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PHP variable scope, reference, object reference, transfer_PHP tutorial

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Release: 2016-07-13 17:14:11
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This article introduces the variable scope, reference, object reference, and transfer of PHP. Friends in need can refer to it.

Variable scope
The scope of a variable is the context in which it is defined (that is, its effective scope). Most PHP variables have only a single scope. This single scope span also includes files introduced by include and require. For example:

The code is as follows Copy code
代码如下 复制代码
$a = 1;
include 'b.inc';
?>
$a = 1;

include 'b.inc';

?>
 代码如下 复制代码

$a = 1; /* global scope */

function Test()
{
echo $a; /* reference to local scope variable */
}

Test();
?>

This variable $a will take effect in the included file b.inc. However, in user-defined functions, a local function scope will be introduced. Any variables used inside a function will be restricted to the local function scope by default. For example:

The code is as follows Copy code

$a = 1; /* global scope */

function Test()

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

echo $a; /* reference to local scope variable */

}

Test();
 代码如下 复制代码
function test(&$a)
{
$a=$a+100;
}
$b=1;
echo $b;//输出1
echo "
";
test($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
?>
?> This script will produce no output because the echo statement refers to a local version of the variable $a, and it is not assigned a value within this scope. You may notice that PHP's global variables are a little different from C language. In C language, global variables automatically take effect in functions unless overridden by local variables. This may cause some problems, someone may accidentally change a global variable. Global variables in PHP must be declared global when used in functions. The meaning of quoting in PHP is: different names access the same variable content. It is different from pointers in C language. The pointer in C language stores the address where the content of the variable is stored in memory Variable reference
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$a="ABC";<🎜> $b =&$a;<🎜> echo $a;echo "rn";//Output here: ABC<🎜> echo $b;echo "rn";//Output here: ABC<🎜> $b="EFG";<🎜> echo $a;echo "rn";//The value of $a here becomes EFG, so EFG<🎜> is output echo $b;echo "rn";//Output EFG<🎜> here ?>
Pass-by-reference call of function I won’t go into details about call by address. The code is given directly below
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"; test($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 Copy code
 代码如下 复制代码

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
echo "
";
$a=5;
$a=test();//这条语句会输出 $b的值 为2
echo "
";
$a=&test();//这条语句会输出 $b的值 为3
echo "
";
$a=5;
$a=test();//这条语句会输出 $b的值 为6
?>

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

echo "
";
 代码如下 复制代码

class a{
public $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 6
?>


Explain below: In this way, $a=test(); actually does not get a function reference return, which is no different from an ordinary function call. As for the reason: this is a PHP regulation.
 代码如下 复制代码

class a{
public $abc="ABC";
}
$b=new a;
$c=$b;
$d = clone$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
echo $b->abc;//这里输出DEF
$d->abc="111";
echo $d->abc;//这里输出111
echo $b->abc;//这里输出DEF,说明clone后的副本$d对象的改变并没有影响$b对象
?>

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 not understand it for a long time To explain using the above example, 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 the function. In fact, the reference return of the function is mostly used in objects Object reference
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class a{<🎜> public $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 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 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 . For example, the following example
The code is as follows Copy code
class a{<🎜> public $abc="ABC";<🎜> }<🎜> $b=new a;<🎜> $c=$b;<🎜> $d = clone$b;<🎜> echo $b->abc;//Output ABC here echo $c->abc;//Output ABC here $b->abc="DEF"; echo $c->abc;//Output DEF here echo $b->abc;//Output DEF here $d->abc="111"; echo $d->abc;//The output here is 111 echo $b->abc;//DEF is output here, indicating that changes to the $d object after cloning do not affect the $b object ?>

The role of quotation
If the program 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. Otherwise, use the default of php5 Way. In addition, in php5
For transferring large arrays, 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 ($a);
var_dump($a);//这里输出null
var_dump($b);//这里输出int 1
?>
Copy code

$a = 1;
$b = &$a;
unset ($a);
var_dump($a);//null is output here
var_dump($b);//Int 1 is output here
?>


Won’t unset $b, just $a.

global quote
When you declare a variable with global $var you actually create a reference to the global variable. That is the same as doing this:
$var =& $GLOBALS["var"];
?>
This means that, for example, unset $var will not unset a global variable.
$this
In an object method, $this is always a reference to the object that calls it.

//Another little interlude 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", which means that unless a write operation occurs, it points to the same address. The variable or object is

Will not be copied.

In layman’s terms

1: If there is the following code

$a="ABC";

$b=$a;

In fact, at this time, $a and $b both point to the same memory address, rather than $a and $b occupying different memories

2: If you add the following code to the above code
 代码如下 复制代码



my favorite moviesite


echo "my favorite movie site is :";
echo $_GET['favmovie'];
echo "
";
$movierate= 5 ;
echo "my favorite movie rating for this movie is ";
echo  $movierate;
?>

$a="ABC";

$b=$a;

$a="EFG";

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 and automatically produce a data copy of $a for $b and re-apply for a piece of memory for storage

Learn variable passing in php!

1 Pass variables using url

Example

The code is as follows
 代码如下 复制代码


 
  find my favorite movie
 
 
  //$myfavmovie=urlencode("life of brian");
echo "";
 echo "click here to see more information about my favorite movie!";
 echo "
";
 ?>
 

Copy code
my favorite moviesite echo "my favorite movie site is :"; <🎜> echo $_GET['favmovie']; <🎜> echo "
"; $movierate= 5 ; echo "my favorite movie rating for this movie is "; echo $movierate; ?> Save this file as movie1.php. Use $_GET['favmovie'] to receive the variables passed by the url! Write to another file and save as moviesite.php
The code is as follows Copy code
find my favorite movie //$myfavmovie=urlencode("life of brian");<🎜> echo ""; echo "click here to see more information about my favorite movie!"; echo ""; ?>

www.bkjia.comtruehttp: //www.bkjia.com/PHPjc/629033.htmlTechArticleThis article introduces the variable scope, reference, object reference, and transfer of PHP. Friends in need can refer to it. Variable scope The scope of a variable is the context in which it is defined (that is, its...

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