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 $a;
}
foo(bar()); // Produces fatal error as of PHP 5.0.5, strict standards notice
// as of PHP 5.1.1, and notice as of PHP 7.0.0
foo($a = 5); // Expression, not variable
foo(5); // Produces fatal error
?>← What References Are Not
Returning References →
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