When defining a function in PHP, what does it mean to add the "&" symbol before the function name?

In PHP, the & symbol passes a reference to a variable rather than a copy. Reference means accessing the same variable content with different names. This is not like C's pointers, which are symbol table aliases. Note that in PHP, the variable name and the variable content are different, so the same content can have different names. The closest analogy is Unix's filenames and the files themselves - the variable names are the directory entries, and the variable contents are the files themselves. References can be thought of as tight connections in a Unix file system.
PHP’s references allow you to use two variables to point to the same content. Meaning, when you do:

<?php 
$a =&$b 
?>

It means $a and $b point to the same variable.

Note: $a and $b are exactly the same here. This does not mean that $a points to $b or vice versa, but that $a and $b point to the same place.

The same syntax can be used in functions, which return references, and in the new
operator
(PHP 4.0.4 and later):

<?php 
$bar =& new fooclass(); 
$foo =& find_var ($bar); 
?>

Note : Not using the & operator causes a copy of the object to be generated. If you use $this in a class, it will apply to the current instance of that class. Assignment without & will copy the instance (e.g. object) and $this will be applied to the copy, which is not always the desired result. Due to performance and memory consumption issues, usually you only want to work on one instance.

Although you can use the @ operator to turn off any

error messages
in a constructor, such as with @new, this has no effect when using the &new statement. This is a limitation of the Zend engine and will cause a parsing error. The second thing a reference does is pass a variable by reference. This is accomplished by creating a local variable within the function and that variable references the same content in the calling scope. For example:

<?php 
function foo (&$var) { 
$var++; 
} 
$a=5; 
foo ($a); 
?>

will change $a to 6. This is because in function foo the variable $var points to the same thing that $a points to. See Passing by Reference for a more detailed explanation.

The third thing a reference does is reference return.

What a reference is not
As mentioned before, a reference is not a pointer. This means that the following structure will not produce the effect you expect:

<?php 
function foo (&$var){ 
$var =& $GLOBALS["baz"]; 
} 
foo($bar); 
?>

This will cause the $var variable in the foo function to be bound to $bar when the function is called, but then be bound to $bar. Rebound to $GLOBALS["baz"]. It is not possible to bind $bar to another variable within the function call scope through the reference mechanism, because there is no variable $bar in function foo (it is represented as $var, but $var only has the variable content and no call symbol table name-to-value binding).

Passing by reference

You can pass a variable to a function by reference so that the function can modify the value of its parameter. The syntax is as follows:

<?php 
function foo (&$var) { 
$var++; 
} 
$a=5; 
foo ($a); 
// $a is 6 here 
?>

Note that there are no reference symbols in the function call - only in the function definition. The function definition alone is enough for parameters to be passed correctly by reference.

The following can be passed by reference:

Variables, such as foo($a)
New statements, such as foo(new foobar())
References returned from functions, such as:

<?php 
function &bar() { 
$a = 5; 
return $a; 
} 
foo(bar()); 
?>

For detailed explanation, see reference return.

Any other

expression
cannot be passed by reference, and the result is undefined. For example, the following example of passing by reference is invalid:

<?php 
function bar(){ // Note the missing & 
$a = 5; 
return $a; 
} 
foo(bar()); 
foo($a = 5) // 表达式，不是变量 
foo(5) // 常量，不是变量 
?>

These conditions are available in PHP 4.0.4 and later versions.

Reference return

Reference return is used when you want to use a function to find which variable the reference should be bound to. When returning a reference, use this syntax:

<?php 
function &find_var ($param){ 
/* ...code... */ 
return $found_var; 
} 
$foo =& find_var ($bar); 
$foo->x = 2; 
?>

In this example the properties of the object returned by the find_var function will be set (Translator: Refers to $foo->x = 2; statement) instead of copying, just like not using reference syntax.

Note: Different from parameter passing, the ampersand must be used in both places here - to indicate that a reference is returned instead of a usual copy. It also indicates that $foo is bound as a reference, not Usual assignment.

Unreference
When you unset a reference, you just break the binding between the variable name and the variable content. This does not mean that the variable contents are destroyed. For example:

<?php 
$a = 1; 
$b =& $a; 
unset ($a); 
?>

will not unset $b, just $a.

An analogy between this and Unix's unlink call may help to understand.

Reference positioning
Many PHP syntax structures are implemented through the reference mechanism, so everything above about reference binding also applies to these structures. Some constructs, such as pass-by-reference and return, have already been mentioned above. Other structures that use references are:
global reference
When declaring a variable with global $var, a reference to the global variable is actually established. That is the same as doing:

<?php 
$var =& $GLOBALS["var"]; 
?>

这意味着，例如，unset $var 不会 unset 全局变量。
$this
在一个对象的方法中，$this 永远是调用它的对象的引用。

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