$var[1] = new Base();
$var[2] =& $var[1];
Will it make any difference to my script if I use $var[2] = $var[1]; instead of $var[2] =& $var[1];
As it will eventually pass refrence to $var[2] when i write $var[2] = $var[1];
2 Answers
$var[2] =& $var[1];
Will assign a reference of the second array element (not the value) to the third element:
class Base{};
$var[1] = new Base();
$var[2] =& $var[1];
$var[1] = 'foo';
var_dump($var);
prints
array(2) {
[1]=>
&string(3) "foo"
[2]=>
&string(3) "foo"
}
I think you only want both entries point to the same object and as objects are passed by reference, $var[2] = $var[1]; is the correct way:
class Base{};
$var[1] = new Base();
$var[2] = $var[1];
$var[1]->foo = 'bar';
var_dump($var);
prints
array(2) {
[1]=>
object(Base)#1 (1) {
["foo"]=>
string(3) "bar"
}
[2]=>
object(Base)#1 (1) {
["foo"]=>
string(3) "bar"
}
}
and if you do $var[1] = 'foo' afterwards it gives you:
array(2) {
[1]=>
string(3) "foo"
[2]=>
object(Base)#1 (1) {
["foo"]=>
string(3) "bar"
}
}
Comments
You can understand by an example
$value1 = "Hello";
$value2 =& $value1; /* $value1 and $value2 both equal "Hello". */
$value2 = "Goodbye"; /* $value1 and $value2 both equal "Goodbye". */
echo($value1);
echo($value2);
let take another example
$a = array('a'=>'a');
$b = $a; // $b is a copy of $a
$c = & $a; // $c is a reference of $a
$b['a'] = 'b'; // $a not changed
echo $a['a'], "\n";
$c['a'] = 'c'; // $a changed
echo $a['a'], "\n";
output is a c
1 Comment
Poonam Bhatt
Thanks for reply. What you have explained is true for array and variable. But my question is for class and its object. how will it behave in case of object assignment.
$var[1] = $var[2] = new Base();?