Detailed tutorial on reference-to-value transfer of PHP variables

Detailed tutorial on reference-to-value transfer of PHP variables_PHP tutorial

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Release： 2016-07-13 16:56:17

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

Reference of the variable

PHP’s references allow you to use two variables to point to the same content

代码如下	复制代码
$a="ABC"; $b =&$a; echo $a;//这里输出:ABC echo $b;//这里输出:ABC $b="EFG"; echo $a;//这里$a的值变为EFG 所以输出EFG echo $b;//这里输出EFG ?>

Function call by address
I won’t go into details about call by address. The code is given directly below

The code is as follows

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代码如下	复制代码
function test(&$a) { $a=$a+100; } $b=1; echo $b;//输出１ test($b); //这里$b传递给函数的其实是$b的变量内容所处的内存地址，通过在函数里改变$a的值　就可以改变$b的值了 echo " "; echo $b;//输出101

function test(&$a)

{

$a=$a+100;

}
$b=1;

echo $b;//output 1

代码如下

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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

test($b); //What $b is passed to the function here is actually the memory address where the variable content of $b is located. By changing the value of $a in the function, the value of $b can be changed

echo "
";
echo $b;//Output 101

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

Function reference returns
Let’s look at the code first

The code is as follows

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

return $b;

}

$a=test();//This statement will output that the value of $b is 1

代码如下	复制代码
class a{ var $abc="ABC"; } $b=new a; $c=$b; echo $b->abc;//这里输出ABC echo $c->abc;//这里输出ABC $b->abc="DEF"; echo $c->abc;//这里输出DEF ?>

$a=5; $a=test();//This statement will output that the value of $b is 2 $a=&test();//This statement will output that the value of $b is 3 $a=5; $a=test();//This statement will output that the value of $b is 6

Explain below: In this way, $a=test(); actually does not get a reference return from the function. It is no different from an ordinary function call. As for the reason: This is the regulation of PHP PHP stipulates that what is obtained through $a=&test(); is the reference return of the 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 it is Calling a function using $a=test() only assigns the value of the function to $a, and any changes to $a will not affect $b in the function. When calling a 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 $a variable to the same place That is to say, the effect equivalent to this is produced ($a=&b;), so changing the value of $a also changes the value of $b, so after executing $a=&test(); $a=5; From now on, the value of $b becomes 5 Static variables are used here to let everyone understand the reference return of functions. In fact, the reference return of functions is mostly used in objects Object reference

The code is as follows	Copy code
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 effect of running in PHP5
In PHP5, object copying is achieved by reference. 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 the 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 quotation
If the program is relatively large, there are many variables referencing the same object, and you want to manually clear the object after using it, I personally recommend using the "&" method, and then using $var=null to clear it. Otherwise, use the default of php5 Method. In addition, for transferring large arrays in php5, it is recommended to use the "&" method, after all, it saves memory space.

Unquote
When you unset a reference, you just break the binding between the variable name and the variable's contents. This does not mean that the variable contents are destroyed. For example:

The code is as follows

代码如下	复制代码
$a = 1; $b =& $a; unset ($a); ?>

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$a = 1;
$b =& $a;

unset ($a);

代码如下	复制代码
$var =& $GLOBALS["var"]; ?>

Won’t unset $b, just $a.

global reference

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

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

The code is as follows

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

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

$this

//Another little episode below

The address pointing (similar to a pointer) function in PHP is not implemented by the user himself, but is implemented by the Zend core. The reference in PHP adopts the principle of "copy on write", that is, unless a write operation occurs, it points to the same address. Variables or objects will not be copied.

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 generate a data copy of $a for $b, and re-apply for a piece of memory for storage

http://www.bkjia.com/PHPjc/631609.htmlwww.bkjia.comtruehttp: //www.bkjia.com/PHPjc/631609.htmlTechArticleThe meaning of quoting in PHP is: different names access the same variable content. The same as the pointer in C language There is a difference. Pointers in C language store the contents of variables in memory...