使用OpenGL计算QT中2个点之间的线性插值的最快方法 [英] Fastest Method to compute Linear interpolation between 2 points in QT using OpenGL
问题描述
我正在尝试借助顶点坐标对三角形进行插值.
I'm trying to interpolate a triangle with the help of vertex coordinates.
a
|\
| \
| \
| \
b|_ _ _ \c
我按(b，a)，(a，c)和(c，b)的顺序对顶点进行插值. 这里的a，b和c是具有颜色值的3维坐标.
I'm interpolating the vertices in this order (b,a),(a,c)and (c,b). Here the a,b and c are the 3 dimensional coordinates with a color value.
a = (x1,y1,z1,c1);
b = (x2,y2,z2,c2);
c = (x3,y3,z3,c3);
用于计算计算的结构:
struct pointsInterpolateStruct{
QList<double> x,y,z;
QList<double> r, g, b, clr;
void clear() {
x.clear();
y.clear();
z.clear();
r.clear();
g.clear();
b.clear();
clr.clear();
}
};
插值代码:
QList<double> x,y,z,clrs;
上面提到的列表已用于从包含a，b和c坐标的文件中读取值.
This above mentioned lists has been used to read the values from a file which contains the coordinates of a,b and c.
/**
void interpolate();
@param1 ipts is an object for the point interpolation struct which holds the x,y,z and color
@param2 idx1 is the point A
@param 3idx2 is the point B
@return returns the interpolated values after filling the struct pointsInterpolateStruct
*/
void GLThread::interpolate(pointsInterpolateStruct *ipts,int idx1, int idx2) {
int ipStep = 0;
double delX, imX,iX,delY,imY,iY,delZ,imZ,iZ,delClr,imC,iC;
ipStep = 5; // number of points needed between the 2 points
delX = imX = iX = delY = imY = iY = delZ = imZ = iZ = delClr = imC = iC = 0;
delX = (x.at(idx2) - x.at(idx1));
imX = x.at(idx1);
iX = (delX / (ipStep + 1));
delY = (y.at(idx2) - y.at(idx1));
imY = aParam->y.at(idx1);
iY = (delY / (ipStep + 1));
delZ = (z.at(idx2) - z.at(idx1));
imZ = z.at(idx1);
iZ = (delZ / (ipStep + 1));
delClr = (clrs.at(idx2) - clrs.at(idx1));
imC = clrs.at(idx1);
iC = (delClr / (ipStep + 1));
ipts->clear();
int i = 0;
while(i<= ipStep) {
ipts->x.append((imX+ iX * i));
ipts->y.append((imY+ iY * i));
ipts->z.append((imZ+ iZ * i));
ipts->clr.append((imC + iC * i));
i++;
}
}*
使用OpenGL可视化这些插值点:
所有点都填充到顶点和颜色缓冲区中，我正在使用以下格式绘制它.可视化非常快，即使是较大的点.
All the points are filled to vertices and color buffers and I'm drawing it using the below format. Visualization is very fast even for larger points.
void GLWidget::drawInterpolatedTriangle(void) {
glEnableClientState(GL_COLOR_ARRAY);
glEnableClientState(GL_VERTEX_ARRAY);
glColorPointer(3, GL_FLOAT, 0, clr);
glVertexPointer(3, GL_FLOAT, 0, vrt);
glPushMatrix();
glDrawArrays(GL_POLYGON, 0, vrtCnt);
glPopMatrix();
glDisableClientState(GL_VERTEX_ARRAY);
glDisableClientState(GL_COLOR_ARRAY);
}
}
}
现在一切正常.我正在获得所需的输出.但是问题是，当我尝试对n个三角形(例如n = 40,000)执行相同操作时，即使我在QThread中调用了此函数，应用程序也崩溃了，并且我发现此方法效率不高.该方法需要大量时间进行计算.
Now everything working fine. I'm getting the desired output. But the problem is when I'm trying to do the same for 'n' number of triangles (say n = 40,000), the application gets crashed even if I called this function in a QThread and I found that this method is not an efficient method as it takes lot of time for computation.
请提出一种乐观的方法来执行此过程，以便以良好的性能获得更好的结果.
Please suggest an optimistic way to do this process so that I can achieve better results at good performance.
输出图像:
插值三角形(点视图)
网格视图
多边形视图
推荐答案
检查了应用程序使用的内存后，我发现程序的列表和数组中已存储了大量不需要的数据(即清除列表x，y，z，r，g，b和clrs以pointsInterpolateStruct表示).我已经立即清除了所有不需要的/未使用的数据，并尝试使用较大的三角形运行该应用程序.现在，我可以获得更好的性能.在可视化过程中，我没有做任何更改.
After examining the memory used by the application, I found that there's large number of unwanted data has been stored in the list and arrays in my program (i.e., clearing the list x,y,z,r,g,b and clrs in pointsInterpolateStruct). I have cleared all the unwanted / unused data instantly and tried to run the application with larger triangles. Now I can achieve better performance. I didn't changed anything in visualization process.
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