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run following code:
<?php
$a = array('yes');
$a[] = $a;
var_dump($a);
out put:
array(2) {
[0]=>
string(3) "yes"
[1]=>
array(1) {
[0]=>
string(3) "yes"
}
}
run following code:
<?php
$a = array('no');
$b = &$a;
$a[] = $b;
$a = array('yes');
$a[] = $a;
var_dump($a);
out put:
array(2) {
[0]=>
string(3) "yes"
[1]=>
array(2) {
[0]=>
string(3) "yes"
[1]=>
*RECURSION*
}
}
I have reassigned value of $a, why there are RECURSION circular references?
317
Pass by reference: When variables are passed by reference, use & (ampersand) symbol need to be added before variable argument. For example: function( &$x ). Scope of both global and function variable becomes global as both variables are defined by same reference.
It means you pass a reference to the string into the method. All changes done to the string within the method will be reflected also outside that method in your code. See also: PHP's =& operator.
The $x (single dollar) is the normal variable with the name x that stores any value like string, integer, float, etc. The $$x (double dollar) is a reference variable that stores the value which can be accessed by using the $ symbol before the $x value.
To remove reference you need to call unset. Without unset after $a = array('yes'); $a still bounds with $b and they are still references. So the second part has the same behavior as first one.
Note, however, that references inside arrays are potentially dangerous. Doing a normal (not by reference) assignment with a reference on the right side does not turn the left side into a reference, but references inside arrays are preserved in these normal assignments.
http://php.net/manual/en/language.references.whatdo.php
140
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