## Understanding function for generating sphere ...

Hello,

So I have been looking around for a good function that generates a sphere for OpenGL and I found the function below.

I am not trying to simply cut and paste code in my project so I and trying to understand how it works.

It seems to me that the first three arguments are the x, y, and z coordinates of the center of the sphere and the fourth is the radius.

What I don't understand is the fifth argument; is that the number of sections or "poles" for the sphere being generated?

Source: https://gist.github.com/stuartjmoore/1076642

Code :
```void renderSphere(float cx, float cy, float cz, float r, int p)

{

float theta1 = 0.0, theta2 = 0.0, theta3 = 0.0;

float ex = 0.0f, ey = 0.0f, ez = 0.0f;

float px = 0.0f, py = 0.0f, pz = 0.0f;

GLfloat vertices[p*6+6], normals[p*6+6], texCoords[p*4+4];

if( r < 0 )

r = -r;

if( p < 0 )

p = -p;

for(int i = 0; i < p/2; ++i)

{

theta1 = i * (M_PI*2) / p - M_PI_2;

theta2 = (i + 1) * (M_PI*2) / p - M_PI_2;

for(int j = 0; j <= p; ++j)

{

theta3 = j * (M_PI*2) / p;

ex = cosf(theta2) * cosf(theta3);

ey = sinf(theta2);

ez = cosf(theta2) * sinf(theta3);

px = cx + r * ex;

py = cy + r * ey;

pz = cz + r * ez;

vertices[(6*j)+(0%6)] = px;

vertices[(6*j)+(1%6)] = py;

vertices[(6*j)+(2%6)] = pz;

normals[(6*j)+(0%6)] = ex;

normals[(6*j)+(1%6)] = ey;

normals[(6*j)+(2%6)] = ez;

texCoords[(4*j)+(0%4)] = -(j/(float)p);

texCoords[(4*j)+(1%4)] = 2*(i+1)/(float)p;

ex = cosf(theta1) * cosf(theta3);

ey = sinf(theta1);

ez = cosf(theta1) * sinf(theta3);

px = cx + r * ex;

py = cy + r * ey;

pz = cz + r * ez;

vertices[(6*j)+(3%6)] = px;

vertices[(6*j)+(4%6)] = py;

vertices[(6*j)+(5%6)] = pz;

normals[(6*j)+(3%6)] = ex;

normals[(6*j)+(4%6)] = ey;

normals[(6*j)+(5%6)] = ez;

texCoords[(4*j)+(2%4)] = -(j/(float)p);

texCoords[(4*j)+(3%4)] = 2*i/(float)p;

}

glVertexPointer(3, GL_FLOAT, 0, vertices);

glNormalPointer(GL_FLOAT, 0, normals);

glTexCoordPointer(2, GL_FLOAT, 0, texCoords);

glDrawArrays(GL_TRIANGLE_STRIP, 0, (p+1)*2);

}

}```