Detailed explanation of PHP reference passing by value example

What is a reference?
Quoting in PHP means accessing the same variable content with different names. This is not like a C pointer; instead, the reference is a symbol table alias. Note that in PHP, variable names and variable contents are different, so the same content can have different names. The closest analogy is Unix's filenames and the files themselves - the variable names are the directory entries, and the variable contents are the files themselves. References can be thought of as hardlinks in Unix file systems.
1: Variable reference

Copy code The code is as follows:

$a =100;
$b = &$a;
echo $b; //Output 100 here
echo $a; //Output 100 here, indicating $a, and $b The values are all one hundred.
$b= 200;
echo $a; //Output 200 here
echo $b; //Output 200 here, which shows that they use the same address. Change one and the other will change. > :

function main($a,$b){ $b= $a+100;
return $b;}

main(55,&$b); //The $b here actually passes its memory address to the $b parameter in the function main, and changes the value of the outer $b through the change of the parameter $b. echo $b; //155 will be output here, ?>

3: Object reference passing by value

Object reference

Copy code

The code is as follows:

class club{
var $name="real madrid" ;}
$b=new club;

$c=$b;echo $b->name;//Output real madrid hereecho $c->name ;//Output here real madrid$b->name="ronaldo";echo $c->name;//Output here ronaldo?>

Unreference
When you unset a reference, you just break the binding between the variable name and the variable content. This does not mean that the variable contents are destroyed. For example:

Copy code

The code is as follows:

$a = 'ronaldo'
$b =&$a;
unset ($a);
?>

will not unset $b, just $a. Example, pass by reference

test1.php

Copy code

The code is as follows:

/**

* Pass by reference
The following can be passed by reference:
Variables, such as foo($a)

New statements, such as foo(new foobar()) References returned from functions , for example: */ function foo(&$var)

{

$var++;
}

$a=5;
// Legal
foo($a);
foo(new stdClass());
//Illegal use of
function bar() // Note the missing &
{
$a = 5;
return $a;
}
foo(bar()); // Causes fatal error since PHP 5.0.5
foo($a = 5) // Expression, Not a variable
foo(5) // Causes fatal error

?>

test2.php

Copy code

The code is as follows:

function test(&$a)

{
$a=$a+100;
}

$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. You can change the value of $b by changing the value of $a in the function echo "
";

echo $b;//output 101

/*****************************
*
* It should be noted here that the parameters after call_user_func_array require &
*
* ****************************/

//Do not add the & symbol in front of $b in the above "test($b);", but in the function "call_user_func_array", if you want to pass parameters by reference , you need the & symbol, as shown in the following code:

function a(&$b){
$b++;
}
$c=0;
call_user_func_array(' a',array(&$c));
echo $c;
//Output 1
?>

Reference return
Reference return is used when you want to use a function to find which variable the reference should be bound to. Don't use return references to increase performance, the engine is smart enough to optimize it itself. Only return references if there is a valid technical reason! To return a reference, use this syntax

Copy the code The code is as follows:

function &test()
{
static $b=0;//Declare a static variable
$b=$b+1;
echo $b;
return $b;
}

$a=test();//This statement will output that the value of $b is 1
$a=5;
$a=test();//This statement will output $ The value of b is 2

$a=&test();//This statement will output the value of $b is 3 Here the memory address of the $b variable in $b and the memory of the $a variable are returned The address points to the same place
$a=5; //The value of the $b variable in return $b has been changed

$a=test();//This statement will output $ The value of b is 6
?>

Explanation:
What $a=test(); gets in this way is not actually a reference return of the function, which is different from ordinary functions There is no difference in calling. As for the reason: This is the regulation of PHP
PHP stipulates that the reference return of the function is obtained through $a=&test();
As for what is a reference return (the PHP manual says: reference return Used when you want to use a function to find which variable a reference should be bound to. ) This nonsense made me unable to understand it for a long time

Using the above example to explain it is
$a=test. () method calls a function, just assigns the value of the function to $a, and any changes to $a will not affect $b in the function
But when calling a function through $a=&test(), he The function is to point the memory address of the $b variable in return $b and the memory address of the $a variable to the same place
, which produces the equivalent effect ($a=&$b;) so change $ The value of a also changes the value of $b, so after executing
$a=&test();
$a=5;
, the value of $b becomes 5

Static variables are used here to let everyone understand the reference return of functions. In fact, function reference returns are mostly used in objects
Here is an interesting example I saw on oschina:

Copy code The code is as follows:

$a = array('abe','ben','cam');
foreach ($a as $k=>&$n)
$n = strtoupper($n);
foreach ($a as $k=>$n) // notice NO reference here!
echo "$nn";
print_r($a);
?>

will result in:

ABE
BEN
BEN
Array
(
[0] => ABE
[1] => BEN
[2] => BEN
)
Explanation : The loop in the second foreach is as follows:
Array
(
[0] => ABE
[1] => BEN
[2] => ABE
)
Array
(
[0] => ABE
[1] => BEN
[2] => BEN
)
Array
(
[0] => ABE
[1] => BEN
[2] => BEN
)
Because there is no unset($n), it always Pointing to the last element of the array, the first loop in the second foreach changes $n, that is, $a[2] to ABE, the second loop changes it to BEN, and the third loop also changes it to BEN.

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source：php.cn

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Detailed explanation of PHP reference passing by value example_PHP tutorial