On Fri, 2008-01-04 at 18:46 +0000, Alain Williams wrote: > On Fri, Jan 04, 2008 at 01:06:26PM -0500, Robert Cummings wrote: > > <?php > > > > echo 'Foo: '.(0 + '5five')."\n"; > > echo 'Foo: '.((int)'5five')."\n"; > > echo 'Foo: '.(intval( ' 5five' ))."\n"; > > echo 'Foo: '.(sprintf( '%d', '5five' ))."\n"; > > > > ?> > > > > All produce 5. > > But the string 'foobar' converts to a 0 (zero) -- not what is wanted.
If that's not what's wanted then use the more strict type hint that generates E_RECOVERABLE. The conversion of 'foobar' to 0 (zero) is what happens anywhere else when 'foobar' is converted to type int. Why change that now? Cheers, Rob. -- ........................................................... SwarmBuy.com - http://www.swarmbuy.com Leveraging the buying power of the masses! ........................................................... -- PHP Internals - PHP Runtime Development Mailing List To unsubscribe, visit: http://www.php.net/unsub.php
