glutPassiveMotionFunc 问题

Question

我对 glut 和 opengl 有点陌生，我正在尝试让相机随着鼠标的移动而移动，但是当试图让鼠标在屏幕上的位置时，我假设你想要将 x 和 y 传递给的方法应该只是在 glutPassiveMotionFunc() 参数中被引用。但是在尝试为函数提供 CameraMove 方法时出现错误。我知道我传递的方法有误，但我不确定如何传递。

void helloGl::CameraMove(int x, int y)
{
oldMouseX = mouseX;
oldMouseY = mouseY;

// get mouse coordinates from Windows
mouseX = x;
mouseY = y;

// these lines limit the camera's range
if (mouseY < 60)
    mouseY = 60;
if (mouseY > 450)
    mouseY = 450;

if ((mouseX - oldMouseX) > 0)       // mouse moved to the right
    angle += 3.0f;`enter code here`
else if ((mouseX - oldMouseX) < 0)  // mouse moved to the left
    angle -= 3.0f;
}




void helloGl::mouse(int button, int state, int x, int y)
{
switch (button)
{
    // When left button is pressed and released.
case GLUT_LEFT_BUTTON:

    if (state == GLUT_DOWN)
    {
        glutIdleFunc(NULL);

    }
    else if (state == GLUT_UP)
    {
        glutIdleFunc(NULL);
    }
    break;
    // When right button is pressed and released.
case GLUT_RIGHT_BUTTON:
    if (state == GLUT_DOWN)
    {
        glutIdleFunc(NULL);
        //fltSpeed += 0.1;
    }
    else if (state == GLUT_UP)
    {
        glutIdleFunc(NULL);
    }
    break;
case WM_MOUSEMOVE:

    glutPassiveMotionFunc(CameraMove);

    break;

default:
    break;
}
}

Answer 1

假设 helloGl 是 class。那么答案是，你不能。函数与方法不同。问题是 glutPassiveMotionFunc() 期望：

void(*func)(int x, int y)

但是您要提供的是：

void(helloGl::*CameraMove)(int x, int y)

换句话说一个thiscall. This doesn't work because a thiscall basically has an additional hidden argument in contrast to a cdecl。总之，您可以将 CameraMove() 想象成：

void CameraMove(helloGl *this, int x, int y)

如您所见，这不一样。因此解决方案是将 CameraMove() 移出您的 helloGl class 或使方法静态化。

glutPassiveMotionFunc 问题

glutPassiveMotionFunc problems

c++

opengl

glut

freeglut