PHP 7.0.6 Released

Assignment Operators

The basic assignment operator is "=". Your first inclination might be to think of this as "equal to". Don't. It really means that the left operand gets set to the value of the expression on the right (that is, "gets set to").

The value of an assignment expression is the value assigned. That is, the value of "$a = 3" is 3. This allows you to do some tricky things:

<?php

$a 
= ($b 4) + 5// $a is equal to 9 now, and $b has been set to 4.

?>

In addition to the basic assignment operator, there are "combined operators" for all of the binary arithmetic, array union and string operators that allow you to use a value in an expression and then set its value to the result of that expression. For example:

<?php

$a 
3;
$a += 5// sets $a to 8, as if we had said: $a = $a + 5;
$b "Hello ";
$b .= "There!"// sets $b to "Hello There!", just like $b = $b . "There!";

?>

Note that the assignment copies the original variable to the new one (assignment by value), so changes to one will not affect the other. This may also have relevance if you need to copy something like a large array inside a tight loop.

An exception to the usual assignment by value behaviour within PHP occurs with objects, which are assigned by reference in PHP 5. Objects may be explicitly copied via the clone keyword.

Assignment by Reference

Assignment by reference is also supported, using the "$var = &$othervar;" syntax. Assignment by reference means that both variables end up pointing at the same data, and nothing is copied anywhere.

Example #1 Assigning by reference

<?php
$a 
3;
$b = &$a// $b is a reference to $a

print "$a\n"// prints 3
print "$b\n"// prints 3

$a 4// change $a

print "$a\n"// prints 4
print "$b\n"// prints 4 as well, since $b is a reference to $a, which has
              // been changed
?>

As of PHP 5, the new operator returns a reference automatically, so assigning the result of new by reference results in an E_DEPRECATED message in PHP 5.3 and later, and an E_STRICT message in earlier versions.

For example, this code will result in a warning:

<?php
class {}

/* The following line generates the following error message:
 * Deprecated: Assigning the return value of new by reference is deprecated in...
 */
$o = &new C;
?>

More information on references and their potential uses can be found in the References Explained section of the manual.

User Contributed Notes

straz at mac dot nospam dot com
12 years ago
This page really ought to have table of assignment operators,
namely,

See the Arithmetic Operators page (http://www.php.net/manual/en/language.operators.arithmetic.php)
Assignment    Same as:
$a += $b     $a = $a + $b    Addition
$a -= $b     $a = $a - $b     Subtraction
$a *= $b     $a = $a * $b     Multiplication
$a /= $b     $a = $a / $b    Division
$a %= $b     $a = $a % $b    Modulus

See the String Operators page(http://www.php.net/manual/en/language.operators.string.php)
$a .= $b     $a = $a . $b       Concatenate

See the Bitwise Operators page (http://www.php.net/manual/en/language.operators.bitwise.php)
$a &= $b     $a = $a & $b     Bitwise And
$a |= $b     $a = $a | $b      Bitwise Or
$a ^= $b     $a = $a ^ $b       Bitwise Xor
$a <<= $b     $a = $a << $b     Left shift
$a >>= $b     $a = $a >> $b      Right shift
Robert Schneider
1 year ago
Be aware of assignments with conditionals. The assignment operator is stronger as 'and', 'or' and 'xor'.

<?php
$x
= true and false;   //$x will be true
$y = (true and false); //$y will be false
?>
Peter, Moscow
5 years ago
Using $text .= "additional text"; instead of $text =  $text ."additional text"; can seriously enhance performance due to memory allocation efficiency.

I reduced execution time from 5 sec to .5 sec (10 times) by simply switching to the first pattern for a loop with 900 iterations over a string $text that reaches 800K by the end.
Hayley Watson
8 years ago
bradlis7 at bradlis7 dot com's description is a bit confusing. Here it is rephrased.

<?php
$a
= 'a';
$b = 'b';

$a .= $b .= "foo";

echo
$a,"\n",$b;?>
outputs

abfoo
bfoo

Because the assignment operators are right-associative and evaluate to the result of the assignment
<?php
$a
.= $b .= "foo";
?>
is equivalent to
<?php
$a
.= ($b .= "foo");
?>
and therefore
<?php
$b
.= "foo";
$a .= $b;
?>
asc at putc dot de
9 months ago
PHP uses a temporary variable for combined assign-operators (unlike JavaScript), therefore the left-hand-side (target) gets evaluated last.

Input:
$a += $b + $c;

Meaning:
$a = ($b + $c) + $a;

Not:
$a = $a + ($b + $c);

This can be important if the target gets modified inside the expression.

$a = 0;
$a += (++$a) + (++$a); // yields 5 (instead of 4)
Hayley Watson
8 years ago
You could also take adam at gmail dot com's xor-assignment operator and use the fact that it's right-associative:

$a ^= $b ^= $a ^= $b;
ma dot bx dot ar at gamil dot com
1 year ago
Document says:
"An exception to the usual assignment by value behaviour within PHP occurs with objects, which are assigned by reference in PHP 5. Objects may be explicitly copied via the clone keyword."

But it's not very accurate! Considering this code:
<?php
$a
= new StdClass;
$b = $a;

$a = new StdClass;

var_dump ($a, $b);
?>

Output:
object(stdClass)#2 (0) {
}
object(stdClass)#1 (0) {
}
Note: #2 and #1 means two different objects.

But this code:
<?php
$a
= new StdClass;
$b = &$a;

$a = new StdClass;

var_dump ($a, $b);
?>

Output will be:

object(stdClass)#2 (0) {
}
object(stdClass)#2 (0) {
}

Note: Still pointing to the same object.

And this shows that that exception is not valid, PHP assignment for objects still makes a copy of variable and does not creates a real reference, albeit changing an object variable members will cause both copies to change.
So, I would say assignment operator makes a copy of 'Object reference' not a real object reference.
bradlis7 at bradlis7 dot com
10 years ago
Note whenever you do this

<?php
$a
.= $b .= "bla bla";
?>

it comes out to be the same as the following:

<?php
$a
.= $b."bla bla";
$b .= "bla bla";
?>

So $a actually becomes $a and the final $b string. I'm sure it's the same with numerical assignments (+=, *=...).
haubertj at alfredstate dot edu
4 years ago
[[   Editor's note: You are much better off using the foreach (array_expression as $key => $value) control structure in this case   ]]

When using

<php
while ($var = current($array) {
#do stuff
next($aray)
?>

to process an array, if current($array) happens to be falsy but not === false it will still end the loop.  In such a case strict typing must be used.

Like this:

<php
while (($var = current($array)) !== FALSE) {
#do stuff
next($aray)
?>

Of course if your array may contain actual FALSE values you will have to deal with those some other way.
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