1

I'm working on some logic, using fill clocks to drive my command blocks.

I usually end up having to extend my circuit, and rather than adjust the fill clock command, I usually just put in some redstone dust.

My question is, does this add more load than extending the fill clock?

I know that when making a fill clock, you must use a solid block in place of the redstone block, so that lighting updates are not triggered.

Does redstone dust trigger a lighting update when it's switched?

  • I think redstone dust turning on and off will trigger lighting updates, but I'm not sure. It wouldn't be hard to test if redstone emits light. – MBraedley Dec 25 '14 at 18:17
  • 2
    As a side note, if you decided to go on with the dust anyway, you might get some undesired results from your circuit since fill clocks run at 20 cycles/sec whereas redstone dust is limited by the "redstone tick" rate at 10/sec. – SpectralFlame Dec 25 '14 at 18:28
  • Technically, fill clocks don't have to run at 20Hz, it's just it's the typical implementation does. – MBraedley Dec 25 '14 at 18:59
  • @SpectralFlame Redstone dust can activate command block 20 times a second, but its activation order is random depending on location so not very convenient when you need to run various commands in certain order. – QbsidianH20 Dec 25 '14 at 20:05
2

Yes it does

This article goes into quite a lot of depth on the matter but the main point is that while redstone dust does not trigger lighting updates, it does trigger redstone propagation updates, which are just as bad! Wherever possible, redstone dust should be eliminated from clock circuitry, or any circuits that are running very frequently. Use fill strips instead to power all your command blocks.

  • 1
    Oh, wow, that's ugly. I should go back and modify some of my old answers. – MBraedley Dec 25 '14 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.