What is pass by reference in php?

PHP variables, by default, are passed by value. Once PHP variables are passed by value, the variable scope, defined at the function level is linked within the scope of the function. Modification of either of the variables has no impact.

Let’s see an example:

<?php

// Function used for assigning new 
// value to $string variable and 
// printing it 
function print_string($string)
{
  $string = "W3docs" . "\n";
  
  // Print $string variable 
  print($string);
}

// Driver code 
$string = "Global w3docs" . "\n";
print_string($string);
print($string);
?>

The output will look as follows:

  Function w3docs
  Global w3docs

For passing variables by reference, it is necessary to add the ampersand (&) symbol before the argument of the variable.

An example of such a function will look as follows: function( &$x ).

The scope of the global and function variables becomes global. The reason is that they are defined by the same reference. Hence, after changing the global variable, the variable that is inside the function also gets changed.

Here is a clear example:

<?php

// Function applied to assign a new value to 
// $string variable and printing it 
function print_string(&$string)
{
  
  $string = "Function w3docs \n";
  
  // Print $string variable 
  print($string);
}

// Driver code 
$string = "Global w3docs \n";
print_string($string);
print($string);
?>

The output will be:

  Function w3docs 
  Function w3docs

As a rule, PHP variables start with the $ sign followed by the variable name.

While assigning a text value to a variable, putting quotes around the value is necessary.

In contrast to other programming languages, PHP doesn’t have a command to declare a variable. It is generated at the moment a value is first assigned to it.

In PHP, variables are containers to store data.

The PHP variables may have both short( for instance, x and y) and more descriptive names (fruitname, age, height, and so on).

When it comes to references, things get the tiniest bit more complicated - you need to be able to accept parameters by reference and also return values by reference. This is done with the reference operator, &.

Marking a variable as "passed by reference" is done in the function definition, not in the function call. That is:

function multiply(&$num1, &$num2) {

is correct, whereas

$mynum = multiply(&5, &10);

is wrong. This means that if you have a function being used hundreds (thousands?) of times across your project, you only need edit the function definition to make it take variables by reference. Passing by reference is often a good way to make your script shorter and easier to read - the choice is not always driven by performance considerations. Consider this code:

<?php
    function square1($number) {
        return $number * $number;
    }

    $val = square1($val);

    function square2(&$number) {
        $number = $number * $number;
    }

    square2($val);
?>

The first example passes a copy of $val in, multiplies the copy, then returns the result which is then copied back into $val. The second example passes $val in by reference, and it is modified directly inside the function - hence why square2($val); is all that is required in place of the first example's copying.

One key thing to remember is that a reference is a reference to a variable . If you define a function as accepting a reference to a variable, you cannot pass a constant into it. That is, given our definition of square2(), you cannot call the function using square2(10);, as 10 is not a variable, so it cannot be treated as a reference.

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What is pass by value and pass by reference in PHP?

Is PHP object pass by reference?

Is PHP array pass by reference?

Why do we pass by reference?