LÖVE

• Sometimes, the red circles shift to the right. I'm not sure if my eyes are playing tricks or there is a bug.
  Fixed thanks to pgimeno. Thank you.
• The text looks awful... How do I fix it?
  Fixed thanks to Andiac028. Thank you.
• Any general advice for improvements will be appreciated.

PackerMan wrote: ↑Mon Dec 07, 2020 6:05 am The text looks awful... How do I fix it?

Andlac028 wrote: ↑Mon Dec 07, 2020 5:03 pm Hi and welcome to the forums!

PackerMan wrote: ↑Mon Dec 07, 2020 6:05 am The text looks awful... How do I fix it?

You are scaling default font (which has size 12) five times and that's why it looks so awful. You can create new font in love.load (with bigger size, eg: ourFont=love.graphics.newFont(size) and set it in love.draw (eg: love.graphics.setFont(ourFont)) and remove scaling from love.graphics.print and that will make the text normal.

PackerMan wrote: ↑Tue Dec 08, 2020 12:25 pm Edit: I did just that and now it looks crisp. If only I could figure out why the bullets shift.

  for i,v in ipairs(bullets) do  -- this sets i to 1, then to 2, then to 3
    v:update(dt) -- when i = 1, we're assuming that this marks it for deletion
    if v.dead then -- will be true when i = 1
      table.remove(bullets, i)  -- this deletes bullets[1] BUT it also shifts bullets[2] to bullets[1] and bullets[3] to bullets[2]
      score = score + 1
    end
    ... -- irrelevant part
    -- Now, on the next loop, i will be = 2, but because bullet[2] became bullet[1],
    -- that one won't be updated! That's what causes the shift.
  end

  for i = #bullets, 1, -1 do  -- this sets i to 3, then to 2, then to 1
    local v = bullets[i]
    v:update(dt) -- when i = 3, we're assuming that this marks it for deletion
    if v.dead then -- will be true when i = 3
      table.remove(bullets, i)  -- this deletes bullets[3], shifting everything above index 3,
                                -- and leaves bullets[2] and bullets[1] unmodified
      score = score + 1
    end
    ...
    -- Now, on the next loop, i will be = 2 and everything will Just Work™.
  end

  local i = 1
  while i <= #bullets do
    local v = bullets[i]
    v:update(dt) -- when i = 1, we're assuming that this marks it for deletion

    if v.dead then -- will be true when i = 1
      table.remove(bullets, i)  -- this deletes bullets[1], shifting bullets[2] to bullets[1] and bullets[3] to bullets[2]
      -- we don't increment i in this branch, therefore i will still be 1 on the next loop, therefore bullets[1] will be correctly updated
      score = score + 1
    else
      i = i + 1 -- No deletion - increment i
    end
    ...
  end

pgimeno wrote: ↑Wed Dec 09, 2020 2:49 pm The easiest solution is to loop backwards:

PackerMan wrote: ↑Thu Dec 10, 2020 4:18 am But looping backwards will result in a nil loop whenever a bullet dies, right?? Since the end of the table is moving faster than the counter.

1. i = 5. Bullet b5 is checked, it does not need to be deleted.
2. i = 4. Bullet b4 is checked, it needs to be deleted. You delete position 4, so the array becomes: 1:b1, 2:b2, 3:b3, 4:b5.
3. i = 3. Bullet b3 is checked, it needs to be deleted. You delete position 3, so the array becomes: 1:b1, 2:b2, 3:b5.
4. i = 2. Bullet b2 is checked, it needs to be deleted. You delete position 2, so the array becomes: 1:b1, 2:b5.
5. i = 1. Bullet b1 is checked, it needs to be deleted. You delete position 1, so the array becomes: 1:b5.

PackerMan wrote: ↑Thu Dec 10, 2020 4:18 am Edit: First I tried the "in place" solution of decrementing i by one everytime a bullet is removed, didn't work. I tried the reverse for loop and it threw me an error about trying to call something with number on the for loop line. So I went with the while loop and it made the gameplay looks clean.