Passing by value and passing by reference in PHP_PHP Tutorial

There are value passing and reference passing in PHP. The latter requires the use of the address character & to identify the variable. Examples of using assignment operations are as follows:

1. Value transfer

1) Basic data types:

01
02
$a = 1;
03
$b = $a;
04
$b += 2;
05
 
06
echo "$a = ".$a."
";
07
echo "$b = ".$b."
";
08
 
09
/* Output:
10
         $a = 1
11
         $b = 3
12
*/
13
?>
Note: The statement $b = $a is used here to assign the value of $a to $b. That is, the value assigned at this time is 1.

After executing statement $b += 2:

So, $b becomes 3, but the value of $a does not change.

2) Reference data type

Person class is as follows:

01
class Person {
02
private $_name;
03
 
04
Public function setName($name) {
05
            $this->_name = $name;
06
}
07
 
08
Public function getName() {
09
          return $this->_name;
10
}
11
 
12
Public function toString() {
13
           return "name is ".$this->_name;
14
}
15
}
Initialize two Person class objects and set properties:

1
$p1 = new Person();
2
$p1->setName("person1");
3
$p2 = new Person();
4
$p2->setName("person2");
Do the following:

1
$p3 = $p1;
Test:

1
echo "$p1's ".$p1->toString().".
";
2
echo "$p3's ".$p3->toString().".
";
3

4
/* Output:
5
$p1's name is person1.
6
$p3's name is person1.
7
*/
Note: For the statement $p3 = $p1, the value assigned at this time is 0x000a. That is, both $p1 and $p3 refer to the same object.

2. Pass by reference

1) Basic data types

01
02
$a = 1;
03
$b = &$a;
04
$b += 2;
05
 
06
echo "$a = ".$a."
";
07
echo "$b = ".$b."
";
08
 
09
/* Output:
10
         $a = 3
11
         $b = 3
12
*/
13
?>
Note: The statement $b = &$a is used here to pass by reference: That is, the value assigned at this time is 0x0001.


So after $b += 2,

So the values ​​of $a and $b are both 3.

2) Reference data type

If the following operation is performed to replace the previous $p3 = $p1

1
$p3 = &$p1;
Then do the following:

1
$p3 = $p2;
The test results are as follows:

1
echo "$p1's ".$p1->toString().".
";
2
echo "$p3's ".$p3->toString().".
";
3

4
/* Output:
5
$p1's name is person2.
6
$p3's name is person2.
7
*/
Note: It is very clear that the result information of $p3 has been printed, because the $p3 = $p2 statement has been executed, but why is the result information of $p1 printed different from before?

is because after the $p3 = &$p1 statement is executed, the value assigned at this time is 0x0001.

Immediately after executing the $p3 = $p2 statement

That's why the above result information is printed. This is the address character & at work.

The complete code is as follows:

1) Reference data type

01
02
class Person {
03
        private $_name;
04
                                                 05
          public function setName($name) {
06
                 $this->_name = $name;
07
}
08
                                                 09
          public function getName() {
10
               return $this->_name;
11
}
12
                                                 13
         public function toString() {
14
                return "name is ".$this->_name;
15
}
16
}
17
                                                 18
$p1 = new Person();
19
$p1->setName("person1");
20
$p2 = new Person();
21
$p2->setName("person2");
22
 
23
$p3 = $p1;
24
// $p3 = &$p1;
25
// $p3 = $p2;
26
echo "$p1's ".$p1->toString().".
";
27
echo "$p3's ".$p3->toString().".
";
28
?>
2) Basic data types



1

www.bkjia.com