Question

我正在用Java编写Raspberry项目，并使用Pi4j库。我不会使用轮询，所以例如我使用这个代码：

public class ListenGpioExample {

  public static void main(String args[]) throws InterruptedException {
    System.out.println("<--Pi4J--> GPIO Listen Example ... started.");

    // create gpio controller
    final GpioController gpio = GpioFactory.getInstance();

    // provision gpio pin #02 as an input pin with its internal pull down resistor enabled
    final GpioPinDigitalInput myButton = gpio.provisionDigitalInputPin(RaspiPin.GPIO_02, PinPullResistance.PULL_DOWN);

    // create and register gpio pin listener
    myButton.addListener(new GpioPinListenerDigital() {
        @Override
        public void handleGpioPinDigitalStateChangeEvent(GpioPinDigitalStateChangeEvent event) {
            // display pin state on console
            System.out.println(" --> GPIO PIN STATE CHANGE: " + event.getPin() + " = " + event.getState());
        }

    });

    System.out.println(" ... complete the GPIO #02 circuit and see the listener feedback here in the console.");

    // keep program running until user aborts (CTRL-C)
    for (;;) {
        Thread.sleep(500);
    }

    // stop all GPIO activity/threads by shutting down the GPIO controller
    // (this method will forcefully shutdown all GPIO monitoring threads and scheduled tasks)
    // gpio.shutdown();   <--- implement this method call if you wish to terminate the Pi4J GPIO controller        
}}

我的问题是：存在无限循环的事实仍然是一种轮询形式？

因为我不明白是否需要无限循环。

Answer 1

不，这不是民意调查。此示例使用事件侦听器，该事件侦听器将在GPIO Pin

中发生状态更改时触发

正如您可以阅读Pi4J library page：

侦听器实现不基于GPIO硬件中断州民意调查。

for(;;)旨在使程序保持活动状态，直到用户停止它为止（正如您在评论中所读到的那样）

如何避免使用覆盆子轮询？

1 个答案: