Re: echo count(false); == 1 ?!

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very nice explanation!


On Wed, Nov 27, 2013 at 8:49 PM, Sebastian Krebs <krebs.seb@xxxxxxxxx>wrote:

> 2013/11/27 Daevid Vincent <daevid@xxxxxxxxxx>
>
> >
> >
> > > -----Original Message-----
> > > From: Aziz Saleh [mailto:azizsaleh@xxxxxxxxx]
> > > Sent: Wednesday, November 27, 2013 10:15 AM
> > > To: Daevid Vincent
> > > Cc: php-general@xxxxxxxxxxxxx
> > > Subject: Re:  echo count(false); == 1 ?!
> > >
> > > On Wed, Nov 27, 2013 at 1:04 PM, Daevid Vincent <daevid@xxxxxxxxxx>
> > wrote:
> > >
> > > > Really? 1?? I would have expected 0 or false or something other than
> > > > positive. *sigh*
> > > >
> > > > $ php -a
> > > > php > echo count(false);
> > > > 1
> > > >
> > > >
> > > > :-\
> > > >
> > > http://us3.php.net/count
> > >
> > > The manual is a great place to figure out why things happen a certain
> > way.
> >
> > The manual page does not explain WHY that logic is used and even
> > inconsistent since null returns 0. It only says that it does return 1.
> >
>
> Actually it does, but on a different page [1], because at the end it
> behaves like "count((array) $foo)"
>
> > For any of the types: integer, float, string, boolean and resource,
> converting a value to an array results in an array with a single element
> with
> > index zero and the value of the scalar which was converted. In other
> words, (array)$scalarValue is exactly the same as array($scalarValue).
>
> And some lines below
>
> > Converting NULL to an array results in an empty array.
>
>
> [1] php.net/language.types.array.php#language.types.array.casting
>
>
>
> >
> >
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> >
> >
>
>
> --
> github.com/KingCrunch
>

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