如何连续更新过期窗口？ [英] How to update glut window continuously?

问题描述

我有一个真正的机器人，命令我的虚拟机器人在开放gl。我想显示我的主机器人（真正的机器人）在奴隶（虚拟一个在开放gl）的每一个运动在线，所以我需要连续更新我的过剩的窗口，实际上只要真正的机器人移动我的虚拟一个动作，所有这些运动应该在线。

I have a real robot that is ordering my virtual robot in open gl. I want show every movement of my master robot(real robot) in slave (virtual one in open gl) online, so i need to update my glut window continuously, actually as long as real robot moves my virtual one moves too, and all these movement should be online.

我从master获取数据总是获取数据功能，但我不知道如何更新窗口。

I get data from master always with get data function, but I dont know how I should update the window.

这是我的代码：

* ** * ** * ** ** ** * ** ** ** * ** * **** /

********************************************/

  void OnIdle(void){  
    initSocket();

  printf("\n  Defining Step Time Parameters and Initial Conditions for solving Dynamic equations\n");

  xi=0;
  xf=0.1;
  printf("\n    end value x         : %f ",xf); 
  i=0;  yi[i]=0; 
  i++;yi[i]=-1.570796;
  i++;yi[i]=-1.570796;
  i++;yi[i]=0;
  i++;yi[i]=0;
  i++;yi[i]=0;
  ndata=2; fi=1;

  double counter=0.1;

  Eqdifp(v1,v2,v3,v4,v5,v6,xi,xf,yi,ndata,p,fi);


  for(int i=0;i<50;i++)
    //while(1)
  {

      getData();

      printf("\n");
      for(int i=0;i<6; i++)
      {

          printf("%d = %.3f\n", i,drecvbuf[i]);
      }
      printf("\n");

   yi[0]=v1[ndata];
   yi[1]=v2[ndata];
   yi[2]=v3[ndata];
   yi[3]=v4[ndata];
   yi[4]=v5[ndata];
   yi[5]=v6[ndata];
    printf("my nadata %f\n",v1[ndata]);
    counter=counter+0.1;

    Eqdifp(v1,v2,v3,v4,v5,v6,xi,xf,yi,ndata,p,fi);
    glutPostRedisplay();
 }
  }
/////////////////////////////////////////////////////
  int main(int argc, char **argv)
{
    glutInit(&argc, argv);
    glutInitDisplayMode(GLUT_DOUBLE | GLUT_RGBA | GLUT_DEPTH);
    glutInitWindowSize(900,500);
    int u=glutCreateWindow("3DOF robot");
    myinit();
    createMenu();
    glutIdleFunc (OnIdle);
    glutDisplayFunc(Display);
    glutReshapeFunc(reshape);
    glutKeyboardFunc(KeyDown);

    glutMainLoop(); 

    System::Timers::Timer^ aTimer = gcnew System::Timers::Timer( 100 );

      // Hook up the Elapsed event for the timer.
    aTimer->Elapsed += gcnew System::Timers::ElapsedEventHandler( OnTimedEvent );

      // Set the Interval to 2 seconds (2000 milliseconds).
    aTimer->Enabled = true;
    return 0;

  }

推荐答案

在更新后调用glutPostRedisplay，它调度窗口重新绘制（当然使用GLUT的显示func）一旦它返回消息队列，我想。

You can call glutPostRedisplay after the update, which schedules the window to be redrawn (using GLUT's display func, of course) as soon as it returns to the message queue, I think.

但是，如果您在无限循环中连续轮询机器人数据，这将无法工作，因为这会持续阻止程序。你应该做的是使用一个计时器在短时间内安排机器人更新，这样在这些更新之间程序可以返回到主事件循环并重绘窗口。或者你可以调用一些函数，它告诉框架访问事件循环。你的代码示例并不能真正解释你现在做什么（或者我不熟悉你调用的函数）。

But this won't work if you are continously polling the robot data in an infinite loop as this continously blocks the program. What you should do is use a timer to schedule the robot update in short intervals, so that between these updates the program can return to the main event loop and redraw the window. Or you can call some function, which tells the framework to visit the event loop. Your code sample doesn't really explain how you do it at the moment (or I'm just not familiar with the functions you call).

这篇关于如何连续更新过期窗口？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！