변수 유효영역

변수의 유효영역은 변수가 정의된 환경을 말한다. 대부분의 경우 모든 PHP 변수는 한군데의 유효영역만을 갖는다. 이 한군데의 유효영역은 include되거나 require된 파일로도 확장된다. 예를 들면:

<?php
$a 
1;
include 
'b.inc';
?>

위 예제코드에서는 include된 b.inc 스크립트안에서도 $a 변수가 사용가능하다. 하지만, 사용자-선언 함수에서는 로컬 함수 유효영역이 적용된다. 함수내에서 사용되는 모든 변수는 기본값으로 로컬 변수 유효영역 안으로 제한된다. 예를 들면:

<?php
$a 
1/* global scope */ 

function test()

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


test();
?>

위 스크립트에서 echo문이 $a의 로컬 버전을 참조하고, 이 영역 안에서 값을 지정되지 않았기 때문에 아무것도 출력되지 않는다. C에서 전역변수는 특별히 로컬 선언으로 덮어쓰지 않는이상은 자동적으로 함수안에서 사용가능하다는 점에서 C 언어와 약간 차이가 있다는 것에 주의해야 할것이다. 이런 생각으로 부주의하게 전역변수를 변경하려한다면 문제가 될것이다. PHP에서 전역변수가 함수내에서 계속적으로 사용이 된다면 함수안에서 global로 선언해야 합니다.

global 키워드

우선, global의 사용 예제입니다:

Example #1 global 사용하기

<?php
$a 
1;
$b 2;

function 
Sum()
{
    global 
$a$b;

    
$b $a $b;


Sum();
echo 
$b;
?>

위 스크립느는 "3"를 출력할것이다. $a$b를 함수내에서 global로 선언함으로써, 각 변수에 대한 모든 참조는 전역 버전으로 참조될것이다. 함순에서 조작되는 전역변수의 수는 제한이 없다.

전역 유효영역의 변수에 접근할수 있는 두번째 방법이 특별 PHP-선언 $GLOBALS 배열을 사용하는 것이다. 이전 예제코드는 다음과 같이 다시 작성할 수 있습니다:

Example #2 global 대신 $GLOBALS 사용하기

<?php
$a 
1;
$b 2;

function 
Sum()
{
    
$GLOBALS['b'] = $GLOBALS['a'] + $GLOBALS['b'];


Sum();
echo 
$b;
?>

$GLOBALS 배열은 전역변수명이 key가 되는 연관배열이고 배열의 원소 값이 그 변수의 내용이 된다. $GLOBALS이 어떻게 모든 유효영역에서 존재하는지 주의하라. 이유는 $GLOBALS이 슈퍼전역변수이기 때문이다. 아래에 슈퍼전역변수의 파워를 설명하는 예제코드를 보였다:

Example #3 자동 전역과 영역을 보여주는 예제

<?php
function test_global()
{
    
// 대부분의 예약 변수는 "자동 전역"이 아니기에,
    // 함수 내부 영역에서 사용하려면 'global'이 필요합니다.
    
global $HTTP_POST_VARS;

    echo 
$HTTP_POST_VARS['name'];
    
    
// 자동 전역은 어떠한 영역에서도 사용할 수 있고,
    // 'global'이 필요하지 않습니다. 자동 전역은
    // PHP 4.1.0부터 사용할 수 있고, HTTP_POST_VARS는
    // 배제되었습니다.
    
echo $_POST['name'];
}
?>

정적 변수 사용하기

변수 유효영역의 또 다른 중요한 기능이 static 변수이다. 정적(static) 변수는 로컬 함수 영역에서만 존재한다. 그러나 프로그램이 그 영역을 떠나지 않으면 그 값을 잃지 않는다. 다음 예제를 생각해 봅시다:

Example #4 정적 변수의 필요성을 보여주는 예제

<?php
function test()
{
    
$a 0;
    echo 
$a;
    
$a++;
}
?>

이 함수는 매번 호출될때마다 $a0으로 설정하고 "0"를 출력한다. $a++ 는 변수를 증가시키지만 함수에서 빠져나가면 $a 변수는 사라지게되므로 아무 가치가 없다. 현재 카운트 값을 잃지 않는 유용한 카운트 함수를 만들려면, $a 변수를 static으로 선언한다.

Example #5 정적 변수의 사용 예제

<?php
function test()
{
    static 
$a 0;
    echo 
$a;
    
$a++;
}
?>

처음 함수를 호출할 때만 $a가 초기화 되고, test() 함수가 호출될때마다 $a 값을 출력하고 그 값이 증가합니다.

정적 변수는 또한 재귀함수를 다루는 한 방법을 제공한다. 재귀함수는 자기 자신을 호출하는 함수를 말한다. 재귀함수는 무한히 실행될수 있기 때문에 재귀함수를 작성할때는 주의가 필요하다. 재귀를 벗어나는 방법을 반드시 갖고 있어야 한다. 다음과 같은 단순 재귀함수는 10까지 카운트한다. 정적 변수 $count는 멈춰야 할 때는 안다.

Example #6 재귀 함수에서 정적 변수

<?php
function test()
{
    static 
$count 0;

    
$count++;
    echo 
$count;
    if (
$count 10) {
        
test();
    }
    
$count--;
}
?>

Note:

정적 변수는 위 예제처럼 선언해야 합니다. 이 변수에 표현식의 결과를 할당하려 할 경우는 해석 오류를 발생합니다.

Example #7 정적 변수 선언하기

<?php
function foo() {
    static 
$int 0;          // 적합
    
static $int 1+2;        // 오류 (표현식이기에)
    
static $int sqrt(121);  // 오류 (역시 표현식이기에)

    
$int++;
    echo 
$int;
}
?>

전역 변수와 정적 변수의 참조

PHP 4를 작동하는 Zend Engine 1은 staticglobal참조를 통한 변수 변경자로 구현합니다. 예를 들어, 실제 전역 변수를 global 키워드를 사용하여 함수 영역 내부로 가져올 경우, 그 전역 변수의 참조를 생성합니다. 이로 인해 다음 예제에서 처럼 원하지 않은 동작을 할 수 있습니다:

<?php
function test_global_ref() {
    global 
$obj;
    
$obj = &new stdclass;
}

function 
test_global_noref() {
    global 
$obj;
    
$obj = new stdclass;
}

test_global_ref();
var_dump($obj);
test_global_noref();
var_dump($obj);
?>

위 예제코드를 실행하면 다음과 같은 결과가 유도된다.


NULL
object(stdClass)(0) {
}

이와 비슷한 동작이 static 절에서도 발생한다. 참조가 정적으로 저장되지 않는것이다:

<?php
function &get_instance_ref() {
    static 
$obj;

    echo 
'Static object: ';
    
var_dump($obj);
    if (!isset(
$obj)) {
        
// Assign a reference to the static variable
        
$obj = &new stdclass;
    }
    
$obj->property++;
    return 
$obj;
}

function &
get_instance_noref() {
    static 
$obj;

    echo 
'Static object: ';
    
var_dump($obj);
    if (!isset(
$obj)) {
        
// Assign the object to the static variable
        
$obj = new stdclass;
    }
    
$obj->property++;
    return 
$obj;
}

$obj1 get_instance_ref();
$still_obj1 get_instance_ref();
echo 
"\n";
$obj2 get_instance_noref();
$still_obj2 get_instance_noref();
?>

위 예제코드를 실행하면 다음과 같은 결과가 유도된다.


Static object: NULL
Static object: NULL

Static object: NULL
Static object: object(stdClass)(1) {
["property"]=>
int(1)
}

위 예제 코드는 정적 변수에 대한 참조를 지정할때, &get_instance_ref()함수가 두번째로 호출되는 때에 기억되지 않는다는 것을 보여준다.

add a note add a note

User Contributed Notes 42 notes

up
152
warhog at warhog dot net
16 years ago
Some interesting behavior (tested with PHP5), using the static-scope-keyword inside of class-methods.

<?php

class sample_class
{
  public function
func_having_static_var($x = NULL)
  {
    static
$var = 0;
    if (
$x === NULL)
    { return
$var; }
   
$var = $x;
  }
}

$a = new sample_class();
$b = new sample_class();

echo
$a->func_having_static_var()."\n";
echo
$b->func_having_static_var()."\n";
// this will output (as expected):
//  0
//  0

$a->func_having_static_var(3);

echo
$a->func_having_static_var()."\n";
echo
$b->func_having_static_var()."\n";
// this will output:
//  3
//  3
// maybe you expected:
//  3
//  0

?>

One could expect "3 0" to be outputted, as you might think that $a->func_having_static_var(3); only alters the value of the static $var of the function "in" $a - but as the name says, these are class-methods. Having an object is just a collection of properties, the functions remain at the class. So if you declare a variable as static inside a function, it's static for the whole class and all of its instances, not for each object.

Maybe it's senseless to post that.. cause if you want to have the behaviour that I expected, you can simply use a variable of the object itself:

<?php
class sample_class
{ protected $var = 0;
  function
func($x = NULL)
  {
$this->var = $x; }
}
?>

I believe that all normal-thinking people would never even try to make this work with the static-keyword, for those who try (like me), this note maybe helpfull.
up
128
dodothedreamer at gmail dot com
11 years ago
Note that unlike Java and C++, variables declared inside blocks such as loops or if's, will also be recognized and accessible outside of the block, so:
<?php
for($j=0; $j<3; $j++)
{
     if(
$j == 1)
       
$a = 4;
}
echo
$a;
?>

Would print 4.
up
70
HOSSEIN doesn&#39;t want spam at TAKI.IR
12 years ago
Please note for using global variable in child functions:

This won't work correctly...

<?php
function foo(){
   
$f_a = 'a';
   
    function
bar(){
        global
$f_a;
        echo
'"f_a" in BAR is: ' . $f_a . '<br />'// doesn't work, var is empty!
   
}
   
   
bar();
    echo
'"f_a" in FOO is: ' . $f_a . '<br />';
}
?>

This will...

<?php
function foo(){
    global
$f_a;   // <- Notice to this
   
$f_a = 'a';
   
    function
bar(){
        global
$f_a;
        echo
'"f_a" in BAR is: ' . $f_a . '<br />'// work!, var is 'a'
   
}
   
   
bar();
    echo
'"f_a" in FOO is: ' . $f_a . '<br />';
}
?>
up
30
Michael Bailey (jinxidoru at byu dot net)
18 years ago
Static variables do not hold through inheritance.  Let class A have a function Z with a static variable.  Let class B extend class A in which function Z is not overwritten.  Two static variables will be created, one for class A and one for class B.

Look at this example:

<?php
class A {
    function
Z() {
        static
$count = 0;       
       
printf("%s: %d\n", get_class($this), ++$count);
    }
}

class
B extends A {}

$a = new A();
$b = new B();
$a->Z();
$a->Z();
$b->Z();
$a->Z();
?>

This code returns:

A: 1
A: 2
B: 1
A: 3

As you can see, class A and B are using different static variables even though the same function was being used.
up
24
andrew at planetubh dot com
13 years ago
Took me longer than I expected to figure this out, and thought others might find it useful.

I created a function (safeinclude), which I use to include files; it does processing before the file is actually included (determine full path, check it exists, etc).

Problem: Because the include was occurring inside the function, all of the variables inside the included file were inheriting the variable scope of the function; since the included files may or may not require global variables that are declared else where, it creates a problem.

Most places (including here) seem to address this issue by something such as:
<?php
//declare this before include
global $myVar;
//or declare this inside the include file
$nowglobal = $GLOBALS['myVar'];
?>

But, to make this work in this situation (where a standard PHP file is included within a function, being called from another PHP script; where it is important to have access to whatever global variables there may be)... it is not practical to employ the above method for EVERY variable in every PHP file being included by 'safeinclude', nor is it practical to staticly name every possible variable in the "global $this" approach. (namely because the code is modulized, and 'safeinclude' is meant to be generic)

My solution: Thus, to make all my global variables available to the files included with my safeinclude function, I had to add the following code to my safeinclude function (before variables are used or file is included)

<?php
foreach ($GLOBALS as $key => $val) { global $$key; }
?>

Thus, complete code looks something like the following (very basic model):

<?php
function safeinclude($filename)
{
   
//This line takes all the global variables, and sets their scope within the function:
   
foreach ($GLOBALS as $key => $val) { global $$key; }
   
/* Pre-Processing here: validate filename input, determine full path
        of file, check that file exists, etc. This is obviously not
        necessary, but steps I found useful. */
   
if ($exists==true) { include("$file"); }
    return
$exists;
}
?>

In the above, 'exists' & 'file' are determined in the pre-processing. File is the full server path to the file, and exists is set to true if the file exists. This basic model can be expanded of course.  In my own, I added additional optional parameters so that I can call safeinclude to see if a file exists without actually including it (to take advantage of my path/etc preprocessing, verses just calling the file exists function).

Pretty simple approach that I could not find anywhere online; only other approach I could find was using PHP's eval().
up
8
zweibieren at yahoo dot com
7 years ago
Take to heart this hard-won rule:
        Declare AT THE TOP any variable that is to be global.
        Both at the top of the FILE
        AND at the top of any FUNCTION where it appears.

Why AT THE TOP? So it is sure to be declared before use. Otherwise a non-global version of the variable will be created and your code will fail.

Why at the top of a FUNCTION? Because otherwise the function will refer only to its local version of the variable and your code will fail.

Why at the top of the FILE? Because someday--a day that you cannot now imagine--you will want to "include" the file. And when you do, instances of the variable outside functions will not go in the global scope and your code will fail. (When the "include" is inside a calling function, variables in the included file go into the scope of the calling function.)

Example file where variable $x is used outside and inside functions:
    |<!DOCTYPE html ...>
    |<html xmlns ...>
    |    <?php global $x; ?>
    |<head>
    |    Some html headers
    |    <?php
   
|        $x = 1;
    |        function
bump_x() {
    |            global
$x;
    |           
$x += 1;
    |        }
    |   
?>
    |</head>
    |<body>
    |    More html
    |    <?php echo $x; bump_x(); ?>
    |    Yet more html.
    |</body>
</html>
up
17
larax at o2 dot pl
16 years ago
About more complex situation using global variables..

Let's say we have two files:
a.php
<?php
   
function a() {
        include(
"b.php");
    }
   
a();
?>

b.php
<?php
    $b
= "something";
    function
b() {
        global
$b;
       
$b = "something new";
    }
   
b();
    echo
$b;
?>

You could expect that this script will return "something new" but no, it will return "something". To make it working properly, you must add global keyword in $b definition, in above example it will be:

global $b;
$b = "something";
up
11
ddarjany at yahoo dot com
14 years ago
Note that if you declare a variable in a function, then set it as global in that function, its value will not be retained outside of that function.  This was tripping me up for a while so I thought it would be worth noting.

<?PHP

foo
();
echo
$a; // echoes nothing

bar();
echo
$b; //echoes "b";

function foo() {
 
$a = "a";
  global
$a;
}

function
bar() {
  global
$b;
 
$b = "b";
}

?>
up
11
php at keith tyler dot com
11 years ago
Sometimes a variable available in global scope is not accessible via the 'global' keyword or the $GLOBALS superglobal array. I have not been able to replicate it in original code, but it occurs when a script is run under PHPUnit.

PHPUnit provides a variable "$filename" that reflects the name of the file loaded on its command line. This is available in global scope, but not in object scope. For example, the following phpUnit script (call it GlobalScope.php):

<?php
print "Global scope FILENAME [$filename]\n";
class
MyTestClass extends PHPUnit_Framework_TestCase {
  function
testMyTest() {
    global
$filename;
    print
"Method scope global FILENAME [$filename]\n";
    print
"Method scope GLOBALS[FILENAME] [".$GLOBALS["filename"]."]\n";
  }
}
?>

If you run this script via "phpunit GlobalScope.php", you will get:

Global scope FILENAME [/home/ktyler/GlobalScope.php]
PHPUnit 3.4.5 by Sebastian Bergmann.

Method scope global FILENAME []
Method scope GLOBALS[FILENAME] []
.

You have to -- strange as it seems -- do the following:

<?php
$GLOBALS
["filename"]=$filename;
print
"Global scope FILENAME [$filename]\n";
class
MyTestClass extends PHPUnit_Framework_TestCase {
  function
testMyTest() {
    global
$filename;
    print
"Method scope global FILENAME [$filename]\n";
    print
"Method scope GLOBALS[FILENAME] [".$GLOBALS["filename"]."]\n";
  }
}
?>

By doing this, both "global" and $GLOBALS work!

I don't know what it is that PHPUnit does (I know it uses Reflection) that causes a globally available variable to be implicitly unavailable via "global" or $GLOBALS. But there it is.
up
9
dexen dot devries at gmail dot com
5 years ago
If you have a static variable in a method of a class, all DIRECT instances of that class share that one static variable.

However if you create a derived class, all DIRECT instances of that derived class will share one, but DISTINCT, copy of that static variable in method.

To put it the other way around, a static variable in a method is bound to a class (not to instance). Each subclass has own copy of that variable, to be shared among its instances.

To put it yet another way around, when you create a derived class, it 'seems  to' create a copy of methods from the base class, and thusly create copy of the static variables in those methods.

Tested with PHP 7.0.16.

<?php

require 'libs.php';
require
'setup.php';

class
Base {
    function
test($delta = 0) {
        static
$v = 0;
       
$v += $delta;
        return
$v;
    }
}

class
Derived extends Base {}

$base1 = new Base();
$base2 = new Base();
$derived1 = new Derived();
$derived2 = new Derived();

$base1->test(3);
$base2->test(4);
$derived1->test(5);
$derived2->test(6);

var_dump([ $base1->test(), $base2->test(), $derived1->test(), $derived2->test() ]);

# => array(4) { [0]=> int(7) [1]=> int(7) [2]=> int(11) [3]=> int(11) }

# $base1 and $base2 share one copy of static variable $v
# derived1 and $derived2 share another copy of static variable $v
up
7
danno at wpi dot edu
21 years ago
WARNING!  If you create a local variable in a function and then within that function assign it to a global variable by reference the object will be destroyed when the function exits and the global var will contain NOTHING!  This main sound obvious but it can be quite tricky you have a large script (like a phpgtk-based gui app ;-) ).

example:

<?php
function foo ()
{
   global
$testvar;

  
$localvar = new Object ();
  
$testvar = &$localvar;
}

foo ();
print_r ($testvar);   // produces NOTHING!!!!
?>

hope this helps someone before they lose all their hair
up
9
Anonymous
10 years ago
It will be obvious for most of you: changing value of a static in one instance changes value in all instances.

<?php

   
class example {
        public static
$s = 'unchanged';
       
        public function
set() {
           
$this::$s = 'changed';
        }
    }

   
$o = new example;
   
$p = new example;

   
$o->set();

    print
"$o static: {$o::$i}\n$p static: {$p::$i}";

?>

Output will be:

$o static: changed
$p static: changed
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5
pogregoire##live.fr
6 years ago
writing : global $var; is exactely the samething that writing : $var =& $GLOBALS['var'];
It creates a reference on $GLOBALS['var'];

<?php
$var
=1;
function
teste_global(){
    global
$var;
    for (
$var=0; $var<5; $var++){

    }
}

teste_global();
var_dump($var);// return : int(5).
?>
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3
shaman_master at list dot ru
2 years ago
Cycle not create inner scope:
<?php
foreach (range(1, 3) as $step) {
    echo
sprintf('%d: %s  ', $step, isset($a) ? 'yes' : 'no');
    if (! isset(
$a)) {
       
$a = 1;
    }
}
// 1: no 2: yes 3: yes
?>
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5
Randolpho
18 years ago
More on static variables:

A static variable does not retain it's value after the script's execution. Don't count on it being available from one page request to the next; you'll have to use a database for that.

Second, here's a good pattern to use for declaring a static variable based on some complex logic:

<?php
 
function buildStaticVariable()
  {
     
$foo = null;
     
// some complex expression or set of
      // expressions/statements to build
      // the return variable.
     
return $foo;
  }

  function
functionWhichUsesStaticVar()
  {
      static
$foo = null;
      if(
$foo === null) $foo = buildStaticVariable();
     
// the rest of your code goes here.
 
}
?>

Using such a pattern allows you to separate the code that creates your default static variable value from the function that uses it. Easier to maintain code is good. :)
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3
gried at NOSPAM dot nsys dot by
6 years ago
In fact all variables represent pointers that hold address of memory area with data that was assigned to this variable. When you assign some variable value by reference you in fact write address of source variable to recepient variable. Same happens when you declare some variable as global in function, it receives same address as global variable outside of function. If you consider forementioned explanation it's obvious that mixing usage of same variable declared with keyword global and via superglobal array at the same time is very bad idea. In some cases they can point to different memory areas, giving you headache. Consider code below:

<?php

error_reporting
(E_ALL);

$GLOB = 0;

function
test_references() {
    global
$GLOB; // get reference to global variable using keyword global, at this point local variable $GLOB points to same address as global variable $GLOB
   
$test = 1; // declare some local var
   
$GLOBALS['GLOB'] = &$test; // make global variable reference to this local variable using superglobal array, at this point global variable $GLOB points to new memory address, same as local variable $test

   
$GLOB = 2; // set new value to global variable via earlier set local representation, write to old address

   
echo "Value of global variable (via local representation set by keyword global): $GLOB <hr>";
   
// check global variable via local representation => 2 (OK, got value that was just written to it, cause old address was used to get value)

   
echo "Value of global variable (via superglobal array GLOBALS): $GLOBALS[GLOB] <hr>";
   
// check global variable using superglobal array => 1 (got value of local variable $test, new address was used)
   
   
echo "Value ol local variable \$test: $test <hr>";
   
// check local variable that was linked with global using superglobal array => 1 (its value was not affected)
   
   
global $GLOB; // update reference to global variable using keyword global, at this point we update address that held in local variable $GLOB and it gets same address as local variable $test
   
echo "Value of global variable (via updated local representation set by keyword global): $GLOB <hr>";
   
// check global variable via local representation => 1 (also value of local variable $test, new address was used)
}

test_references();
echo
"Value of global variable outside of function: $GLOB <hr>";
// check global variable outside function => 1 (equal to value of local variable $test from function, global variable also points to new address)
?>
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4
Jonathan Kenigson
8 years ago
Just a note about static properties declared at class level:

class Test_Class {
  static $a = 0;
  public function ReturnVar(){
    return $this->a;
  }
  }
  $b = new Test_Class();
  echo $b->ReturnVar();

Will not output "0"  because $a is declared static. Changing "static" to "public" or "private" will produce the output "0".
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3
kouber at php dot net
17 years ago
If you need all your global variables available in a function, you can use this:

<?php
function foo() {
 
extract($GLOBALS);
 
// here you have all global variables

}
?>
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8
Stephen Dewey
13 years ago
For nested functions:

This is probably obvious to most people, but global always refers to the variable in the global (top level) variable of that name, not just a variable in a higher-level scope. So this will not work:
<?php

// $var1 is not declared in the global scope

function a($var1){

    function
b(){
        global
$var1;
        echo
$var1; // there is no var1 in the global scope so nothing to echo
   
   
}

   
b();
}

a('hello');

?>
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3
eduardo dot ferron at zeion dot net
11 years ago
There're times when global variables comes in handy, like universal read only resources you just need to create once in your application and share to the rest of your scripts. But it may become quite hard to track with "variables".
up
2
moraesdno at gmail dot com
12 years ago
Use the superglobal array $GLOBALS is faster than the global keyword. See:

<?php
//Using the keyword global
$a=1;
$b=2;
function
sum() {
    global
$a, $b;
   
$a += $b;
}

$t = microtime(true);
for(
$i=0; $i<1000; $i++) {
    
sum();
}
echo
microtime(true)-$t;
echo
" -- ".$a."<br>";

//Using the superglobal array
$a=1;
$b=2;
function
sum2() {
   
$GLOBALS['a'] += $GLOBALS['b'];
}

 
$t = microtime(true);
for(
$i=0; $i<1000; $i++) {
    
sum2();
}
echo
microtime(true)-$t;
echo
" -- ".$a."<br>";
?>
up
2
jakub dot lopuszanski at nasza-klasa dot pl
12 years ago
If you use __autoload function to load classes' definitons, beware that "static local variables are resolved at compile time" (whatever it really means) and the order in which autoloads occur may impact the semantic.

For example if you have:
<?php
class Singleton{
  static public function
get_instance(){
     static
$instance = null;
     if(
$instance === null){
       
$instance = new static();
     }
     return
$instance;
  }
}
?>

and two separate files A.php and B.php:
class A extends Singleton{}
class B extends A{}

then depending on the order in which you access those two classes, and consequently, the order in which __autoload includes them, you can get strange results of calling B::get_instance() and A::get_instance().

It seems that static local variables are alocated in as many copies as there are classes that inherit a method at the time of inclusion of parsing Singleton.
up
1
simon dot barotte at gmail dot com
5 years ago
To be vigilant, unlike Java or C++, variables declared inside blocks such as loops (for, while,...) or if's, will also be recognized and accessible outside of the block, the only valid block is the BLOCK function so:

<?php
for($j=0; $j<5; $j++)
{
     if(
$j == 1){
       
$a = 6;
     }
}

echo
$a;
?>

Would print 6.
up
1
info AT SyPlex DOT net
18 years ago
Some times you need to access the same static in more than one function. There is an easy way to solve this problem:

<?php
 
// We need a way to get a reference of our static
 
function &getStatic() {
    static
$staticVar;
    return
$staticVar;
  }

 
// Now we can access the static in any method by using it's reference
 
function fooCount() {
   
$ref2static = & getStatic();
    echo
$ref2static++;
  }

 
fooCount(); // 0
 
fooCount(); // 1
 
fooCount(); // 2
?>
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-1
admin at shenjinpeng dot cn
1 year ago
<?php

function foo(){
    static
$a= 0;
   
$bar = function(){
        global
$a ;
        return ++
$a;
    };
   
    return
$bar();
}

echo
foo(); // 1
echo foo(); // 2

function foo1(){
   
$bar = function(){
        static
$a = 0;
        return ++
$a;
    };
   
    return
$bar();
}

echo
foo1() ; // 1
echo foo1() ; // 1

function foo2(){
    static
$a = 0;
   
$bar = function() use (&$a){
        return ++
$a;
    };
   
    return
$bar();
}

echo
foo2() ; // 1
echo foo2() ; // 2

function foo3($func){
    static
$a = 0;
    return
$func($a);
}

echo
foo3(fn(&$a) => ++$a ); // 1
echo foo3(fn(&$a) => ++$a ); // 2

?>
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-1
2505036090 at qq dot com
1 year ago
About assignment of object:
```php
<?php
function aaa() {
    global
$obj;
   
$cl = new stdclass;
   
$obj = &$cl; //  assign By Ref
   
$obj->name = 'jack';
   
$obj = NULL; // the change of $obj leads to the change of $cl
   
var_dump($cl); // output: NULL
}
aaa();
var_dump($obj); // output: NULL

function bbb() {
    global
$obj;
   
$cl = new stdclass;
   
$obj = $cl; // general assignment of object
   
$obj->name = 'jack';
   
$obj = NULL; // the value of $obj changes to NULL, but the value of $cl  is still original;
   
var_dump($cl); //output:  object(stdClass)#2 (1) { ["name"]=> string(4) "jack" }
}
bbb();
var_dump($obj); // output: NULL
?>
```
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-1
atesin () 6m4i1 ! c0m
1 year ago
i hadn't found info on this so i had to make some tests...

inside a function, when globalling a variable that doesn't exist in the parent scope, it will set to null... i mean...

<?php

function f()
{
  global
$g;
 
var_dump($g); // prints 'NULL' with no log message, means if $g is unset globally then here IS SET to NULL (*)
 
var_dump($l); // prints 'NULL' and logs the 'undefined variable l' notice in this line
 
  // (*) notice until the variable $g IS SET to null here, isset($g) will still return false
  //     because for isset() the var has to be set and not null (see doc for details)...
  //     maybe the only way to find if a variable is actually unset is throught the log warnings
}

var_dump($g); // prints 'NULL' and logs the 'undefined variable g' in this line
var_dump($g); // same as above but in this line, means the variable is still unset
f();

?>

.
knowing this could be useful in a case like mine... consider something like this...

<?php

//--- account.php ---

$user = 'john doe';

//--- main.php ---

function welcome()
{
  global
$user;
  echo
"welcome $user"; // if $user is not set in parent scope it just prints "welcome "

  // the right way afaik
 
if ( isset($user) )
    echo
"welcome $user";
}

if (
login_ok() )
 
incude 'account.php';

welcome();

?>
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-1
Duke dot Bouvier at NOPSAM dot Gmail dot com
1 year ago
The recursion example has some excess code:

"$count--" is superfluous.
"$count++" can be built into the loop (but be sure to use pre-increment not a post-incremenet

Thus:

<?php
function test()
{
    static
$count = 1;

    echo
$count." ";

    if (++
$count < 10) {
       
test();
    }
}
?>

Though with a static, all you are really doing is creating a loop but with lots of function calls pushed onto the stack.

A great things about recursive functions is that you don't actually have to have any static data.  You simply supply an argument to pass the value (and can use a default value for the argument for your initialisation if desired.)

<?php
function test($count = 1)
{
    echo
$count." ";

    if (
$count < 10) {
       
test(++$count);
    }

}
test()
?>
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-1
jtunaley at gmail dot com
1 year ago
It's worth noting that block statements without a control structure also don't affect variable scope.

<?php
{
   
$a = 4;
}
echo
$a; // Outputs "4"
up
0
jake dot tunaley at berkeleyit dot com
3 years ago
Beware of using $this in anonymous functions assigned to a static variable.

<?php
class Foo {
    public function
bar() {
        static
$anonymous = null;
        if (
$anonymous === null) {
           
// Expression is not allowed as static initializer workaround
           
$anonymous = function () {
                return
$this;
            };
        }
        return
$anonymous();
    }
}

$a = new Foo();
$b = new Foo();
var_dump($a->bar() === $a); // True
var_dump($b->bar() === $a); // Also true
?>

In a static anonymous function, $this will be the value of whatever object instance that method was called on first.

To get the behaviour you're probably expecting, you need to pass the $this context into the function.

<?php
class Foo {
    public function
bar() {
        static
$anonymous = null;
        if (
$anonymous === null) {
           
// Expression is not allowed as static initializer workaround
           
$anonymous = function (self $thisObj) {
                return
$thisObj;
            };
        }
        return
$anonymous($this);
    }
}

$a = new Foo();
$b = new Foo();
var_dump($a->bar() === $a); // True
var_dump($b->bar() === $a); // False
?>
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1
Anonymous
14 years ago
I was pondering a little something regarding caching classes within a function in order to prevent the need to initiate them multiple times and not clutter the caching function's class properties with more values.

I came here because I remembered something about references being lost. So I made a test to see if I could pull what I wanted to off anyway. Here's and example of how to get around the references lost issue. I hope it is helpful to someone else!

<?php
class test1{}
class
test2{}
class
test3{}

function
cache( $class )
{
    static
$loaders = array();
   
   
$loaders[ $class ] = new $class();

   
var_dump( $loaders );
}
print
'<pre>';
cache( 'test1' );
cache( 'test2' );
cache( 'test3' );

?>
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-2
iphstich at gmail dot com
2 years ago
It seems that static variables defined within a class method or static function are stored separately for the class and each extended class of it.

It's sort of like, every time you extend a class, functions with static variables are re-writen, and the static variable re-created. Even if the function is defined as final!

<?php

class A {
    function
x () { static $x = 0; return ++$x; }
    static function
y () { static $y = 0; return ++$y; }
    final static function
z () { static $z = 0; return ++$z; }
}

class
B extends A { }
class
C extends B { }

// As methods, A and B have different values for x
$a = new A();
echo
$a->x(); // 1
$b = new B();
echo
$b->x(); // 1
echo $b->x(); // 2

// And different instances of C will have the same values for x
$c1 = new C();
$c2 = new C();
$c3 = new C();
echo
$c1->x(); // 1
echo $c2->x(); // 2
echo $c3->x(); // 3

// The same behavior applies to static functions
echo A::y(); // 1
echo B::y(); // 1
echo B::y(); // 2
echo C::y(); // 1
echo C::y(); // 2
echo C::y(); // 3

// The same behavior still applies to FINAL static functions
echo A::z(); // 1
echo B::z(); // 1
echo B::z(); // 2
echo C::z(); // 1
echo C::z(); // 2
echo C::z(); // 3
?>

Tested with version 7.0.13
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0
nino dot skopac at gmail dot com
6 years ago
Interesting behavior in PHP 5.6.12 and PHP 7 RC3:

<?php
class Foo {
    public function
Bar() {
        static
$var = 0;
       
        return ++
$var;
    }
}

$Foo_instance = new Foo;

print
$Foo_instance->Bar(); // prints 1
print PHP_EOL;

unset(
$Foo_instance);

$Foo_instance2 = new Foo;

print
$Foo_instance2->Bar(); // prints 2
print PHP_EOL;
?>

How can a 2 be printed, since we unseted the whole instance before?

Consider a similar example:

<?php
class Foo {
    public static
$var = 0;
   
    public static function
Bar() {
        return ++
self::$var;
    }
}

$Foo_instance = new Foo;

print
$Foo_instance->Bar(); // prints 1
print PHP_EOL;

unset(
$Foo_instance);

$Foo_instance2 = new Foo;

print
$Foo_instance2->Bar(); // prints 2
print PHP_EOL;
?>

No idea why is this happening.
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0
alan
16 years ago
Using the global keyword inside a function to define a variable is essentially the same as passing the variable by reference as a parameter:

<?php
somefunction
(){
   global
$var;
}
?>

is the same as:

<?php
somefunction
(& $a) {

}
?>

The advantage to using the keyword is if you have a long list of variables  needed by the function - you dont have to pass them every time you call the function.
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0
jameslee at cs dot nmt dot edu
17 years ago
It should be noted that a static variable inside a method is static across all instances of that class, i.e., all objects of that class share the same static variable.  For example the code:

<?php
class test {
    function
z() {
        static
$n = 0;
       
$n++;
        return
$n;
    }
}

$a =& new test();
$b =& new test();
print
$a->z();  // prints 1, as it should
print $b->z();  // prints 2 because $a and $b have the same $n
?>

somewhat unexpectedly prints:
1
2
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-1
Ray.Paseur often uses Gmail
8 years ago
Variable "Visibility" in PHP Object Oriented Programming is documented here:
http://php.net/manual/en/language.oop5.visibility.php
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-1
pedro at worcel dot com
12 years ago
Another way of working with a large ammount of global variables could be the following.

<?php

$var
= "3";
$smarty = new Smarty();

function
headers_set_404() {
extract($globals);

echo
$var . "<br />";
print_r($smarty);

return;

}

?>

Regards,
Droope
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-1
Hayley Watson
4 years ago
static variables are implicitly initialised to NULL if no explicit initialisation is made.

<?php
function foo()
{
  static
$v;
  echo
gettype($v);
}

foo();
?>

will echo NULL without complaining that $v is undefined.

In short: "static $v;" is equivalent to "static $v = null;".
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-2
nullhility at gmail dot com
13 years ago
Like functions, if you declare a variable in a class, then set it as global in that class, its value will not be retained outside of that class either.

<?php
class global_reference
{
    public
$val;
   
    public function
__construct () {
        global
$var;
       
$this->val = $var;
    }
   
    public function
dump_it ()
    {
       
debug_zval_dump($this->val);
    }
   
    public function
type_cast ()
    {
       
$this->val = (int) $this->val;
    }
}
$var = "x";
$obj = new global_reference();
$obj->dump_it();
$obj->type_cast();
echo
"after change ";
$obj->dump_it();
echo
"original $var\n";
?>

The work-around is of course changing the assignment in the constructor to a reference assignment as such:

<?php
   
//....
       
$this->val = &var;
   
//....
?>

If the global you're setting is an object then no reference is necessary because of the way PHP deals with objects. If you don't want to reference to the same object however you can use the clone keyword.

<?php
//...
   
global $Obj;
   
$this->obj_copy = clone $Obj;
//...
?>

[EDIT BY danbrown AT php DOT net:  Merged all thoughts and notes by this author into a single note.]
up
-2
Ganlv
6 years ago
<?php
$var
= 1;
function
foo() {
   
$var = &$GLOBALS['var'];
   
var_dump($var);
}
function
bar() {
    global
$var; // they are the same.
   
var_dump($var);
}
foo();
bar();
var_dump($var);
?>

In a function, 'global $var;' is to declare a local variant, and the local $var has the same reference to the global $var.

<?php
$var
= 1;
function
foo() {
    global
$var;
    unset(
$var);               // unset local $a, the global $a is still there.
   
var_dump($var);            // Undefined variable: var
   
var_dump($GLOBALS['var']); // this is ok.
}
foo();
var_dump($var);                // this is ok.
?>

<?php
$var
= 1;
function
bar() {
    global
$var;
    unset(
$GLOBALS['var']);    // unset global $a, the local $a is still here.
   
var_dump($var);            // this is ok.
   
var_dump($GLOBALS['var']); // Undefined index: var
}
foo();
var_dump($var);                // Undefined variable: var
?>

'unset($var);' is like 'var = NULL;'(var is a pointer) in the C language, instead of 'free(var);'
up
-1
ketansheladiya1
2 years ago
As per PHP standards
The new operator returns a reference automatically, so assigning the result of new by reference is not allowed as of PHP 7.0.0, results in an E_DEPRECATED message as of PHP 5.3.0, and an E_STRICT message in earlier versions.
https://www.php.net/manual/en/language.operators.assignment.php

so you can use following code to check

<?php
function &get_instance_ref() {
    static
$obj;

    echo
'Static object: ';
   
var_dump($obj);
    if (!isset(
$obj)) {
       
// Assign a reference to the static variable
       
$obj = new stdclass;
    }
   
$obj->property++;
    return
$obj;
}

function &
get_instance_noref() {
    static
$obj;

    echo
'Static object: ';
   
var_dump($obj);
    if (!isset(
$obj)) {
       
// Assign the object to the static variable
       
$obj = new stdclass;
    }
   
$obj->property++;
    return
$obj;
}

$obj1 = get_instance_ref();
$still_obj1 = get_instance_ref();
echo
"\n";
$obj2 = get_instance_noref();
$still_obj2 = get_instance_noref();
?>
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-2
Semyon Naitur
5 years ago
function f(){
    global $a; // global $a is declared, local reference is created
    $a = 'a';  // global $a is set
    unset($a); // local reference is unset, global $a remains set
    $a = 'b';  // local $a is declared and set
}
f();
echo $a; // prints 'a'

function f(){
    global $a; // global $a is declared, local reference is created
    $a = 'a';  // global $a is set
    unset($a); // local reference $a is unset, global var $a remains set
    global $a; // local reference is created again
    $a .= 'b';
}
f();
echo $a; // prints 'ab'
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