In-depth understanding of assignments and references in PHP

In-depth understanding of assignments and references in PHP

Let's look at the following questions:

<?php $a = 10;//assigns a constant value to a variable, allocates a memory space for a $b = $a;//variable is assigned to the variable, is not copy a copy, B also allocated memory space? $c = & $a;//references do not allocate space for C, and C and a are shared with one part of the space.?>

What is your answer to the question in the middle? Before today, my answer is to allocate memory space for B. Because that's what I understand:

& when assigning a value, consider a variable to define an alias, and add a reference to the memory space. Changing one of these will affect the other references. When using unset (), only a reference to the variable memory space is broken, and the memory space is not freed.

and = Assignment is different, it will re-open a memory space to store a copy of the original variable. The changes between the two do not affect each other.

This is confirmed by the following procedure:

<?php $a = 10;//assigns a constant value to a variable, allocates a memory space for a $b = $a;//variable is assigned to the variable, is not copy a copy, B also allocated memory space? $c = & $a;//references do not allocate space for C, and C and a share a space. $a = 5;echo $c;//Output 5, because A and C are pointing to the same memory space as the echo Php_eol;echo $b;//Because B is a copy, the operation on a does not affect B, the output 10?>

What if

$b = $a;//After a  and  b do not make any changes, remain consistent

There is such a problem, if = after the assignment, two variables have not changed, if it is two copies, is not too wasteful memory?

This is actually avoided in PHP.

　　When you assign a variable to a new variable in PHP, the new variable is not immediately allocated a memory space, but a reference to the memory space is added. A new variable is allocated a memory space when the original variable or new variable makes any changes.

<?php $a = 1; $b = $a; echo $a;//Before this, B is a shared memory space with a. $a = 2;//a made a change, at which point B will have its own space?>

Each PHP variable exists in a variable container called "Zval". A Zval variable container that includes two bytes of extra information in addition to the type and value of the variable. The first is "Is_ref", which is a bool value that identifies whether the variable belongs to a reference collection (Referenceset). With this byte, the PHP engine can differentiate between normal and reference variables, and since PHP allows users to use the custom reference by using &, there is an internal reference counting mechanism in the Zval variable container to optimize memory usage. The second extra byte is "RefCount", which represents the number of variables (also known as symbols) that point to the Zval variable container. When the value of "RefCount" is 1 , the value of "Is_ref" is always FALSE .

After installing Xdebug, using Xdebug_debug_zval (), you can see the ZVAL structure:

As follows:

　　

<?php $a = 1; $b = $a; echo $a;//Before this, B is a shared memory space with a. Xdebug_debug_zval (' B '); $a = 2;//a made a change, at which point B will have its own space xdebug_debug_zval (' B '); >

Output:

B:

(refcount=2, is_ref=0),

Int

1

B:

(Refcount=1, is_ref=0),

Int

1

As can be seen from the above results, the reference count is 2 before a change, and when a is changed, the reference count of B becomes 1 because B re-allocates space

In-depth understanding of assignments and references in PHP