Detailed explanation of php reference return and dereference

Detailed explanation of php reference return and dereference_PHP tutorial

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Release： 2016-07-21 15:07:28

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1. Reference return
Reference return is used when you want to use a function to find which variable the reference should be bound to. Don't use return references to increase performance, the engine is smart enough to optimize it itself. Only return references if there is a valid technical reason! To return a reference, use this syntax:

Copy the code The code is as follows:

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.
?>

The above is the explanation given by the PHP Manual and it is easy to understand

Copy code The code is as follows:

 function &test(){
 static $b = 0; // Declare a static variable 
 $b = $b+1;
 echo $b."
";
 return $b;
 }
 $a = test(); //The output value of $b is: 1
 $a = 5;
 $a = test(); //The output value of $b is: 2

 $a = &test() ; //The value of $b is: 3 **Note**
 $a = 5; //The value of $b becomes 5
 $a = test(); //The value of $b is output is: 6 **Note**
 ?>

$a = test() Although when the function is defined, it is a reference return method, but if When a function is called in this ordinary situation, its function is the same as an ordinary function, so the result is 1, 2

$a = &test() This method of calling is a reference Returning is similar to $a = &$b, and then the second sentence sets $a = 5, which is equivalent to changing the variable $b = 5. The 6 obtained in the last sentence is easy to understand!
Different from parameter passing, the ampersand must be used in both places here - indicating that a reference is returned, not the usual copy, and also indicating that $a is bound as a reference, not the usual Assignment.

Note: If you try to return a reference from a function like this: return ($this->value);, this will not work because you are trying to return the result of an expression. Not a reference variable. You can only return reference variables from functions - there is no other way. If code attempts to return the result of a dynamic expression or the new operator, an E_NOTICE error is issued starting with PHP 4.4.0 and PHP 5.1.0.

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

Copy code The code is as follows:

$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.