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

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