Detailed explanation of php reference (&)_PHP tutorial

The meaning of reference in PHP is: different names access the same variable content. There is a difference between
and pointers in C language. The pointer in C language stores the address of the variable's content in memory
Variable reference
PHP's reference allows you to use two variables to point to the same content

Copy code The code is as follows:

$a="ABC";
$b =&$ a;
echo $a;//Output here: ABC
echo $b;//Output here: ABC
$b="EFG";
echo $a;//Here $a The value becomes EFG so output EFG
echo $b;//Output EFG here
?>

Pass-by-reference call of function
Pass I won’t say much about the address call. The code is given directly below:

Copy the 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. By changing the value of $a in the function, the value of $b can be changed
echo "
";
echo $b;//Output 101

It should be noted that if test(1); is used here, an error will occur. Think about the reason yourself.
The reference return of the function
Look at the code first

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 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

Explain below:
In this way $a=test(); what you get is not actually a function reference return, which is no different from an ordinary function call. As for the reason: This is PHP regulations
PHP stipulates that what is obtained through $a=&test(); is the reference return of a function
As for what is a reference return (the PHP manual says: reference return is used when you want to use a function to find a reference Which variable should be bound to. ) I couldn’t understand this nonsense for a long time
Using the above example to explain it is that
$a=test() calls the function, which just assigns the value of the function to $a, and $a does anything The changes will not affect $b in the function
. If the function is called through $a=&test(), its function is to compare the memory address of the $b variable in return $b with the memory address of the $a variable. Pointing to the same place
produces the equivalent effect ($a=&b;), so changing the value of $a also changes the value of $b, so after executing
$a=&test() ;
$a=5;
After that, the value of $b becomes 5
The static variable is used here to let everyone understand the reference return of the function. In fact, the reference return of the function is mostly used in objects
Object reference

Copy code The code is as follows:

class a{
var $abc="ABC";
}
$b=new a;
$c=$b;
echo $b->abc;//output here ABC
echo $c->abc;//Output ABC here
$b->abc="DEF";
echo $c->abc;//Output DEF here
?>

The above code is the running effect in PHP5
In PHP5, the copying of objects is achieved through references. In the above column, $b=new a; $c=$b; is actually equivalent to $b=new a; $c=&$b;
The default in PHP5 is to call objects by reference, but sometimes you may want to Create a copy of an object and hope that changes to the original object will not affect the copy. For this purpose, PHP defines a special method called __clone.
The role of reference
If the program is relatively large, there are many variables referencing the same object, and you want to manually clear the object after using it, I personally recommend using the "&" method, and then using $var=null to clear it. Other times, use php5. The default method. In addition, for transferring large arrays in php5, it is recommended to use the "&" method, after all, it saves memory space.

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 = 1;
$ b =& $a;
unset ($a);
?>

Won’t unset $b, just $a.

global reference
When you declare a variable with global $var you actually create a reference to the global variable. In other words, it is the same as doing this:

Copy the code The code is as follows:

$ var =& $GLOBALS["var"];
?>

This means, for example, that unset $var will not unset a global variable.
$this
In a method of an object, $this is always a reference to the object that calls it.

//Here’s another little episode
The pointing (similar to pointer) function of the address in PHP is not implemented by the user himself, but is implemented by the Zend core. The reference in PHP uses "write The principle of "time-copy" is that unless a write operation occurs, variables or objects pointing to the same address will not be copied.
In layman terms
1: If there is the following code
$a="ABC";
$b=$a;
In fact, $a and $b point to the same point at this time The memory address is not that $a and $b occupy different memories
2: If you add the following code to the above code
$a="EFG";
Because $a and $b The data in the pointed memory needs to be rewritten. At this time, the Zend core will automatically determine and automatically produce a data copy of $a for $b, and re-apply for a piece of memory for storage.

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