About global variables in PHP

Hello everyone, I read the difference between the $_GLOBALS super global array and the global variable defined by global from the Internet. The difference is as follows:

1. $GLOBALS['var'] is the external global variable itself

2. global $var is a reference or pointer to an external $var variable with the same name, and is not a real assignment

So I tried it, the code is as follows:

<code><?php 
$var1 = 1; 
function test(){ 
    global $var1;                       //变量的引用
    unset($GLOBALS['var1']);            //销毁变量本身
    echo $var1;
} 

test(); 
?></code>

According to Xiaobai's thinking, this is this: Since it has been unset($GLOBALS['var1']);, the variable itself is destroyed, and the reference to the variable has no meaning, so the result cannot be output.

But I ran it, and the result was 1, so I couldn’t understand it as a novice. I would like to ask someone to explain it to me. Thank you in advance!

Reply content:

Hello everyone, I read the difference between the $_GLOBALS super global array and the global variable defined by global from the Internet. The difference is as follows:

1. $GLOBALS['var'] is the external global variable itself

2. global $var is a reference or pointer to an external $var variable with the same name, and is not a real assignment

So I tried it, the code is as follows:

<code><?php 
$var1 = 1; 
function test(){ 
    global $var1;                       //变量的引用
    unset($GLOBALS['var1']);            //销毁变量本身
    echo $var1;
} 

test(); 
?></code>

According to Xiaobai's thinking, this is this: Since it has been unset($GLOBALS['var1']);, the variable itself is destroyed, and the reference to the variable has no meaning, so the result cannot be output.

But I ran it, and the result was 1, so I couldn’t understand it as a novice. I would like to ask someone to explain it to me. Thank you in advance!

In PHP, the function inside is always a private variable. Global generates an alias variable in the function that points to the external variable of the function, rather than a simple reference or pointer to the external variable with the same name

In fact, it can be simply understood as the problem of the address pointer of the variable.

1. $GLOBALS['var1'] and the external $var1 are the same pointer, pointing to the memory address where the value 1 is stored

2. global $var1 is a copy pointer of the external $val1 pointer, also pointing to the memory address where the value 1 is stored

3. No matter $GLOBALS['var1'] or global $var1, because the value memory address pointed to is the same, the purpose of modifying the value of the external variable can be obtained.
   So:

unset($GLOBALS['var1']) The operation also destroys the external $var1
test function global $val1; unset($val1) will not destroy the external $var1
Look at the code below, you will have a clearer view

<code><?php 
$var1 = 1; 
function test(){ 
    global $var1;
    unset($var1);                     
} 
test(); 
echo $var1;// 此处输出1</code>

<code><?php 
$var1 = 1; 
function test(){ 
    unset($GLOBALS['var1']);
} 
test(); 
echo $var1;// 此处报错PHP Notice:  Undefined variable: var1</code>

<code>global $var1;
</code>

is approximately equal to

<code>$var1 = $GLOBALS['var1'];</code>