Detailed explanation of the role of reference (& symbol) in php_PHP tutorial

PHP reference (that is, adding an ampersand in front of a variable, function, object, etc.)

Quotation in PHP means: different names access the same variable content.
and C language The pointers in are different. The pointer in C language stores the address where the content of the variable is stored in memory

Variable reference

PHP’s reference allows you to use two variables to point to the same content


$a="ABC";
$b =&$a;
echo $a;//Output here: ABC
echo $b;//Output here :ABC
$b="EFG";
echo $a;//The value of $a here becomes EFG, so EFG is output
echo $b;//EFG is output here
?>

Function call by address
I won’t go into details about the call by address. Below is the code directly.

function test(&$a)
{
$a =$a+100;
}
$b=1;
echo $b;//Output 1
test($b); //What $b is passed to the function here is actually $ The memory address where the variable content of b is located. You can change the value of $b by changing the value of $a in the function
echo "
";
echo $b;//output 101

It should be noted that if you test(1); here, an error will occur. Think about the reason yourself

The reference return of the function
Look at the code first

function &test ()
{
static $b=0;//Declare a static variable
$b=$b+1;
echo $b;
return $b;
}

$a=test();//This statement will output the value of $b as 1
$a=5;
$a=test();//This statement will Output the value of $b as 2

$a=&test();//This statement will output the value of $b as 3
$a=5;
$a=test() ;//This statement will output the value of $b as 6

Explanation below:
In this way, $a=test(); actually does not get the reference return of the function, which is different from ordinary There is no difference in function calls. As for the reason: This is the regulation of PHP
PHP stipulates that the reference return of the function is obtained through $a=&test();
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 it is
$a When calling a function using =test(), it just assigns the value of the function to $a, and any changes to $a will not affect $b in the function
How about calling the function through $a=&test() , its function is to point the memory address of the $b variable in return $b and the memory address of the $a variable to the same place
, which produces the equivalent of this effect ($a=&b;) so change $ 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

Object reference


class a{
var $abc="ABC";
}
$b=new a;
$c=$b;
echo $b->abc;//Output ABC here
echo $c->abc;//Output ABC here
$b->abc="DEF";
echo $c->abc;//Output DEF here
?> ;

The above code is the running effect in PHP5
In PHP5, the copying of objects is achieved through references. In the above column, $b=new a; $c=$b; is actually equivalent to $b=new a; $c=&$b;
The default in PHP5 is to call objects by reference, but sometimes you may want to Create a copy of an object and hope that changes to the original object will not affect the copy. For this purpose, PHP defines a special method called __clone.

The role of reference
If the program It is relatively large, there are many variables referencing the same object, and you want to clear it manually after using the object. I personally recommend using the "&" method, and then using $var=null to clear it. Other times, use the default method of php5. . In addition, for the transfer of large arrays in php5, it is recommended to use the "&" method, after all, it saves memory space.

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:

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

Will not unset $b, just $a.

global reference
When declaring a variable with global $var, a reference to the global variable is actually established. This means , for example, unset $var will not unset global variables.

$this
In a method of an object, $this is always a reference to the object that calls it.

//Here’s another little episode
The pointing (similar to pointer) function of the address in PHP is not implemented by the user himself, but is implemented by the Zend core. The reference in PHP uses "write The principle of "time-copy" is that unless a write operation occurs, variables or objects pointing to the same address will not be copied.

In layman terms
1: If there is the following code

$a="ABC";
$b=$a;

In fact, both $a and $b point to the same memory address at this time. It’s not that $a and $b occupy different memories

2: If you add the following code to the above code

$a="EFG";

Since the data in the memory pointed to by $a and $b needs to be rewritten, at this time the Zend core will automatically determine and automatically produce a data copy of $a for $b, and re-apply for a piece of memory for storage.

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