2008/7/10 TSa <[EMAIL PROTECTED]>: > So, does it make sense to define ++ and -- for nums as > $x++ meaning $x += $x.step?
No. > Or should these operators floor the value to an Int and step the result? No. 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. All of this is IMESHO, of course, but I feel rather strongly on this issue. " ++ " means " += 1 ". -- Mark J. Reed <[EMAIL PROTECTED]>
