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