OpenGL编程学习笔记——交互与直线算法

交互

键盘交互（glutKeyboardFunc）

例子：实现小方块的移动

代码

#include 

int xd = 0, yd = 0;

void myDisplay()
{
	int p1[] = { 10 + xd,10 + yd };
	int p2[] = { 20 + xd,10 + yd };
	int p3[] = { 20 + xd,0 + yd };
	int p4[] = { 10 + xd,0 + yd };
	glClear(GL_COLOR_BUFFER_BIT);
	glColor3f(0.8, 0.5, 0.6);
	glPointSize(5);
	//glBegin(GL_POINTS);
	//glBegin(GL_LINES);
	//glBegin(GL_LINE_STRIP);
	//glBegin(GL_LINE_LOOP);
	//glBegin(GL_POLYGON);//多边形
	//glBegin(GL_TRIANGLES);//三角形
	glBegin(GL_QUADS);//四边形
	glVertex2iv(p1);
	glVertex2iv(p2);
	glVertex2iv(p3);
	glVertex2iv(p4);
	glEnd();
	glFlush();
}

void myKeyBorad(unsigned char key, int x, int y) 
{
	switch (key) 
	{
	case 'w': yd++; break;
	case 's': yd--; break;
	case 'a': xd--; break;
	case 'd': xd++; break;
	}
	glutPostRedisplay();
}

void init()
{
	glClearColor(1.0, 1.0, 1.0, 0.0);
	glMatrixMode(GL_PROJECTION);
	glLoadIdentity();
	gluOrtho2D(-20, 30, -20, 30);
}

int main(int argc, char* argv[])
{
	glutInit(&argc, argv);
	glutInitDisplayMode(GLUT_RGB | GLUT_SINGLE);
	glutInitWindowPosition(300, 100);
	glutInitWindowSize(600, 500);
	glutCreateWindow("key interaction");

	init();
	glutDisplayFunc(myDisplay);
	glutKeyboardFunc(myKeyBorad);//键盘交互
	glutMainLoop();
	return 0;
}
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鼠标交互（glutMouseFunc）

例子：实现小方块的移动

代码

#include 

GLint w = 600, h = 500;

GLint xd = 0, yd = 0;

void myDisplay()
{
	glClear(GL_COLOR_BUFFER_BIT);
	glColor3f(0.87, 0.56, 0.4);
	glPointSize(2);
	//glBegin(GL_POINTS);
	//glBegin(GL_LINES);
	//glBegin(GL_LINE_STRIP);
	//glBegin(GL_LINE_LOOP);
	//glBegin(GL_TRIANGLES);//三角形
	//glBegin(GL_QUADS);//四边形
	glBegin(GL_POLYGON);//多边形
	glVertex2i(10 + xd, 10 + yd);
	glVertex2i(10 + xd,100 + yd);
	glVertex2i(100 + xd,100 + yd);
	glVertex2i(100 + xd,10 + yd);
	glEnd();
	glFlush();
}

void mouseMotion(GLint button, GLint state, GLint x, GLint y)
{
	if (button == GLUT_LEFT_BUTTON && state == GLUT_DOWN)//当鼠标左键按下去的时候
	{
		xd = x;//改变全局变量xd，改变矩形的位置
		yd = h - y;//注意这里是h-y
		glutPostRedisplay();
	}
}

void myKeyBorad(unsigned char key, int x, int y)
{
	switch (key)
	{
	case 'w': yd++; break;
	case 's': yd--; break;
	case 'a': xd--; break;
	case 'd': xd++; break;
	}
	glutPostRedisplay();
}

void init()
{
	glClearColor(1.0, 1.0, 1.0, 0.0);
	glMatrixMode(GL_PROJECTION);
	glLoadIdentity();
	gluOrtho2D(0, w, 0, h);
}

int main(int argc, char* argv[])
{
	glutInit(&argc, argv);
	glutInitDisplayMode(GLUT_RGB | GLUT_SINGLE);
	glutInitWindowPosition(300, 100);
	glutInitWindowSize(w, h);
	glutCreateWindow("Mouse Motion");

	init();
	glutDisplayFunc(myDisplay);
	//glutKeyboardFunc(myKeyBorad);
	glutMouseFunc(mouseMotion);
	glutMainLoop();
	return 0;
}
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直线算法

DDA数值微分线段算法

方法论

1. 起始点$(x_1,y_1)$，终点$(x_n,y_n)$
2. $\Delta x=|x_n-x_1|,\Delta y=|y_n-y_1|$
3. 比较$\Delta x$和$\Delta y$的步长
4. steps=$\Delta x$和$\Delta y$的较大者
5. $stepX=\frac{\Delta x}{steps},stepY=\frac{\Delta y}{steps}$

代码

#include 
#include

GLint w = 600, h = 500;
GLint xd = 0, yd = 0;

void myDDA(GLfloat x1,GLfloat y1,GLfloat xn,GLfloat yn)
{
	float dx = fabs(xn - x1);//注意这里需要绝对值
	float dy = fabs(yn - y1);//注意这里abs会报错
	float steps;
	if (dx > dy)
		steps = dx;
	else
		steps = dy;
	float stepX = dx / steps;
	float stepY = dy / steps;
	glBegin(GL_POINTS);
	for (int i=0; i < (int)steps; i++) {
		glVertex2f(x1, y1);
		x1 += stepX;
		y1 += stepY;
	}
}

void myDisplay()
{
	glClear(GL_COLOR_BUFFER_BIT);
	glColor3f(0.87, 0.56, 0.4);
	glPointSize(1);
	myDDA(1.5, 3.8, 189.8, 267.5);//调用myDDA
	glEnd();
	glFlush();
}

void init()
{
	glClearColor(1.0, 1.0, 1.0, 0.0);
	glMatrixMode(GL_PROJECTION);
	glLoadIdentity();
	gluOrtho2D(-10, w, 0, h);
}

int main(int argc, char* argv[])
{
	glutInit(&argc, argv);
	glutInitDisplayMode(GLUT_RGB | GLUT_SINGLE);
	glutInitWindowPosition(300, 100);
	glutInitWindowSize(w, h);
	glutCreateWindow("DDA application");

	init();
	glutDisplayFunc(myDisplay);
	glutMainLoop();
	return 0;
}
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效果

Bresenham直线算法

方法论

1. 输入起始点$(x_1,y_1)$，终点$(x_n,y_n)$
2. $dx=x_n-x_1,dy=y_n-y_1$
3. 计算$2dx$和$2dy$
4. 计算$P_0=2dy-dx$
5. 循环
   如果$P_k>0$，选上面点$(x_k+1,y_k+1),P_{k+1}=P_k+2dy-2dx$
   如果$P_k<0$，选下面点$(x_k+1,y_k),P_{k+1}=P_k+2dy$

中点划线法

方法论

1. 输入起始点$(x_1,y_1)$，终点$(x_n,y_n)$
2. $\Delta x=|x_n-x_1|,\Delta y=|y_n-y_1|$
3. 计算$m=\frac{\Delta x}{\Delta y}$ 和 $b=y_1-m{x}_1$
4. 计算$d=F(x_1,y_1+0.5)=y_1+0.5-m{x}_1-b$
5. 循环
   如果$d>0$，$x++,(x,y),d+=-m$
   如果$d<0$，$x++,y++,(x,y),d+=1-m$