References in PHP, '&' explained-PHP Tutorial-php.cn

References in PHP, '&' explained

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Release： 2016-07-25 08:52:01

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$a =& $b
?>

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This means $a and $b point to the same variable. Note: $a and $b are exactly the same here. It’s not 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):

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

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Note: No & operator is used 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 the constructor, such as @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 that references do is pass variables 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:

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

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will make $a become 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 that references do is reference returns.

What a quote is not As mentioned before, references are not pointers. This means that the following construct will not produce the effect you expect:

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

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This will cause the $var variable in the foo function to be bound to $bar when the function is called, but then re-bound 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).

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

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

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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 statement, such as foo(new foobar())

Reference returned from function, for example:

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

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See reference return for detailed explanation.

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

function bar() // Note the missing &
{
$a = 5;
return $a;
}
foo(bar());< ;/p>
foo($a = 5) // Expression, not a variable
foo(5) // Constant, not a variable
?>

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These conditions are available in PHP 4.0.4 and later versions.

Quote 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:

function &find_var ($param)
{
return $found_var;
}
$foo =& find_var ($bar) ;
$foo-> x = 2;
?>

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In this example, the properties of the object returned by the find_var function will be set (Translator: referring to the $foo-> x = 2; statement) instead of copied, just like without using reference syntax .

Note: Different from parameter passing, the ampersand must be used in both places here - to indicate that a reference is returned, not a usual copy, and also to indicate that $foo is bound as a reference, not a normal copy. Assignment.

Unquote 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:

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

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will not unset $b, Just $a.

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

Quote positioning Many PHP syntax structures are implemented through the reference mechanism, so everything mentioned 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 using references are:

global quote When declaring a variable with global $var you actually create a reference to the global variable. That is the same as doing:

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

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This means that, for example, unset $var will not unset global variables .

$this In an object method, $this is always a reference to the object that calls it. Articles you may be interested in: Detailed introduction to php reference passing by value Understand the difference between passing by value and passing by reference in PHP through examples Look at the efficiency issues of php address reference through examples About the problem of changing the variable value of the php reference address