Libevent 疑问

echo1
class Libevent
{
    /**
     * base event
     *
     * @var null
     */
    public $baseEvent = NULL;

    public function __construct()
    {
        $this->baseEvent = event_base_new();
    }

    public function add($fd, $flag = EV_READ, $callback, $args = [])
    {
        $event = event_new();

        if (!$event) {
            exit("event_new faild\n");
        }

        $flag = $flag == EV_READ ? EV_READ | EV_PERSIST : EV_WRITE | EV_PERSIST;

        if (!event_set($event, $fd, $flag, $callback, null)) {
            exit("event_set faild\n");
        }

        if (!event_base_set($event, $this->baseEvent)) {
            exit("event_base_set faild\n");
        }

        if (!event_add($event)) {
            exit("event_add faild\n");
        }

        //$this->arr[] = $event;
    }

    public function loop()
    {
        $ret = event_base_loop($this->baseEvent);

        if ($ret == 1) {
            exit("no events were registered\n");
        } else if ($ret == -1) {
            exit("event_base_loop faild\n");
        }
    }

}

function read($fd)
{
    $buffer = fread($fd, 65536);

    $buffer = trim($buffer);

    echo $buffer ."\n";
}

$libevent = new Libevent();

$libevent->add(STDIN, EV_READ, 'read');

$libevent->loop();

@walkor 大大, 我对workerman 的libevent 进行了拆解, 测试,

我运行中,注释掉 $this->arr[] = $event; 这行, 命令行会提示 no events were registered,

当我不注释时正常运行, 请问下, 这是怎么一回事, 求解答。

php 的版本是 php5.5.34的

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1个回答

walkor 打赏

$event 变量是函数内部局部变量,出了函数作用域就释放了,所以有问题。
加上$this->arr[] = $event; 这句将局部变量存储了起来,就不会被释放,所以没问题

  • echo1 2016-11-11

    @1 , event_base_set($event, $this->baseEvent), 这句话不是将 $event 关联到$this->baseEvent么, 为啥外面的变量还能影响的到 libevent的执行?

  • walkor 2016-11-11

    这个要看event扩展源码是怎么处理的

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