在分叉的子级上调用rand/mt_rand会产生相同的结果


Calling rand/mt_rand on forked children yields identical results

我正在编写一个需要在PHP中执行并发任务的脚本。

我做了一个小测试,结果很奇怪。我正在使用pcntl_fork生成一个子项。父进程只等待子进程完成。

我生成了5个孩子,每个孩子都运行一个函数,生成一个随机数(秒)并睡那么长时间。出于某种原因,所有的孩子都会生成相同的数字。

下面是一个代码示例:

private $_child_count = 0;
private function _fork_and_exec($func)
{
    $cid = ++$this->_child_count;
    $pid = pcntl_fork();
    if ($pid){  // parent
        return $pid;
    } else {    // child
        $func($cid);
        //pcntl_waitpid(-1, $status);
        exit;
    }
}
public function parallel_test()
{
    $func = function($id){
        echo 'child ' . $id . ' starts'."'n";
        $wait_time = mt_rand(1,4);
        echo 'sleeping for '.$wait_time."'n";
        sleep($wait_time);
        echo 'child ' . $id . ' ends'."'n";
    };
    $children = [];
    for ($i=0; $i<5; $i++){
        $children[] = $this->_fork_and_exec($func) ."'n";
    }
    pcntl_wait($status);
    echo 'done' ."'n";
    exit;
}

示例输出:

child 1 starts
sleeping for 1
child 2 starts
sleeping for 1
child 3 starts
sleeping for 1
child 4 starts
sleeping for 1
child 5 starts
sleeping for 1
child 1 ends
child 2 ends
child 3 ends
child 4 ends
child 5 ends
done

提前感谢

这是因为所有子级都以相同的状态开始(fork()重复代码和数据段)。由于rand和mt_rand是伪随机生成器,它们都将生成相同的序列。

您将不得不重新初始化随机生成器,例如使用进程/线程ID或从/dev/urandom读取几个字节。

我真的认为你应该看看pthreads,它提供了与基于Posix Threads的PHP兼容的多线程。

简单如

class AsyncOperation extends Thread {
    public function __construct($arg) {
        $this->arg = $arg;
    }
    public function run() {
        if ($this->arg) {
            echo 'child ' . $this->arg . ' starts' . "'n";
            $wait_time = mt_rand(1, 4);
            echo 'sleeping for ' . $wait_time . "'n";
            sleep($wait_time);
            echo 'child ' . $this->arg . ' ends' . "'n";
        }
    }
}
$t = microtime(true);
$g = array();
foreach(range("A","D") as $i) {
    $g[] = new AsyncOperation($i);
}
foreach ( $g as $t ) {
    $t->start();
}

输出

child B starts
sleeping for 3
child B ends
child C starts
sleeping for 3
child C ends
child A starts
sleeping for 4
child A ends
child D starts
sleeping for 4
child D ends