oukei, stopping to ask anymore :)) Andrey
----- Original Message ----- From: "Andi Gutmans" <[EMAIL PROTECTED]> To: "Andrey Hristov" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]> Sent: Saturday, December 21, 2002 7:20 PM Subject: Re: [PHP-DEV] Interesting result > Again, as it is undefined in PHP the question "Why" in itself is wrong :) > > Andi > > At 04:43 PM 12/21/2002 +0200, Andrey Hristov wrote: > > >----- Original Message ----- > >From: "Andi Gutmans" <[EMAIL PROTECTED]> > >To: "Andrey Hristov" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]> > >Sent: Saturday, December 21, 2002 4:17 PM > >Subject: Re: [PHP-DEV] Interesting result > > > > > > > It doesn't matter because the behavior here is undefined just like in C. > > > You can't use a variable and it's post/pre increment value in the same > > > expression. > > > > > > >Two years ago I've been told by a professort that a+++a+++a depends on the > >language and the compiler. Yesterday i found this page : > >http://www.blueshoes.org/en/developer/syntax_exam/ > >If a=1 -> > >Java : 6 > >PHP : 6 > >gcc 2.95 : 3 > > > >I know that is "on the edge" and I will never use it in real script by I was > >curious why the > >reference change the result. > > > >Best wishes, > >Andrey > > > > > > > > At 03:26 PM 12/21/2002 +0200, Andrey Hristov wrote: > > > > Hi, > > > >i got an interesting case : > > > > > > > ><?php > > > >$a = 1; > > > >var_dump($a + $a++); > > > > > > > > > > > >$a = 1; > > > >$x = &$a; > > > >var_dump($a + $a++); > > > > > > > >?> > > > >Result > > > >int(2) > > > >int(3) > > > > > -- PHP Development Mailing List <http://www.php.net/> To unsubscribe, visit: http://www.php.net/unsub.php
