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Reference return and dereference in php_PHP tutorial

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Jul 14, 2016 am 10:08 AM

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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:

[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.

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

[php]

function &test(){

static $b = 0; //Declare a static variable

$b = $b+1;

echo $b."
";

return $b;

}

$a = test(); //The value of output $b is: 1

$a = 5;

$a = test(); //The value of output $b is: 2

$a = &test(); //The value of output $b is: 3 **Note**

$a = 5; //The value of $b becomes 5

$a = test(); //The output value of $b is: 6 **Note**

$a = test() Although the function is defined in a reference return method, if the function is called in this ordinary situation, its function will be the same as an ordinary function, so the results are 1 and 2

$a = &test() This method of calling is to return by reference, which is similar to $a = &$b. Then the second sentence sets $a = 5, which is equivalent to setting the variable $b = 5. The last sentence The obtained 6 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. www.2cto.com

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 rather than a referenced 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. 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:

[php]

$a = 1;

$b =& $a;

unset($a);

Won’t unset $b, just $a.

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

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