What References Are Not
Returning References
References Explained
PHP Manual

## Passing by Reference

You can pass a variable by reference to a function so the function can modify the variable. The syntax is as follows:

`<?php
function foo(&$var)
{
$var++;
}
$a=5;
foo($a);
// $a is 6 here
?>`

> Note: There is no reference sign on a function call - only on function definitions. Function definitions alone are enough to correctly pass the argument by reference. As of PHP 5.3.0, you will get a warning saying that "call-time pass-by-reference" is deprecated when you use & in foo(&$a);. And as of PHP 5.4.0, call-time pass-by-reference was removed, so using it will raise a fatal error.

The following things can be passed by reference:

- Variables, i.e. foo($a)
- New statements, i.e. foo(new foobar())

- References returned from functions, i.e.:

  `<?php
  function foo(&$var)
  {
  $var++;
  }
  function &bar()
  {
  $a = 5;
  return $a;
  }
  foo(bar());
  ?>`
  See more about returning by reference.

No other expressions should be passed by reference, as the result is undefined. For example, the following examples of passing by reference are invalid:

`<?php
function foo(&$var)
{
$var++;
}
function bar() // Note the missing &
{
$a = 5;
return $a;
}
foo(bar()); // Produces fatal error since PHP 5.0.5
foo($a = 5); // Expression, not variable
foo(5); // Produces fatal error
?>`
These requirements are for PHP 4.0.4 and later.

What References Are Not
Returning References
References Explained
PHP Manual