a=b//c VS a=math.floor(b/c)

Comments

• hgy29 Maintainer

  August 2022 Accepted Answer

  // is an integer divide, it justs truncates of any fractional part, whereas math.floor will round to the lower integer and math.ceil to upper integer

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• keszegh Member

  August 2022

  in this case in the docs it should be made more clear:
  https://wiki.gideros.rocks/index.php/Integer_Divide_Operator
  in particular here:
  a=b//c -- faster than a=math.floor(b/c)
  could be replaced by
  a=b//c -- faster than a=math.floor(b/c) and the same if b/c is positive
  
  btw then i guess if i need math.floor(b/c) also for negative numbers, then i cannot make it faster using // somehow?
  

  Longtitle Productions, creator of
  Fragmenter - animated loop machine and IKONOMIKON - the memory game

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• hgy29 Maintainer

  August 2022

  Yes, it will be equivalent to math.floor for positive numbers and math.ceil for negative numbers. If you want an equivalent to math.floor for negative numbers, then you could try something like

  a=b//c if a>b/c then a-=1 end

  but it may not be more efficient than math.floor

  Likes: keszegh

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• keszegh Member

  August 2022

  yes, probably not much faster or at least it does not matter if it is called a few times per frame.

  Longtitle Productions, creator of
  Fragmenter - animated loop machine and IKONOMIKON - the memory game

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