Analysis of global and $GLOBALS[] in PHP language Part 2

Analysis of global and $GLOBALS[] in PHP language Part 2_PHP tutorial

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Release： 2016-07-21 15:21:36

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Let’s borrow the example from the previous article:

PHP code

Copy code The code is as follows:

 
< ?php 
// Example 1 
function test_global() { 
global $var1, $var2; 
$var2 =& $var1; 
} 
function test_globals() { 
$GLOBALS['var3'] =& $GLOBALS['var1']; 
} 
$var1 = 5; 
$var2 = $var3 = 0; 
test_global(); 
print $var2 .”n”; 
test_globals(); 
print $var3 .”n”; 
?> 

The execution result is:
0
5
How could this be? Shouldn’t it be 2 5s? How could there be 1 0 and 1 5?

Well, let’s keep the above questions and analyze $GLOBALS in depth And the principle of global!
We all know that variables are actually the "codenames" of the corresponding physical memory in the code
Quoting the explanation of $GLOBALS in the PHP manual:
Global variable: $GLOBALS, note: $GLOBALS Applicable in PHP 3.0.0 and later versions.
An array consisting of all defined global variables. The variable name is the index into the array. This is a "superglobal", or can be described as an automatic global variable.
That is to say, $var1 and $GLOBALS['var1'] in the above code refer to the same variable, not 2 different variables!
Let’s analyze what global has done?
Quoting the explanation of global in the PHP manual:
If a reference is assigned to a variable declared as global inside a function, the reference is only visible inside the function. This can be avoided by using the $GLOBALS array.
We all know that the variables generated by functions in PHP are private variables of the function, so the variables generated by the global keyword certainly cannot escape this rule. Why do we say this? Look at the following code:
PHP Code

Copy code The code is as follows:

 
// Example 2 
function test( ) { 
global $a; 
unset($a); 
} 
$a = 1; 
test(); 
print $a; 
?> 

The execution result is:
1
Why is 1 output? Hasn’t $a been unset? Unset failed? PHP bug?
No, in fact, unset works. It unsets $a in the test function. You can add
print $a;
to the function test() to test!
Go back to Example 1 above and look at the code "$var2 =& $var1;" in test_global. The above is a reference assignment operation, that is, $var2 will point to the physical memory address pointed to by var1, so After Example 1 executes the test_global function, the change in the variable only has an effect locally in the function. The physical memory address pointed to by $var2 outside the function does not change, it is still itself.
At this point, you can understand why Example 1 is executed After finishing, $var2 is 0, and $var3 is 5!
So we come to the conclusion that the difference between global and $GLOBALS[] in a function is:
global generates an alias variable in the function that points to the external variable of the function, instead of the real external variable of the function. Once it changes If the address of the alias variable is pointed to, some unexpected situations will occur, such as Example 1.
$GLOBALS[] is indeed called an external variable, and it will always be consistent inside and outside the function
You can compare the following two Liezi deepens his impression:
global:

Copy code The code is as follows:

 
 function myfunction(){ 
global $bar; 
unset($bar); 
} 
$bar="someting"; 
myfunction(); 
echo $bar; 
?> 

Output: someting
$GLOBALS[]:

Copy code The code is as follows:

 
function foo() 
{ 
unset($GLOBALS['bar']); 
} 
$bar = "something"; 
foo(); 
echo $bar; 
?> 

Output: empty
After understanding according to the above ideas, will you encounter the following situation? Feeling dizzy again?

Copy code The code is as follows:

 
$a = 1; 
$b = 2; 
function Sum() 
{ 
global $a, $b; 
$b = $a + $b; 
} 
Sum() ; 
echo $b; 
?> 

The output will be "3". Global variables $a and $b are declared in the function, and all reference variables of any variable will point to the global variables.
Why is it not 2? Doesn’t it have no effect outside the function? Please note that $b is not modified by reference in the function, but the modified $b points to the value of physical memory, so the external input is 3.