If a global variable in PHP references the address of a local variable PHPcode $ B & quot; dasdf & quot; functionabc () {$ a & quot; abc & quot ;; global $ B; $ B & amp; $ a; echo $ B. & quot; inner; & quo if a global variable of PHP references the address of a local variable
PHP code
$b = "dasdf"; function abc(){ $a = "abc"; global $b; $b = &$a; echo $b."inner;"; } abc(); echo $b.";";
When the function outputs B internally, it is abc, but after leaving the function, the output B is restored to the initial value of dasdf. what is the problem? It is reasonable to say that the address of a local variable is referenced. after the local variable is destroyed, the value of B should be blank.
------ Solution --------------------
In the manual:
Http://www.php.net/manual/zh/language.references.whatdo.php#example-251
If a global variable is assigned to a reference within a function, the reference is only visible within the function. You can avoid this by using the $ GLOBALS array.
The reason is as follows:
Use global $ var as the abbreviation of $ var = & $ GLOBALS ['var. Therefore, assigning other references to $ var only changes the reference of local variables.
------ Solution --------------------
$ B = "dasdf ";
Function abc (){
$ A = "abc ";
Global $ B; // This is a reference to $ GLOBALS ['B '].
$ B = & $;
$ GLOBALS ['BB'] = & $;
Echo $ B. "inner ;";
}
Abc (); // abcinner;
Echo $ B. ";"; // dasdf;
Echo $ bb; // abc
------ Solution --------------------
Remember the two concepts of php. all variables are pointers, and all variables with different scopes are different variables.
