Re: Whacky increment/assignment logic with $foo++ vs ++$foo
| From: | Ralph Deffke | Date: | Sat, 03 Oct 2009 13:57:33 +0000 |
| Subject: | Re: Whacky increment/assignment logic with $foo++ vs ++$foo | ||
| References: | 1 2 3 4 5 6 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-298517@lists.php.net to get a copy of this message | ||
Ben,
might be intersting to consider that in ur c axample u r working with a pure
memory position, while php works with references. thry it with pointers it
I'm pretty shure u get the same result as in PHP.
I'm not shure, because I don't work in perl, but doesn't per work on
references as well ?
ralph_deffke@yahoo.de
"Ben Dunlap" <bdunlap@agentintellect.com> wrote in message
news:7997e80e0910021458h20ebd75dtfc51f9264f35171b@mail.gmail.com...
> mind-blowing. What the heck /is/ supposed to happen when you do this:
>
> $a = 2;
> $a = $a++;
> echo $a;
>
> Seems like any way you slice it the output should be 3. I guess what's
... and, in fact, that /is/ how C behaves. The following code:
int a = 2;
a = a++;
printf("a = [%d]\n", a);
Will output "a = [3]". At least on Ubuntu 9 using gcc 4.3.3.
So I retract my initial terse reply and apologize for misunderstanding
your question.
Ben