为什么GlutPostRedisplay和sleep函数在这段代码中不起作用?

时间:2015-05-26 06:31:51

标签: c opengl sleep glut

我试图在这个项目中实现usb和cpu之间的数据传输。数据传输显示为从计算机的一个组件移动到另一个组件的小矩形。

在下面的代码中,GlutPostRedisplay不起作用。

此外,有人可以告诉我使用的sleep()是否正确,因为显示中调用的函数不能同步工作。从不执行casing()。在fisrtscreen()之后,它直接跳转到opened()并且operation()不起作用。

这段代码的错误是什么?

void operate()
{
    URLTEXTX = 200;
    URLTEXTY = 950;
    displayString(READUSB,1);
    //southbrigde to northbrigde
    bottom(488.0,425.0,380.0);
    back(488.0,188.0,380.0);
    top(188.0,380.0,550.0);
    //northbridge to cpu
    front(230.0,350.0,595.0);
    top(345.0,600.0,650.0);
    //read from usb
    back(700.0,625.0,465.0);
    bottom(625.0,460.0,385.0);
    back(620.0,525.0,390.0);
    sleep(1);

    URLTEXTX = 200;
    URLTEXTY = 950;
    displayString(WRITEUSB,1);
    //cpu to northbridge
    bottom(350.0,650.0,595.0);
    back(350.0,230.0,600.0);
    //northbridge to southbridge
    bottom(188.0,550.0,380.0);
    front(188.0,488.0,380.0);
    top(483.0,380.0,425.0);
    //write to usb
    front(525.0,625.0,385.0);
    top(625.0,385.0,460.0);
    front(620.0,700.0,460.0);
    sleep(1);

    URLTEXTX = 200;
    URLTEXTY = 950;
    displayString(READDVD,1);
    //read from dvd
    back(600.0,560.0,810.0);
    bottom(570.0,810.0,600.0);
    back(560.0,525.0,610.0);
    //ram to northbridge
    back(450.0,230.0,580.0);
    //northbridge to cpu
    front(230.0,350.0,595.0);
    top(345.0,600.0,650.0);
    sleep(1);

    URLTEXTX = 200;
    URLTEXTY = 950;
    displayString(WRITEDVD,1);
    //cpu to northbridge
    bottom(350.0,650.0,595.0);
    back(350.0,230.0,600.0);
    //northbridge to ram
    front(230.0,450.0,580.0);
    //write to dvd
    front(525.0,570.0,600.0);
    top(570.0,600.0,800.0);
    front(560.0,600.0,800.0);
    sleep(1);

    URLTEXTX = 200;
    URLTEXTY = 950;
    displayString(READHD,1);
    //read from hard disc
    back(640.0,560.0,300.0);
    top(560.0,300.0,530.0);
    back(560.0,525.0,530.0);
    //ram to northbridge
    back(450.0,230.0,580.0);
    //northbridge to cpu
    front(230.0,350.0,595.0);
    top(345.0,600.0,650.0);
    sleep(1);

    URLTEXTX = 200;
    URLTEXTY = 950;
    displayString(WRITEHD,1);
    //cpu to northbridge
    bottom(350.0,650.0,595.0);
    back(350.0,230.0,600.0);
    //northbridge to ram
    front(230.0,450.0,580.0);
    //write to hard disc
    front(525.0,560.0,530.0);
    bottom(560.0,530.0,300.0);
    front(560.0,640.0,300.0);
    sleep(1);
}
void front(GLfloat x1,GLfloat x2,GLfloat y1)//to move in forward direction                        
{
    GLfloat i;
    for(i=x1;i<=x2;i++)
    {
        drawbit(i,x1+5,y1,y1-5);
        glutPostRedisplay();
    }

}
void back(GLfloat x1,GLfloat x2,GLfloat y1)//to move in backward direction
{
    GLfloat i;
    for(i=x1;i>=x2;i--)
    {
        drawbit(i,i-5,y1,y1-5);
        glutPostRedisplay();
    }

}
void top(GLfloat x1,GLfloat y1,GLfloat y2)//to move in upward direction
{
    GLfloat i;
    for(i=y1;i<=y2;i++)
    {
        drawbit(x1,x1+5,i,i+5);
        glutPostRedisplay();
    }
}
void bottom(GLfloat x1,GLfloat y1,GLfloat y2)//to move in downward direction
{
    GLfloat i;
    for(i=y1;i>=y2;i--)
    {
        drawbit(x1,x1-5,i,i-5);
        glutPostRedisplay();
    }
}
void drawbit(GLfloat x1,GLfloat x2,GLfloat y1,GLfloat y2)
{
        glBegin(GL_POLYGON);
        glColor3f(1.0,1.0,1.0);
        glVertex2f(x1,y1);
        glVertex2f(x2,y1);
        glVertex2f(x2,y2);
        glVertex2f(x1,y2);
        glEnd();
        glFlush();
}

void display()
{
    glClear(GL_COLOR_BUFFER_BIT);
    firstscreen();  //introduction to the project 
    sleep(3);
    glClear(GL_COLOR_BUFFER_BIT);
    casing();   //cpu case
    sleep(2);
    glClear(GL_COLOR_BUFFER_BIT);
    opened();  //when cpu case is opened shows internal components
    sleep(1);
    operate(); //data transfer between various components
}

1 个答案:

答案 0 :(得分:0)

问题与此类似:Pausing in OpenGL successively

glutPostRedisplay只需在glut中设置一个标志即可在下一个循环中调用显示回调。它实际上并没有画任何东西。

我怀疑你之后的功能是glutSwapBuffers。如果没有双缓冲,几何体将直接绘制到屏幕上(尽管&#34;绘制&#34; GPU的命令是缓冲的,您需要glFlush)。这通常会导致闪烁,因为您会看到后来被更近的几何体覆盖的事物(因为深度缓冲区)。双缓冲通过渲染到屏幕外缓冲区然后一次显示结果来解决这个问题。确保将GLUT_DOUBLE传递给glutInit,以便您拥有后台缓冲区。

当您sleep()时,您的应用程序无法捕获和处理事件。让我们说你要关闭窗口。在睡眠恢复之前,整个事情都会没有反应。睡眠仍然很重要,因此您不必占用CPU。我将这些概念分开。

  1. 使用idle函数循环/轮询,直到您的延迟时间结束。然后拨打glutPostRedisplay。如果您要进行双重缓冲,请将glutSwapBuffers添加到display
  2. 写一个叫做睡眠的framerate limiter,这样你就不会生成周期。
  3. 设置延迟后绘制不同内容的简单方法是编写一个小型状态机......

    int state = STATE_INIT;
    float timer = 0.0f;
    
    void idle()
    {
        //insert framerate limiter here
    
        //calculate time since last frame, perhaps using glutGet(GLUT_ELAPSED_TIME)
        float deltaTime = ...
    
        timer -= deltaTime;
        if (timer < 0.0f)
        {
            switch (state)
            {
            case STATE_INIT:
                state = STATE_DRAW_FIRST_THING;
                timer = 123.0f;
            ...
            }
            glutPostRedisplay();
        }
    }
    
    void display()
    {
        ...
        if (state == STATE_DRAW_FIRST_THING)
        {
            ...
        }
        ...
        glutSwapBuffers();
    }
    

    随着你的应用变得越来越大,我怀疑这将是可维护的,你会想要更强大的东西,但在此之前这是一个良好的开端。

    只需更改void (*currentView)(void);中的idle回调函数即可在display中保存一些硬编码。您可能想要创建面向对象的state machine。除了布尔状态,您可能希望查看动画和关键帧插值。而不是硬编码所有东西,在文件中存储几何,关键帧和状态序列是分离代码和数据的好方法。如果您使用库,XML非常适合使用。