Passing by ReferencePodręcznik PHPPoprzedniRozdział 14. References ExplainedNastępnyPassing by Reference You can pass variable to function by reference, so that function could modify its arguments. The syntax is as follows: function foo (&$var) { $var++; } $a=5; foo ($a); // $a is 6 here Note that there's no reference sign on function call - only on function definition. Function definition alone is enough to correctly pass the argument by reference. Following things can be passed by reference: Variable, i.e. foo($a) New statement, i.e. foo(new foobar()) Reference, returned from a function, i.e.: function &bar() { $a = 5; return $a; } foo(bar()); See also explanations about returning by reference. Any other expression should not be passed by reference, as the result is undefined. For example, the following examples of passing by reference are invalid: function bar() // Note the missing & { $a = 5; return $a; } foo(bar()); foo($a = 5) // Expression, not variable foo(5) // Constant, not variable These requirements are for PHP 4.0.4 and later. PoprzedniSpis treściNastępnyWhat References Are NotPoczątek rozdziałuReturning References