PHP Returning References

Answer 1

I believe it's referring to byref arguments not functions. For example this:

function doStuff(&$value1, &$value2) {
    ...
}

is an acceptable use of byref because the doStuff() function has to return 2 values. If it only doStuff() only needed to affect one value, it would be more elegant to have the function return it, by value, of course.

Answer 2

The bolded part means it's useful if you want to keep a reference to a variable, instead of the value of this variable.

The example about returning references, on php.net, explains it pretty well, IMO.

Answer 3

Na. You can't pass a reference to a function name. When passing a variable by reference, if you change it's value in your function, it's value will also be changed outside of the function.

For example :

function test(&$var) {
    $var = strtolower($var);
}

function second_test($var) {
    $var = strtolower($var);
}

$var = 'PHP';

second_test($var);
echo $var;

echo "\r\n";

test($var);
echo $var;

This will display :

PHP
php

As the second_test method doesn't have the variable passed by reference, it's updated value is only updated inside the function. But the test method as the variable passed by reference. So it's value will be updated inside and outside of this function.