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($a)
- New statements, i.e. foo(new foobar())
-
References returned from functions, i.e.:
See more about returning by reference.<?php
function foo(&$var)
{
$var++;
}
function &bar()
{
$a = 5;
return $a;
}
foo(bar());
?>
No other expressions should be passed by reference, as the result is undefined. For example, the following examples of passing by reference are invalid:
<?php
function foo(&$var)
{
$var++;
}
function bar() // Note the missing &
{
$a = 5;
return $a;
}
foo(bar()); // Produces fatal error since PHP 5.0.5
foo($a = 5); // Expression, not variable
foo(5); // Produces fatal error
?>
Коментарии
By removing the ability to include the reference sign on function calls where pass-by-reference is incurred (I.e., function definition uses &), the readability of the code suffers, as one has to look at the function definition to know if the variable being passed is by-ref or not (I.e., potential to be modified). If both function calls and function definitions require the reference sign (I.e., &), readability is improved, and it also lessens the potential of an inadvertent error in the code itself. Going full on fatal error in 5.4.0 now forces everyone to have less readable code. That is, does a function merely use the variable, or potentially modify it...now we have to find the function definition and physically look at it to know, whereas before we would know the intent immediately.beware unset() destroys references
$x = 'x';
change( $x );
echo $x; // outputs "x" not "q23" ---- remove the unset() and output is "q23" not "x"
function change( & $x )
{
unset( $x );
$x = 'q23';
return true;
}
<?php
// PHP >= 5.6
// Here we use the 'use' operator to create a variable within the scope of the function. Although it may seem that the newly created variable has something to do with '$x' that is outside the function, we are actually creating a '$x' variable within the function that has nothing to do with the '$x' variable outside the function. We are talking about the same names but different content locations in memory.
$x = 10;
(function() use ($x){
$x = $x*$x;
var_dump($x); // 100
})();
var_dump($x); // 10
// Now the magic happens with using the reference (&). Now we are actually accessing the contents of the '$y' variable that is outside the scope of the function. All the actions that we perform with the variable '$y' within the function will be reflected outside the scope of this same function. Remembering this would be an impure function in the functional paradigm, since we are changing the value of a variable by reference.
$y = 10;
(function() use (&$y){
$y = $y*$y;
var_dump($y); // 100
})();
var_dump($y); // 100
?>
Within a class, passing array elements by reference which don't exist are added to the array as null. Compared to a normal function, this changes the behavior of the function from throwing an error to creating a new (null) entry in the referenced array with a new key.
<?php
class foo {
public $arr = ['a' => 'apple', 'b' => 'banana'];
public function normalFunction($key) {
return $this->arr[$key];
}
public function &referenceReturningFunction($key) {
return $this->arr[$key];
}
}
$bar = new foo();
$var = $bar->normalFunction('beer'); //Notice Error. Undefined index beer
$var = &$bar->referenceReturningFunction('beer'); // No error. The value of $bar is now null
var_dump($bar->arr);
/**
[
"a" => "apple",
"b" => "banana",
"beer" => null,
],
*/
?>
This is in no way a "bug" - the framework is performing as designed, but it took careful thought to figure out what was going on. PHP7.3
Parameters passed by references can have default values.
You can find out if a variable was actually passed by using func_num_args():
<?php
function refault( & $ref = 'Do I have to be calculated?'){
echo 'NUM ARGS: '. func_num_args()."\n";
echo "ORI VALUE: {$ref}\n";
if( func_num_args() > 0 ) $ref = 'Yes, expensive to calculate result: ' . sleep(1);
else $ref = 'No.';
echo "NEW VALUE: {$ref}\n";
}
$result = 'Do I have to be calculated?';
refault( $result );
echo "RESULT: {$result}\n";
// NUM ARGS: 1
// ORI VALUE: Do I have to be calculated?
// NEW VALUE: Yes, expensive to calculate result: 0
// RESULT: Yes, expensive to calculate result: 0
refault();
// NUM ARGS: 0
// ORI VALUE: Do I have to be calculated?
// NEW VALUE: No.
?>