Detailed explanation of phpCOW mechanism

Copy-on-Write (also abbreviated as COW), as the name suggests, actually copies a copy of the memory for modification when writing. COW was first used in *nix systems to optimize thread and memory usage, and was later widely used in various programming languages, such as C++'s STL, etc. In the PHP kernel, COW is also the main memory optimization method. In the previous discussion about variables and memory, reference counting plays a crucial identification role in the destruction and recycling of variables. The purpose of reference counting is to enable COW to operate normally, thereby achieving optimal use of memory.

The role of copy-on-write

The following is a piece of code:

<?php var_dump(memory_get_usage());//先打印出当前内存情况
$arr = array_fill(0, 100000, &#39;tioncico&#39;);//生成一个0-100000键的数组
var_dump(memory_get_usage());//打印内存
$arr_copy = $arr;//把数组赋值给另一个
var_dump(memory_get_usage());//打印内存
$j=1;
foreach($arr_copy as $i) {//循环遍历该数组键值查看内存情况
    $j += count($i);
}
var_dump(memory_get_usage());//打印内存

In other words, even if we don't use references, PHP variables are passing values ​​and assigning values, all pointing to the same memory, but what happens if the value of $arr_copy changes? <?php

var_dump(memory_get_usage());
//$tipi = array_fill(0, 3, 'php-internal');
//不用array_fill的原因可自己试着打印下
$tipi[0]='php-internal';
$tipi[1]='php-internal';
$tipi[2]='php-internal';
 
var_dump(memory_get_usage());
 
$copy = $tipi;
 
xdebug_debug_zval('tipi', 'copy');
 
var_dump(memory_get_usage());
 
$copy[0] = '123';
 
xdebug_debug_zval('tipi', 'copy');
 
var_dump(memory_get_usage());

The results are as follows: (Note: The results are under the php5.6 web environment, and the references of php7 are different)

It can be seen that when $arr assigns the value to $arr_copy, the execution memory does not change significantly, and 5443320 memory is not directly increased. Amount

Even in the subsequent foreach traversal, the memory is not increased.

Because when $arr is assigned to $arr_copy, it is not copied in the memory Instead of the entire value of $arr, the value of $arr_copy points to $arr, which is equivalent to when fetching the data of $arr_copy, it still points to the memory where $arr stores the value.

In other words, even if we don’t Using references, PHP variables are passing values ​​and assigning values, all pointing to the same memory, but what happens if the value of $arr_copy changes?

<?php
var_dump(memory_get_usage());
//$tipi = array_fill(0, 3, 'php-internal');
//You can try to print the reason why array_fill is not used
$tipi[0]='php- internal';
$tipi[1]='php-internal' ;
$tipi[2]='php-internal';

var_dump(memory_get_usage());

$copy = $tipi;

xdebug_debug_zval('tipi', 'copy');

var_dump(memory_get_usage());

$copy[0] = '123';

xdebug_debug_zval('tipi', 'copy');

var_dump(memory_get_usage());

The result is as follows: (Note: This result is php5.6 web environment (The reference of php7 is different)

##It can be seen that when the value of $copy[0] changes, php will re-apply memory for $copy[0]. Then assign a new value to it without affecting the memory state of other values. The minimum granularity of copy-on-write is the zval structure. For collections of zval structures (such as arrays and objects, etc.), when memory needs to be copied, complex objects are decomposed into the smallest granularity for processing. This way, when a certain part of a complex object in memory is modified, it is not necessary to "separate" all elements of the object into a memory copy, thus saving memory usage.

(xdebug_debug_zval in the article is a function in the xdebug extension, used to view the reference information of variables)

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