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On 03/16/2012 07:51 PM, Egor Skriptunoff wrote:
2)
Frequently both quotient and remainder are needed.
Let % operator returns pair of values:
a % b == a-floor(a/b)*b, floor(a/b)


The argument for this is that some architectures have a fast divmod implementation. On the other hand, you could do this:

function divmod(q,r)
  local i,f = math.modf(q/r)
  return i, f*r
end

--
- tom
telliamed@whoopdedo.org