检测到堆栈粉碎，中止，OpenGl 自由过剩

Question

所以我有以下用于图形透视投影的代码：

#include <GL/glut.h>
#include <iostream>
#include <unistd.h>
#include <math.h> 
#define UpperBD 5 
#define PI      3.1415926 
#define Num_pts 10
float Xe = 200.0f;
float Ye = 200.0f;
float Ze = 200.0f;
float Rho = sqrt(pow(Xe, 2) + pow(Ye, 2) + pow(Ze, 2));
float D_focal = 20.0f;
typedef struct {
float X[UpperBD];
float Y[UpperBD];
float Z[UpperBD];
} pworld;
typedef struct {
float X[UpperBD];
float Y[UpperBD];
float Z[UpperBD];
} pviewer;
typedef struct {
float X[UpperBD];
float Y[UpperBD];
} pperspective;
void mydisplay()
{
// define x-y coordinate 
float p1x = -1.0f, p1y = 0.0f;
float p2x = 1.0f, p2y = 0.0f;
float p3x = 0.0f, p3y = 1.0f;
float p4x = 0.0f, p4y = -1.0f;
glClear(GL_COLOR_BUFFER_BIT);
glLoadIdentity();
pworld  world;
pviewer viewer;
pperspective perspective;
//define the x-y-z world coordinate 
world.X[0] = 0.0;    world.Y[0] = 0.0;   world.Z[0] = 0.0;    // origin  
world.X[1] = 50.0;   world.Y[1] = 0.0;   world.Z[1] = 0.0;    // x-axis
world.X[2] = 0.0;    world.Y[2] = 50.0;  world.Z[2] = 0.0;    // y-axis    
world.X[3] = 0.0;    world.Y[3] = 0.0;   world.Z[3] = 50.0;   // y-axis 
float sPheta = Ye / sqrt(pow(Xe, 2) + pow(Ye, 2));
float cPheta = Xe / sqrt(pow(Xe, 2) + pow(Ye, 2));
float sPhi = sqrt(pow(Xe, 2) + pow(Ye, 2)) / Rho;
float cPhi = Ze / Rho;
float xMin = 1000.0, xMax = -1000.0;
float yMin = 1000.0, yMax = -1000.0;
for (int i = 0; i <= UpperBD; i++)
{
viewer.X[i] = -sPheta * world.X[i] + cPheta * world.Y[i];
viewer.Y[i] = -cPheta * cPhi * world.X[i]
- cPhi * sPheta * world.Y[i]
+ sPhi * world.Z[i];
viewer.Z[i] = -sPhi * cPheta * world.X[i]
- sPhi * cPheta * world.Y[i]
- cPheta * world.Z[i] + Rho;
}
for (int i = 0; i <= UpperBD; i++)
{
perspective.X[i] = D_focal * viewer.X[i] / viewer.Z[i];
perspective.Y[i] = D_focal * viewer.Y[i] / viewer.Z[i];
if (perspective.X[i] > xMax) xMax = perspective.X[i];
if (perspective.X[i] < xMin) xMin = perspective.X[i];
if (perspective.Y[i] > yMax) yMax = perspective.Y[i];
if (perspective.Y[i] < yMin) yMin = perspective.Y[i];
/*
std::cout << "xMin " << xMin << std::endl;
std::cout << "xMax " << xMax << std::endl;
std::cout << "yMin " << yMin << std::endl;
std::cout << "yMax " << yMax << std::endl;
*/
}
for (int i = 0; i <= UpperBD; i++)
{
if ((xMax - xMin) != 0) perspective.X[i] = perspective.X[i] / (xMax - xMin);
if ((yMax - yMin) != 0) perspective.Y[i] = perspective.Y[i] / (yMax - yMin);
//std::cout << i << perspective.X[i] << perspective.Y[i] << std::endl;
}

glBegin(GL_LINES);
// cross at the display screen 
//glVertex2f(p1x,p1y);  
//glVertex2f(p2x,p2y);
//glVertex2f(p3x, p3y);
//glVertex2f(p4x, p4y);

glClear(GL_COLOR_BUFFER_BIT);
glColor3f(1.0, 0.0, 0.0);
glVertex2f(perspective.X[0], perspective.Y[0]);
glVertex2f(perspective.X[1], perspective.Y[1]);
glColor3f(0.0, 1.0, 0.0);
glVertex2f(perspective.X[0], perspective.Y[0]);
glVertex2f(perspective.X[2], perspective.Y[2]);
glColor3f(0.0, 0.0, 1.0);
glVertex2f(perspective.X[0], perspective.Y[0]);
glVertex2f(perspective.X[3], perspective.Y[3]);
glEnd();
glFlush();
}
int main(int argc, char** argv)
{
glutInit(&argc, argv);
glutCreateWindow("132 transformation pipeline");
glutDisplayFunc(mydisplay);
glutMainLoop();
}

我能够编译此代码，但是每当我运行它时，我都会收到错误"检测到堆栈粉碎"。我不知道为什么会这样。其他人也有同样的代码来工作，但它对我不起作用。我为此使用 ubuntu。任何有关此错误的帮助将不胜感激。

Answer 1

替换：

for (int i = 0; i <= UpperBD; i++)

跟：

for (int i = 0; i < UpperBD; i++)

在所有情况下。