PHP 8.4.0 RC4 available for testing

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. 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
?>

The new operator returns a reference automatically, as such assigning the result of new by reference is an error.

<?php
class C {}

$o = &new C;
?>

The above example will output:

Parse error: syntax error, unexpected 'new' (T_NEW) in …

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

Arithmetic Assignment Operators

Example Equivalent Operation
$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
$a **= $b $a = $a ** $b Exponentiation

Bitwise Assignment Operators

Example Equivalent Operation
$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

Other Assignment Operators

Example Equivalent Operation
$a .= $b $a = $a . $b String Concatenation
$a ??= $b $a = $a ?? $b Null Coalesce
add a note

User Contributed Notes 4 notes

up
139
Peter, Moscow
13 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.
up
60
Robert Schneider
9 years 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
?>
up
30
Hayley Watson
17 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;
?>
up
16
asc at putc dot de
9 years 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)
To Top