Table of Contents
Variable reference
Pass-by-reference call of function
Function reference returns
Object reference
Home Backend Development PHP Tutorial Usage examples of PHP quotation character &_PHP Tutorial

Usage examples of PHP quotation character &_PHP Tutorial

Jul 13, 2016 am 10:33 AM
php Quote

The reference of PHP is to add the & symbol in front of variables, functions, objects, etc. Quoting in PHP means: different names access the same variable content. It is different from the pointer in C language. The pointer in C language stores the address where the content of the variable is stored in the memory.

Variable reference

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

<? 
$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 
?> 
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Pass-by-reference call of function

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

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 "<br />"; 
echo $b;//输出101 
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It should be noted that if test(1); is used here, an error will occur. You have to think about the reason yourself.

Function reference returns

Look at the code first:

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 
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What $a=test(); gets in this way is not actually a function reference return, which is no different from an ordinary function call. As for the reason: it's a PHP provision. 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.)

Using the above example to explain:

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, which produces the equivalent of this effect ( $a=&b;) So changing the value of $a also changes the value of $b, so after executing

$a=&test(); 
$a=5; 
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Afterwards, 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;//这里输出ABC 
echo $c->abc;//这里输出ABC 
$b->abc="DEF"; 
echo $c->abc;//这里输出DEF 
?> 
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In PHP5, object copying 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 the object by reference, but sometimes you may want to create an object A copy of the original object and hopes that changes to the original object will not affect the copy. For such purposes, PHP defines a special method called __clone.

The role of references: If the program is relatively large, there are many variables that refer to 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. At 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.

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); 
?>  
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Won’t unset $b, just $a.

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

<?php 
$var =& $GLOBALS["var"]; 
?>  
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This means that, for example, unset $var will not unset a global variable.

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

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 adopts the principle of "copy-on-write", that is, unless a write operation occurs, the pointer Variables or objects at the same address will not be copied.

In layman’s terms, if there is the following code:

$a="ABC"; 
$b=$a; 
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In fact, $a and $b point to the same memory address at this time, rather than $a and $b occupying different memories.

If you add the following code to the above code:

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

www.bkjia.comtruehttp: //www.bkjia.com/PHPjc/752524.htmlTechArticleThe reference of php is to add symbols in front of variables, functions, objects, etc. Quoting in PHP means: different names access the same variable content. It is different from pointers in C language...
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