ID:               37047
 Updated by:       [EMAIL PROTECTED]
 Reported By:      karoly at negyesi dot net
 Status:           Open
 Bug Type:         Documentation problem
 Operating System: Irrevelant
 PHP Version:      5.1.2
 New Comment:

Heh, funny thing. What static does is two things:
1. When compiling, create "static" variable entry in function's static
symbol table and assign it a value from expression or just null if none
supplied (note no code is executed because the value is constant -
parser just parses it and puts into symbol table). 
2. When "static" statement is reached in runtime, bind that entry to
regular symbol table with the same name.

Now note that when you define static second time, you instantly change
the value in static table to new value. Maybe PHP might notify you on
that, but right now it's just gives you enough rope :)


Previous Comments:
------------------------------------------------------------------------

[2007-08-20 12:22:02] [EMAIL PROTECTED]

No, there is a difference between definition and assignment. The
variable is overwritten by the static-call during compilation. The value
is only assigned when the assignment is actually done.

In Jakub's last example, the code never hits the assignment statement.

In the original report, the "static $storage" definition is done twice,
and during the 2nd time the original $storage static definition is
destroyed. The static fetch in the return statement will therefore not
work. I'd say this should be documented...

It wouldn't be hard to add a notice for this though.

------------------------------------------------------------------------

[2007-08-20 10:42:39] [EMAIL PROTECTED]

If static variables are resolved in compile time then

<?php
function storage($key) {
  return $storage;
  static $storage = array('a' => array('x', 'y'));
}
var_dump(storage('a'));
?>

should give an expected result.

------------------------------------------------------------------------

[2006-04-12 08:31:01] [EMAIL PROTECTED]

It's the same as
<?php
exit;
class Test { }
?>
The class will be still declared, even though there is an exit
statement before the declaration. It doesn't mean that it's "executed",
because there is a big difference between "execution" and "compilation".

------------------------------------------------------------------------

[2006-04-11 22:44:27] karoly at negyesi dot net

Hint. If you doc this please doc everything as well that executes at
compile time. It will be a very interesting handbook page...

------------------------------------------------------------------------

[2006-04-11 22:38:31] [EMAIL PROTECTED]

Sorry. I misread.

You're right (-:

S


------------------------------------------------------------------------

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