What does this signature mean (&) in PHP? - php

Consider:
public function & get($name, $default = null)
Why &?

In PHP's syntax, this means that the function returns a reference instead of a value. For example:
<?php
$foo = 'foo';
function & get_foo_ref ()
{
global $foo;
return $foo;
}
// Get the reference to variable $foo stored into $bar
$bar = & get_foo_ref();
$bar = 'bar';
echo $foo; // Outputs 'bar', since $bar references to $foo.
?>
In the above example, removing the & from the function declaration would make the $foo variable still contain 'foo', since only the value, not the reference was returned from the function.
This was used more often in PHP4, because it did not pass objects by their reference and cloned them instead. Because of this, object variables had to be passed by reference to avoid unwanted cloning. This is no longer the case in PHP5 and references should not be used for this purpose.
However, functions that return references are not completely useless either (or bad practice, when not used for replacing object references).
For example, personally I've used them when creating a script that passes a "path" to an function, which returns reference to variable in that path allowing me to set value to it and read the value. Due to the recursive nature of the function, returning the reference was needed.
By rithiur

It means it returns a reference to the result as opposed to a copy of it.
Returning References in PHP

PHP's pass by reference metacharacter. See Passing by Reference

Related

Passing by reference - how does it work and why is it used?

Take the following code from CodeIgniter's show_error function:
$_error =& load_class('Exceptions', 'core');
The documentation for the load_class function says it acts as a singleton. The function basically takes the given parameters and searches for a class in the appropriate path. It then includes the file if it exists. The function is declared as:
function &load_class(...)
Why does it have the & and what is its purpose? Is $_error declared as such as a result of defining the function like that?
I don't see any point of declaring and using load_class like that. From the source code of load_class(), we can see that it caches loaded objects in an array with the class name as the key. If it is not in the cache, it loads an object given a name, and then stores that object reference into the array. In both cases, it returns the element of the array (by reference).
Returning by reference allows the caller to have a reference to the element of the array. The only things that this allows us to do are:
See later changes to that array element (i.e. the value associated with that key) from the outside reference we have. But this is not applicable, since the load_class function never changes the value associated with a key after it sets it.
Have external code be able to change the element in the array, without the load_class function knowing about it. But this would be a highly dubious practice, to mess with the cache from the outside, and I highly doubt this is something the authors wanted.
So there is no legitimate reason to return by reference. My guess is that it is a leftover from PHP 4, when objects were values, and so assigning or returning an "object value" would copy it. In PHP 5, objects are not values; you can only manipulate them through object references, and assigning or returning an object reference by value never copies the object it points to.
The php documentation seems to explain why you have to uses =& even though the function is marked to return a refrence function &load_class
Returning References
Returning by reference is useful when you want to use a function to
find to which variable a reference should be bound. Do not use
return-by-reference to increase performance. The engine will
automatically optimize this on its own. Only return references when
you have a valid technical reason to do so. To return references, use
this syntax:
<?php class foo {
public $value = 42;
public function &getValue() {
return $this->value;
}
}
$obj = new foo;
$myValue = &$obj->getValue(); // $myValue is a reference to $obj->value, which is 42.
$obj->value = 2;
echo $myValue;
// prints the new value of $obj->value, i.e. 2. ?>
In this example,
the property of the object returned by the getValue function would be
set, not the copy, as it would be without using reference syntax.
Note: Unlike parameter passing, here you have to use & in both places
- to indicate that you want to return by reference, not a copy, and to indicate that reference binding, rather than usual assignment, should
be done for $myValue.
If you are asking what references in general are the documentation explains.

Help me understand PHP variable references and scope

References:
If I pass a variable to a function (e.g. $var), is that supposed to be a copy of a reference to the actual variable (such that setting it null doesn't affect other copies)?
Or is it receiving a reference to what is a new copy of the actual variable (such that setting it to null destroys its copy only)?
If the latter, does this copy objects and arrays in memory? That seems like a good way to waste memory and CPU time, if so.
I think can understand passing by reference (e.g. &$var) correctly by knowing how this works, first.
Scope:
What's the deal with local scope? Am I right in observing that I can declare an array in one function and then use that array in other functions called within that function WITHOUT passing it to them as a parameter?
Similarly, does declaring in array in a function called within a function allow it to be available in the caller?
If not, does scoping work by a call stack or whatever like every bloody thing I've come to understand about programming tells me it should?
PHP is so much fun. :(
If I pass a variable to a function (e.g. $var), is that supposed to be a copy of a reference to the actual variable (such that setting it null doesn't affect other copies)?
Depends on the function. And also how you call it. Look at this example:
http://www.ideone.com/LueFc
Or is it receiving a reference to what is a new copy of the actual variable (such that setting it to null destroys its copy only)?
Again depends on the function
If the latter, does this copy objects and arrays in memory? That seems like a good way to waste memory and CPU time, if so.
Its going to save memory to use a reference, certainly. In php>4 it always uses reference for objects unless you specify otherwise.
What's the deal with local scope? Am I right in observing that I can declare an array in one function and then use that array in other functions called within that function WITHOUT passing it to them as a parameter?
No you can't.
Similarly, does declaring in array in a function called within a function allow it to be available in the caller?
No, it doesn't.
If not, does scoping work by a call stack or whatever like every bloody thing I've come to understand about programming tells me it should?
If you want to use a variable from outside the function, before using it, you'd write global $outsidevar
Concerning your first set of questions:
foo($a);
function foo($b) { echo $b; }
In this case, $a will not be copied to a new variable $b, only because it is passed by value.
This is because PHP uses the copy-on-write concept. PHP will not copy the contents of a variable, unless they are changed. Instead PHP will increment the refcount property of the existing "zval" of $a.
Well, the whole thing is not that trivial, but to answer your question: No, it does not copy the variable, unless you write to it in the function and no, you won't save CPU and Memory by using a reference. In most cases the reference won't change performance at all, but in the worst case it will actually degrade it (because if a not is_ref variant of the variable already exists and a reference is created the value of the variable must be copied to get a zval with is_ref and one without). Optimizing code by using references is no good.
if argument to a function is defined as so "function my_function($variable) {}" then you are getting a copy of the variable and any alterations made to the variable inside your function will not be available to the function caller. you can pass a variable by reference by prepending an ampersand to the argument when defining your function and thus any alterations made to the variable will persist to the function caller, ie "function my_function(&$variable) {}"
function myfunction($var) {
$var = 'World';
}
$var = 'Hello';
myfunction($var);
echo $var; // 'Hello';
Passing a variable by reference
function myfunction(&$var) {
$var = 'World';
}
$var = 'Hello';
myfunction($var);
echo $var; // 'World'

Function to create new, previously undefined variable

Sorry for such a lame title, but I just had no idea what to put there, hope you understand it. Plus, I have no idea if similar question has been asked before, because I don't know the proper keywords for it - therefore couldn't google it too.
Basicly... When looking at preg_match_all();, they got this matches parameter that will create new array defined in function, and give us the ability to access it after function execution.
And the question is.. How can I implement such a feature in my own function? But that it could create single variable and/or array.
Thanks in advance!
preg_match_all() accepts a reference to an array, which in its own scope is called $matches. As seen in the function prototype:
array &$matches
If you call the function and pass in a variable, if it does not already exist in the calling scope it will be created. So in your user-defined function, you accept a parameter by reference using &, then work with it inside your function. Create your outer-scope variable by simply declaring it in your function call, like you the way you call preg_match_all() with $matches.
An example:
function foo(&$bar) {
$bar = 'baz';
}
// Declare a variable and pass it to foo()
foo($variable);
echo $variable; // baz
I think you are referring to function parameters passed by reference, are you not?
function putValInVar(&$myVar, $myVal){
$myVar = $myVal;
}
$myVar = 1;
putValInVar($myVar, 2);
echo $myVar; // outputs '2', but will output '1' if we remove the '&' //
By default function arguments in PHP are passed by value. This means that new variables are created at each function call and those variables will exist only inside the function, not affecting anything outside it.
To specify that an argument should be used by reference the syntax is to append an & before declaring it in the function header. This will instruct PHP to use the passed variable inside the function rather than creating a copy of it.
Exception: Objects are always passed by reference. (Well... Not really, but it's complicated. See the comment thread for more info.)
I think what you are asking for is passing-by-reference. What preg_match_all basically does to "create" an array variable outside its scope is:
function preg_match_all($foo, $bar, & $new_var) {
$new_var = array(1,2,3);
}
The crucial point here is & in the function definition. This allows you to overwrite variables in the outer scope when passed.
Stylistically this should be used with care. Try to return arrays or results instead of doing it via reference passing.
Like this:
$myvariable = runfunction();
function runfunction() {
//do some code assign result to variable (ie $result)
return $result;
}
Or
global $result;
function runfunction() {
global $result;
$result = 'something';
}

What's the benefit of function refrence?

When I reading the code of CodeIgniter,I found some functions written as follows:
function &get_instance()
{
global $CI, $OBJ;
if (is_object($CI))
{
return $CI;
}
return $OBJ->load;
}
I can understand variable refrence,but I can hardly get this through.Is it necessary to use this function style?And any benefit?
Thanks.
In PHP's syntax, this means that the function returns a reference instead of a value. For example:
<?php
$foo = 'foo';
function & get_foo_ref ()
{
global $foo;
return $foo;
}
// Get the reference to variable $foo stored into $bar
$bar = & get_foo_ref();
$bar = 'bar';
echo $foo; // Outputs 'bar', since $bar references to $foo.
?>
In the above example, removing the & from the function decleration would make the $foo variable still contain 'foo', since only the value, not the reference was returned from the function.
This was used more often in PHP4, because it did not pass objects by their reference and cloned them instead. Because of this, object variables had to be passed by reference to avoid unwanted cloning. This is no longer the case in PHP5 and references should not be used for this purpose.
However, functions that return references are not completely useless either (or bad practice, when not used for replacing object references). For example, personally I've used them when creating a script that passes a "path" to an function, which returns reference to variable in that path allowing me to set value to it and read the value. Due to recursive nature of the function, returning the reference was needed.
It is the same concept as variable reference, in fact this "style" is required when you want to return a REFERENCE to a variable and not the value itself.
Explained in the manual here: http://www.php.net/manual/en/functions.returning-values.php
no, there's no benefit. It was necessary for php4, but in php5 it's a bad practice.

Symbol in PHP I've never come across before

I probably should have, but I've never seen this before. Ran into it when looking over the documenation of a Smarty Plugin.
$smarty =& new Smarty;
The =& sign in particular. If you enter it in Google, it gets ignored, just like any other search engine. What is this used for?
Same goes for this function signature:
function connect(&$smarty, $reset = false)
Why the & symbol?
Actually, this code is written to be compatible with PHP 4. The ampersand is useless in PHP 5 (as Tim said - since PHP 5, all objects are passed by reference).
With PHP 4, all variables were passed by value.
If you wanted to pass it by reference, you had to declare a reference assignment :
$ref_on_my_object =& new MyObject();
This code is still accepted with PHP 5 default configuration, but it's better to write :
$ref_on_my_object = new MyObject(); // Reference assignment is implicit
For your second problem, the issue is "almost" the same...
Because PHP lets you declare function arguments (resp. types), and you can't do it for return values.
An accepted, but "not so good" practice is to avoid reference declaration within the function's declaration :
function foo($my_arg) {
// Some processing
}
and to call with a reference...
$my_var;
$result = foo( &$my_var );
// $my_var may have changed because you sent the reference to the function
The ideal declaration would be more like :
function foo( & $my_input_arg ) {
// Some processing
}
then, the call looses the ampersand :
$my_var;
$result = foo( $my_var );
// $my_var may have changed because you sent the reference to the function
It is used for passing values by reference rather than by value which is default in php.
& passes an argument by reference. In this fashion, connect() can manipulate the $smarty object so that the calling function can retrieve the modified object.
Similarly, =& sets a variable by reference.
As Tim said its a reference to a variable. But if you're using a recent version of PHP then all class object are passed by reference anyway. You would still need this if you were passing about arrays, or other builtin types though.
The first example is returning reference, the second is passing reference.
You can read all about it in the PHP manual
& is PHP's reference operator. It's used to return a reference to the object. In this case "new Smarty".
The ampersand will assign a reference to the variable, rather than the value of the object.
One of the primary uses of the ampersand operator is to pass by memory address. This is usually something you do when you want to have a variable changed, but not be returned.
function test_array(&$arr)
{
$varr[] = "test2";
}
$var = array('test');
test_array($var);
print_r($var);
this should output
array( test , test2 );
The purpose of this is usually when you need to pass the actual copy[memory address] you are working with into another function / object. Typically it was used in the past to alleviate a lack of memory and speed up performance, it's a feature from C / C++ and a few other low level languages.

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