HaloO, Mark J. Reed wrote:
For any numeric type of $x, $x++ should mean $x += 1. 3.14 becomes 4.14. -3.14 becomes -2.14 (which indicates that floor() is not involved) . 5/8 becomes 13/8. The step size is irrelevant. If $x is so large that adding 1 gets lost due to the precision, then OK, ++$x == $x.
Hmm, for this last case we could make '$x++' mean '$x += $x.step max 1' so that for large native nums ++ makes progress. If I got Larry correctly then Num is automatically extending precision so that the step size remains 1. Regards, TSa. -- "The unavoidable price of reliability is simplicity" -- C.A.R. Hoare "Simplicity does not precede complexity, but follows it." -- A.J. Perlis 1 + 2 + 3 + 4 + ... = -1/12 -- Srinivasa Ramanujan
