preserve the value of x before it is set again.

local x = 5
local y = x

delay(2, function()
    print(y * 2)
end)

x = 12

Not sure why you are doing this. Also I suggest using coroutine or or something else instead of delay as suggested by @Interaxis.

#12

As the script continues the run, the Y will be overwritten (as x will change on the next run)

#13

You can preserve the value inside of the function so it won’t be changed if you change the variable.

It shouldn’t change though either way.

#14

?? The code outputs 10.

image

what are you trying to do. I’ve never had a reason to use delay or spawn over years of using lua on roblox.

It does not output 10 because the function will lookup x after 2 seconds.

local x = 5
delay(2, function()
    local y = x
    print(y * 2)
end)

x = 12

will output 24
image

2 Likes
#15

Yes apologises, thank you very much! Marked as solution.