about unset

[masterinex]

<?php

$bar = 0 ;


function foo(&$bar)

{



    $bar++;


    echo "Before unset: $bar, ";

    unset($bar);


    //echo " unset: $bar, ";


    $bar = 23;

    echo "after unset : $bar \n";


}



function foo2()

{


    //static $bar;

    $bar++;


    echo "Before unset: $bar, ";

    unset($bar);


    //echo " unset: $bar, ";


    $bar = 23;

    echo "after unset : $bar \n";


}


foo($bar);


foo($bar);


foo($bar);


?>

 

when i run the code i am getting :

 

Before unset 1, after unset:23

Before unset 2, after unset 23

Before unset 3, after unset 24 ;

 

 

should it not be :

 

Before unset 1, after unset:23

Before unset 24, after unset:23

Before unset 24, after unset:23

 

 

because after the first time the function foo is called , the value of bar is 23 and when it enters the foo the second time its value will be incremented from 23  +1 which is 24  which yields the line 

 

Before unset 24, after unset:23

 

 

is my logic correct ?

[requinix]

Not quite. unset($bar) in the function only unsets the $bar variable in the function. The $bar from outside the function still exists. You named the two variables the same but that doesn't make them actually the same.

 

And no, you cannot unset the outside $bar. All you can do with the reference is change its value.

[Jacques1]

I don't think you understand what unset() does.

 

When you unset() the $bar variable in your function, then you completely destroy the reference to the outer $bar variable -- that's the whole point of unset(). There's no longer any connection between the two. Setting a new value to $bar is now purely local and doesn't affect the global $bar variable.

 

I think this will become a lot clearer if you use two different names for the variables.