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c++ - OpenGL,C++ : Large number of points

转载 作者:行者123 更新时间:2023-11-28 03:22:31 27 4
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我目前正在开发一个程序,它将能够可视化平面上沿 vector 场流动的点的演变。我已经完成了我粘贴在下面的第一个版本。当运行带有大量点的程序时,似乎只有最后说的 30000 点被绘制到窗口中。我希望能够画出大约 1000000 点,所以我走了很远。

我试过喜欢数字 if 迭代(Iteration 变量 - 控制点数),在这里它表现得很好。然而,当大幅增加时,第一部分不再绘制。

#include <iostream>
#include <stdio.h>
#include <math.h>
//#include <gsl/gsl_math.h>
//#include <gsl/gsl_sf.h>
#include <GL/freeglut.h>
#include <GL/gl.h>

using namespace std;

//Initial iterations to make the system settle:
int initIter=0;
//Iterations once system has settled:
int Iterations = 100000;
/**Starting point in time-phase-space (t,x,y,vx,vy).
For mathematical reasons the last two components should
always be 0**/
float TPS[5]={0,0.00,0.100,0.00,0.000};
//Timestep:
float dt=0.001;




/**The Step function make one Picard
iteration **/
float * Step(float * Arr){
static float NewTPS[5];
NewTPS[0] = Arr[0]+dt;
NewTPS[1] = Arr[1]+Arr[3]*dt;
NewTPS[2] = Arr[2]+Arr[4]*dt;
//This is the dynamical functions:
NewTPS[3] = -Arr[2];
NewTPS[4] = Arr[1];
return NewTPS;
}




/** This function sets up GLUT plotting
window: **/
void myInit(){
// set the background color
glClearColor(0.0f, 0.0f, 0.0f, 1.00f);

// set the foreground (pen) color
glColor4f(1.0f, 1.0f, 1.0f, 0.04f);

// set up the viewport
glViewport(0, 0, 800, 800);

// set up the projection matrix (the camera)
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
gluOrtho2D(-2.0f, 2.0f, -2.0f, 2.0f);

// set up the modelview matrix (the objects)
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();

//Computing initial iterations:
for (int i=0;i<initIter;i++){
//cout << TPS[1]<<" " << TPS[2] << endl;
float * newTPS2;
newTPS2 = Step(TPS);
//Assigning the values of newTPS2 to TPS:
for (int j=0; j<5;j++){
TPS[j]=*(newTPS2+j);
}
}

// enable blending
//glEnable(GL_BLEND);
//glBlendFunc(GL_SRC_ALPHA, GL_ONE_MINUS_SRC_ALPHA);

// enable point smoothing
//glEnable(GL_POINT_SMOOTH);
//glPointSize(1.0f);
}





/** This function draws a the point that
is passed to it: **/
void Draw(){
// clear the screen
glClear(GL_COLOR_BUFFER_BIT);

// draw some points
glBegin(GL_POINTS);
for (int i = 0; i <Iterations ; i++) {
float * newTPS2;
//cout << TPS[0]<< " " << TPS[1] << " " << TPS[2]<< endl;
newTPS2 = Step(TPS);
//Assigning the values of newTPS to TPS:
for (int j=0; j<5;j++){
TPS[j]=*(newTPS2+j);
}
// draw the new point
glVertex2f(TPS[1], TPS[2]);
}
glEnd();

// swap the buffers
glutSwapBuffers();
//glFlush();
}



int main(int argc, char** argv){
// initialize GLUT
glutInit(&argc, argv);

// set up our display mode for color with alpha and double buffering
glutInitDisplayMode(GLUT_RGBA | GLUT_DOUBLE);


glutInitWindowSize(800, 800);
glutCreateWindow("Trace of 2D-dynamics");
myInit();
// register our callback functions
glutDisplayFunc(Draw);
// glutKeyboardFunc(mykey);

// start the program
glutMainLoop();
return 0;
}

最佳答案

如果您只想为屏幕上的特定像素着色,则根本不应该使用 glVertex。将它们全部放在一个连续的内存块中,从中创建一个纹理并渲染一个覆盖整个屏幕的四边形。这可能比在 OpenGL 中计算它们的位置更快。

关于c++ - OpenGL,C++ : Large number of points,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15051607/

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