PHP changes the type of $a from integer to object. Why?

<?php
$a = 3;
echo 'typeof $a is : ' . gettype($a) . "\n"; // integer

$b = &$a;
echo 'typeof  $b es : ' . gettype($b) . "\n"; // integer

$c = new stdClass;
$c->name = "charles";

$b = $c;
$b->name = "bill";

echo '$c->name : ' . $c->name . "\n";
echo 'typeof $b es : ' . gettype($b) . "\n";


echo 'typeof $a is : ' . gettype($a) . "\n"; // object
echo 'The value of $a is : '开发者_运维百科 . $a->name; // bill
?>

Output:

typeof $a is : integer
typeof  $b is : integer
$c->name : bill
typeof  $b is : object
typeof $a is : object
The value of $a is : bill

It's because you're setting $b to share the same memory address as $a. So when you change $b, $a gets changed as well.

Set $b = $a instead of $b = &$a if the results are undesirable.

• $b is a reference to $a.
• You make $b equal to $c, an object.
• That means $b is now an object...
• ...and since $b is just a reference to $a, $a is an object too.

You tell $b to be a reference to $a:

$b = &$a;

Then you tell $b to refer to the object that is referred to by $c:

$b = $c;

Since $b and $a are "linked" to the same value, both are the same reference to the same object, and $a loses its integer value.

You now have two distinct references to a single stdClass object: one belongs to $c, which is obtained by creating the object and assigning it. The other is created by assigning the reference of $c, by value to $b, so it's copied. This is then shared by linking $b and $a together (assigning by reference).

Here is the php documentation on references, must read: http://php.net/manual/en/language.references.php

Because a is a reference (not a copy - thanks Tomalak) of b, because you use the &, so whenever b changes a will change also. So when later on you make b=c (which currently c is an object) b becomes an object and thus a (the reference of b) becomes one too.

Because $b = &$a refer to $a, so $b and $a are the same thing, and any references to either are interchangeable.

Your problem centers around creating this reference:

$b = &$a; 

When you afterwards assign an object to $b it will be inherited by $a. The assignment does not overwrite the reference. Instead $b is a name alias for $a's content, so all changes that you apply to $b will actually show up in $a instead. $b only retains the reference, not the properties.