Analysis of PHP variable scope and address reference issues

function test()
{
static $b=0;//Declare static variables. If declared outside the function, they will not be used inside the function
$b=$b+1;
echo $b;
}
test();//This statement will output the value of $b as 1
test();//This statement will output the value of $b as 2

Note :static $b=0 This assignment operation will only be executed when the variable is initialized for the first time. Attachment A: Static members and static methods in a class almost always use the class name or self or parent plus: :xxx when calling. Their scope is the same as this, but their declaration is outside the method. Appendix B: The scope in js is: use var aa=‘xxx’; what is declared outside the function is the global variable (regardless of whether it has the modifier var or not). Local variables are declared using var inside a function, and global variables are declared without var. Appendix C: About PHP references PHP reference: Add &. Reference in php before a variable, function or object is to access the contents of the same variable with different names.

1. Variable reference:

$a="ABC";
$b =&$a;
echo $a;//Output here: ABC
echo $b;//Output here: ABC
$ b="EFG";
echo $a;//The value of $a here becomes EFG, so EFG is output
echo $b;//EFG is output here

2. Call by address of function

function test(&$a)
{
$a=$a+100;
} // bbs.it-home.org
$b=1;
echo $b ;//Output 1
test($b); //What $b is passed to the function here is actually the memory address where the variable content of $b is located. The value of $b can be changed by changing the value of $a in the function.
echo "";
echo $b;//Output 101

3. Function reference return

function &test()
{
static $b=0;//Declare a static variable
$b=$b+1;
echo $b;
return $b;
 }
$a=test();//This statement will output the value of $b as 1
$a=5;
$a=test();//This statement will output the value of $b as 2
 $a=&test();//This statement will output the value of $b as 3
$a=5;
$a=test();//This statement will output the value of $b as 6

Analysis: What you get using $a=test() is not actually a reference return from the function. Just copy the function's return value to $a without affecting $b. This call is no different from an ordinary call. Php stipulates: $a=&test() method is the reference return of the function. He pointed the memory address of the $b variable and the memory address of the $a variable to the same place. That is equivalent to $a=&$b;

4. Cancel reference

$a = 1;
$b =& $a;
unset ($a);
echo $b;

Analysis: Unsetting a reference only cancels the binding between the variable name and the variable's content. It does not mean that the content is destroyed, and its value still exists.

5. Global quote: When you declare a variable using global $var, you actually create a reference to the global variable.

Global $val  $var=&$GLOBALS['var'] ;

6. Object reference: In the method of the object, the object called by $this is the reference that calls it. Note: The pointing of addresses in PHP is not implemented by the user himself, but through the zend core. PHP references adopt the principle of "write copy", that is, unless a write operation occurs, the variable pointing to the same address or Objects will not be copied. example:

$a = 1; $b =$a;

$a and $b both point to the same memory address, it is not that $a and $b occupy different memories. If you now execute the sentence $a="dsd": the memory data pointed to by $a and $b need to be rewritten, at this time the zend core will automatically judge. Automatically generate a data copy of $a for $b, and re-apply a piece of memory for storage.

That’s all about PHP variable scope and PHP references. I hope it will be helpful to everyone.