On 04/17/2007 10:37 PM, Sebastian Nohn wrote: $a = 6900000000; $b = "6900000000"; $c = $a.""; $d = (string)$a;
You're converting float to string and THEN trying to output as integer. My patch returns the old behaviour, though you're code is still wrong.
--Wbr, Antony Dovgal
