Hello guys
I ve this code
for i = 1 , 8 do print(i) end
in this way the Sys returns me back 1 2 3 4 5 6 7 8 simultanely.
But my task is to print these numbers every 2 sec… How can I implement a timer in this ForLoop
Okay. Your problem here is that you need to really understand event programming first. If you were programming in sequential coding - things like the old Home Computer BASICs - you would write something like.
for i=1,8 do print(i) delay(2) end
but with event driven programming you don’t work like that. What you do is have an event in the program - which could be clicking on something, a timer elapsing, a collision happening etc. and responding to that, and you sort of ‘synchronise’ all these events together to make the program work.
If you are (doing) say a countdown the normal method would be to create a timer (look at timer.performWithDelay which waits a while and then does something) and print out each time.
But honestly, rather than solve this precise problem I would read up on event programming first.
Yes I know about listener…
My bad english probably made everything more complicated… With “timer” I meant that I was simply searching for something that allowed me to print each numbers every 2 seconds using the For loop instead of print every numbers instantanely
That’s the point ; you can’t. You have to do something like - fire a timer 8 times, each time printing one more than last time.
I know - I think - what you want to do. The problem is that you have to ‘think events’.
Oh I got it.
Sorry then xD… Thanks for the help
You could do it like this, but it’s going to be more efficient to set up a timer that repeats 8 times.
[lua]
for i=1,8 do
local delayIt = function ()
print(i)
end
timer.performWithDelay(2000*i, delayIt)
end
[/lua]
thanks it worked…
Okay. Your problem here is that you need to really understand event programming first. If you were programming in sequential coding - things like the old Home Computer BASICs - you would write something like.
for i=1,8 do print(i) delay(2) end
but with event driven programming you don’t work like that. What you do is have an event in the program - which could be clicking on something, a timer elapsing, a collision happening etc. and responding to that, and you sort of ‘synchronise’ all these events together to make the program work.
If you are (doing) say a countdown the normal method would be to create a timer (look at timer.performWithDelay which waits a while and then does something) and print out each time.
But honestly, rather than solve this precise problem I would read up on event programming first.
Yes I know about listener…
My bad english probably made everything more complicated… With “timer” I meant that I was simply searching for something that allowed me to print each numbers every 2 seconds using the For loop instead of print every numbers instantanely
That’s the point ; you can’t. You have to do something like - fire a timer 8 times, each time printing one more than last time.
I know - I think - what you want to do. The problem is that you have to ‘think events’.
Oh I got it.
Sorry then xD… Thanks for the help
You could do it like this, but it’s going to be more efficient to set up a timer that repeats 8 times.
[lua]
for i=1,8 do
local delayIt = function ()
print(i)
end
timer.performWithDelay(2000*i, delayIt)
end
[/lua]
thanks it worked…