How to remove from a table

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Right, don't remove from within the loop ^....^

I took a different approch! After looking through some similar topics I found an comment from sgeos

I was like O.......O Why didn't I think of that!

So now instead of removing from the list then adding I'm just adding all the sprites up front. It's kind of a waste since not all sprites will be shown at once.

But if figure I get the performance boost by not having to generate new sprites and I don't have to :remove them from screen AND remove them from the array then add them back later so it's a win win in my book.

Here are my results

import "coreLibs/object"
import "coreLibs/graphics"
import "coreLibs/sprites"
import "coreLibs/timer"

import "Rock"

local pd <const> = playdate
local gfx <const> = pd.graphics

local rocks = {}

function addRocks(amount)
	local buffer = 16
	local locx = 164--164
	local locy = 140

	if amount > 0 then
		for i = 0 , amount do
			rocks[i] = Rock(locx,locy)

			if i == amount then
				rocks[i].isLast = true
			end

			rocks[i]:add()

			if locx > 383 then
				locx = 164
				locy+=16
			else
				locx += buffer
			end
			
		end
	end
end

--local x, y

function getLastPosition()
	local x , y = 0,0

	for i = 0, #rocks do
		if rocks[i].isLast == true then
			x , y = rocks[i]:getPosition()

			if x >= 383 then
				x = 164
				y += 16
			else
				x += 16
			end

			rocks[i].isLast = false
			break
		end
	end

	return x, y
end

local function initalize()

	addRocks(250)

end

initalize()


function playdate.update()

	for i = 0 , #rocks do
		rocks[i]:moveUp()
	end

	for i = 0 , #rocks do
		if rocks[i].isCrushed == true then
			local x, y = getLastPosition()

			rocks[i].isCrushed = false
			rocks[i].isLast = true
			rocks[i]:moveTo(x,y)
			rocks[i]:add()
		end
	end

	gfx.sprite.update()
	pd.timer.updateTimers()
end

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