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Re: $i++ problem understanding

Posted by lain on 11/04/81 11:28

Pugi! wrote:
> Currently I am studying PHP.
>
> I can understand the following :
> $a = 10;
> $b = $a++;
> print("$a, $b");
> will print 11, 10 on the screen.

just wrong.
what happens is:
$b = $a;
$a++;

so the above example actually will print ("10, 10"). understanding this
will also help you understanding the other example of yours. $a++ is
called "post increment" which means incrementing AFTER using the value
of this. of course there is also "pre increment" as of ++$a which will
do the opposite: incrementing the value and then use it for the
expression.

 

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