Reference and return of php functions

The references of php functions are the same as the variable references in php, so today we will take a look at some examples of function returns. Friends in need can refer to them. Let’s take a look together.

Look at the code first:

<?php
function &test() {
    static $b=0;//申明一个静态变量
    $b=$b+1;
    echo $b;
    return $b;
}
$a=test();//这条语句会输出　$b的值　为１
$a=5;
$a=test();//这条语句会输出　$b的值　为2
$a=&test();//这条语句会输出　$b的值　为3
$a=5;
$a=test();//这条语句会输出　$b的值　为6
?>

Explain below:

In this way, $a=test(); actually gets It is not a reference return of a function. It is no different from an ordinary function call. As for the reason: this is a regulation of PHP. PHP stipulates that what is obtained through $a=&test(); is a reference return of a function. As for what is a reference return? (The PHP manual says: Reference return is used when you want to use a function to find which variable the reference should be bound to.) This nonsense made me unable to understand it for a long time.

Using the above example to explain, calling a function in the

$a=test() method only assigns the value of the function to $a, and any changes to $a will not occur. It affects $b in the function, and the function is called through $a=&test(). Its function is to point the memory address of the $b variable in return $b and the memory address of the $a variable to the same place. , which produces an effect equivalent to "$a=&$b;" So changing the value of $a also changes the value of $b, so after executing

$a=&test();
$a=5;

, the value of $b became 5.

Static variables are used here to let everyone understand the reference return of the function. In fact, the reference return of the function is mostly used in objects.

Attached is an official PHP example:

Calling a function using $a=test() only assigns the value of the function to $a. Any changes to $a will not affect $b in the function. Calling a function using $a=&test() Well, its function is to point the memory address of the $b variable in return $b and the memory address of the $a variable to the same place, which produces an effect equivalent to "$a=&$b;" So Changing the value of $a also changes the value of $b, so after executing

$a=&test();
$a=5;

, the value of $b becomes 5.

Static variables are used here to let everyone understand the reference return of the function. In fact, the reference return of the function is mostly used in objects

Attached is an official PHP example:

//This is the way how we use pointer to access variable inside the class.
<?php
class talker{
    private $data = &#39;Hi&#39;;
    public function & get(){
        return $this->data;
    }
    public function out(){
        echo $this->data;
    }  
}
$aa = new talker();
$d = &$aa->get();
$aa->out();
$d = 'How';
$aa->out();
$d = 'Are';
$aa->out();
$d = 'You';
$aa->out();
?>
//the output is "HiHowAreYou"

I believe that after reading the case in this article, you have mastered the method of function reference. For more exciting information, please pay attention to other related articles on the PHP Chinese website!

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