How many sprites can you have on screen before slowdown?

Questions
21 
for i=1,#self.zombies do
		local id = self.zombies[i]
		local zp = go.get_position(id)

Calling go.get_position() in a loop is not great because it creates multiple short-lived vector objects, right? Is there any alternative to this currently?

1 Like
22

You’re correct. One thing you could do is to create one v3 per zombie when the zombies are spawned and reuse the same v3 every frame. Something like this:

function init(self)
	math.randomseed(os.time())
	self.zombies = {}
	for i=1,200 do
		local x = math.random(1, 800)
		local y = math.random(1, 480)
		local p = vmath.vector3(x, y, 0)
		local id = factory.create("#factory", p)
		self.zombies[#self.zombies + 1] = {
			id = id,
			pos = p
		}
	end
end

function update(self, dt)
	local pp = go.get_position("/player/player")
	for i=1,#self.zombies do
		local zombie = self.zombies[i]
		local id = zombie.id
		local zp = zombie.pos
		local dir = vmath.normalize(pp - zp)
		zp = zp + dir * 10 * dt
		go.set_position(zp, id)
	end
end

I also recall that someone on Discord shared an extension that optimized for this scenario by working on the v3 components directly.

1 Like
23

And what about the case where the position is changed extraneously to the object? For example through physics.

I will also try to find the extension thanks!

Maybe go.get_position() could have an overload that accepts a vector reference as an argument to store the result.

24

This is correct. I made a similar optimisation to my original code too.

I was also thinking if it might make sense to have a global variable for the player position - if it was to be monitored by other routines? In this case it’s a single zombie factory. what would happen if I had other factories? They too could share the information? Although the best option would be to use a single update function for all factories.