This section provides a tutorial example on how to return references from functions - prefix the function name with & in both definition and call statements.
PHP supports function return references using this syntax:
function &function_name() {
...
return $var_1;
}
$var_2 = &function_name();
When you use a function to return a reference, you need follow these rules:
To declare a function that returns a reference, the function name must be prefixed with '&' in the function definition statement.
If a function is declared to return a reference, it should use the 'return' statement to return a variable.
To call a function that returns a reference, the function name must be prefixed with '&' in the function call operation.
The returning reference from a function should be assigned to another variable.
To help us understand these rules, I wrote this tutorial example:
<?php # FunctionReturnReferences.php
# Copyright (c) 2003 by Dr. Herong Yang. http://www.herongyang.com/
#
function pvrv($arg) {
$arg = "Two";
return $arg;
}
print("\n 1. Pass by value and return by value:\n");
$x = "One";
$y = pvrv($x);
$y = "Three";
print(" x: $x\n");
print(" y: $y\n");
function prrv(&$arg) {
$arg = "Two";
return $arg;
}
print("\n 2. Pass by reference and return by value:\n");
$x = "One";
$y = prrv($x);
$y = "Three";
print(" x: $x\n");
print(" y: $y\n");
function &pvrr($arg) {
$arg = "Two";
return $arg;
}
print("\n 3. Pass by value and return by reference:\n");
$x = "One";
$y = &pvrr($x);
$y = "Three";
print(" x: $x\n");
print(" y: $y\n");
function &prrr(&$arg) {
$arg = "Two";
return $arg;
}
print("\n 4. Pass by reference and return by reference:\n");
$x = "One";
$y = &prrr($x);
$y = "Three";
print(" x: $x\n");
print(" y: $y\n");
?>
If you run this sample script, you should get:
1. Pass by value and return by value:
x: One
y: Three
2. Pass by reference and return by value:
x: Two
y: Three
3.Pass by value and return by reference:
x: One
y: Three
4. Pass by reference and return by reference:
x: Three
y: Three
Test 4. proves that the main code received the reference from prrr() correctly. $y is assigned with the reference of $arg.
But $arg is an alias of $x, because $x is passed as a reference. So $y becomes an alias of $x.
