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);
// $a is 6 here
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Note that there's no reference sign on function call - only on function definition. Function
definition alone is enough to correctly pass the argument by reference.
Following things can be passed by reference:
-
Variable, i.e. foo($a)
-
New statement, i.e. foo(new foobar())
-
Reference, returned from a function, i.e.:
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
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These requirements are for PHP 4.0.4 and later.
