事件循环卡住:NAO C ++ SDK OnFaceDetection示例

时间:2018-10-16 07:10:37

标签: c++ multithreading event-loop nao-robot qi

我在MAC上安装了NAOqi C ++ SDK,并尝试了该SDK中的一些示例。 HelloWorld-Example工作正常,但使用 OnFaceDetection-Example ,在NAO检测到我的脸后,我将通过 qi.eventloop 收到 Error / Warning (错误/警告)

Narongsones-MacBook-Pro:bin Narongsone$ ./onfacedetection --pip 192.168.1.138
[I] 1295 core.common.toolsmain: ..::: starting onfacedetection :::..

[I] 1295 core.common.toolsmain: Connecting to 192.168.1.138:9559...

[I] 1295 qimessaging.session: Session listener created on tcp://0.0.0.0:0
[I] 1295 qimessaging.transportserver: TransportServer will listen on:    tcp://192.168.1.136:64881
[I] 1295 qimessaging.transportserver: TransportServer will listen on: tcp://127.0.0.1:64881
[I] 1295 core.common.toolsmain: Connection with 192.168.1.138:9559 established

[I] 1295 module.example: No face detected

[I] 1295 core.common.toolsmain: onfacedetection is ready... Press CTRL^C to quit

[I] 3843 module.name: 1 face(s) detected.

[I] 4355 qi.eventloop:eventloop:产生更多线程(5)

[I] 4355 qi.eventloop:eventloop:产生更多线程(6)

[I] 4355 qi.eventloop:eventloop:产生更多线程(7)

[I] 4355 qi.eventloop:eventloop:产生更多线程(8)

[I] 4355 qi.eventloop:eventloop:产生更多线程(9)

[I] 4355 qi.eventloop:eventloop:产生更多线程(10)

如果您对问题有任何了解,请帮助我。谢谢!

我的回调函数:

void OnFaceDetection::callback() {
  /** Use a mutex to make it all thread safe. */
  AL::ALCriticalSection section(fCallbackMutex);

  try {
    /** Retrieve the data raised by the event. */
    fFaces = fMemoryProxy.getData("FaceDetected");
    /** Check that there are faces effectively detected. */
    if (fFaces.getSize() < 2 ) {
      if (fFacesCount != 0) {
        qiLogInfo("module.example") << "No face detected" << std::endl;
        fTtsProxy.say("No face detected.");
        fFacesCount = 0;
      }
      return;
    }
    /** Check the number of faces from the FaceInfo field, and check that it has
    * changed from the last event.*/
    if (fFaces[1].getSize() - 1 != fFacesCount) {
      qiLogInfo("module.name") << fFaces[1].getSize() - 1 << " face(s) detected." << std::endl;
      char buffer[50];

      sprintf(buffer, "%d faces detected.", fFaces[1].getSize() - 1);
      fTtsProxy.say(std::string(buffer));

      /** Update the current number of detected faces. */
      fFacesCount = fFaces[1].getSize() - 1;
    }

  }
  catch (const AL::ALError& e) {
    qiLogError("module.name") << e.what() << std::endl;
  }
}

2 个答案:

答案 0 :(得分:0)

正如@AlexandreMazel所提到的那样,该错误只是向您解释系统正在创建太多新线程,因为该函数被频繁调用(最高10x / s),但是由于有语音提示而需要花费几秒钟执行在里面。

您可能想要一个标志或非阻塞互斥锁,以防止该函数多次运行。

答案 1 :(得分:0)

关于更改tts.say,这是一个示例:

通过处理所有文本命令的方法更改tts.say

tts.say(txt)

成为:

if time.time() - self.lastSaidTime > 5.0 or txt != self.lastSaidText:
    self.lastSaidTime = time.time()
    self.lastSaidText = txt
    tts.say(txt)