问题描述
我在 PHP 的不同线程之间共享静态变量时遇到问题.简单地说，我想1.在一个线程中写一个静态变量2.在其他线程中读取并执行所需的过程并清理它.为了测试上述要求，我写了下面的 PHP 脚本.
I have a problem in sharing static variable between different threads in PHP.In simple words I want to1. Write a static variable in one thread2. Read it in other thread and do the required process and clean it.For testing above requirement I have written below PHP script.
<?php
class ThreadDemo1 extends Thread
{
private $mode; //to run 2 threads in different modes
private static $test; //Static variable shared between threads
//Instance is created with different mode
function __construct($mode) {
$this->mode = $mode;
}
//Set the static variable using mode 'w'
function w_mode() {
echo 'entered mode w_mode() funcion';
echo "<br />";
//Set shared variable to 0 from initial 100
self::$test = 100;
echo "Value of static variable : ".self::$test;
echo "<br />";
echo "<br />";
//sleep for a while
sleep(1);
}
//Read the staic vaiable set in mode 'W'
function r_mode() {
echo 'entered mode r_mode() function';
echo "<br />";
//printing the staic variable set in W mode
echo "Value of static variable : ".self::$test;
echo "<br />";
echo "<br />";
//Sleep for a while
sleep(2);
}
//Start the thread in different modes
public function run() {
//Print the mode for reference
echo "Mode in run() method: ".$this->mode;
echo "<br />";
switch ($this->mode)
{
case 'W':
$this->w_mode();
break;
case 'R':
$this->r_mode();
break;
default:
echo "Invalid option";
}
}
}
$trd1 = new ThreadDemo1('W');
$trd2 = new ThreadDemo1('R');
$trd3 = new ThreadDemo1('R');
$trd1->start();
$trd2->start();
$trd3->start();
?>
预期的输出是，run() 方法中的模式:W进入模式 w_mode() 函数静态变量的值:100
Expected output is,Mode in run() method: Wentered mode w_mode() funcionValue of static variable : 100
run() 方法中的模式:R进入模式 r_mode() 函数静态变量的值:100
Mode in run() method: Rentered mode r_mode() functionValue of static variable : 100
run() 方法中的模式:R进入模式 r_mode() 函数静态变量的值:100
Mode in run() method: Rentered mode r_mode() functionValue of static variable : 100
但实际上我得到的输出是，run() 方法中的模式:W进入模式 w_mode() 函数静态变量的值:100
But actually I am getting the output as,Mode in run() method: Wentered mode w_mode() funcionValue of static variable : 100
run() 方法中的模式:R进入模式 r_mode() 函数静态变量的值:
Mode in run() method: Rentered mode r_mode() functionValue of static variable :
run() 方法中的模式:R进入模式 r_mode() 函数静态变量的值:
Mode in run() method: Rentered mode r_mode() functionValue of static variable :
....真的不知道原因.请帮忙.
....Really unaware of the cause. Please help.
推荐答案
静态变量在上下文之间不共享，原因是静态变量有一个类入口作用域，而处理程序是用来管理对象作用域的.
Static variables are not shared among contexts, the reason is that static variables have a class entry scope, and handlers are for managing the object scope.
当一个新线程启动时，静态数据被复制(删除复杂的变量，如对象和资源).
When a new thread is started, statics are copied (removing complex variables, like objects and resources).
静态作用域可以被认为是一种线程本地存储.
The static scope can be thought of as a kind of thread local storage.
此外，如果成员不是静态的……从 pthreads 定义派生的类的所有成员都被认为是公共的.
In addition, where members are not static ... all members of a class derived from a pthreads definition are considered public.
我鼓励您阅读随 pthread 分发的示例，它们也可以在 github 上找到.
I encourage you to read the examples distributed with pthreads, they are available on github too.
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