The "in" operator checks for properties not values. So what are you doing is to check if the array has something at index 3, which it doesn't
Try this: Alert.show((3 in ["a","b","c","d"]).toString()); Cheers Ralf. On Dec 8, 2007 8:37 AM, vagiff <[EMAIL PROTECTED]> wrote: > > > > > > > Try this > Alert.show((3 in [1,2,3]).toString()); > > Gives me false. > if you check for other elements of array, works fine, but last element > of array is never found. > Is this a bug, or am i missing something ? > > -- Ralf Bokelberg <[EMAIL PROTECTED]> Flex & Flash Consultant based in Cologne/Germany
