Regarding geom. shdr (output tri. strip) [Solved]

OpenGL OpenGL: Advanced Coding
1

Hi all,
I have a basic geometry shader that is rendering the tetrahedron triangles as triangle strips. I strippify the triangles as follows. Assuming that the tetrahedron has 4 vertices p0,p1,p2 and p3 with p3 being the apex vertex. A very simple stripification should be this sequence p0p1p3p2p0p1 so I render the geometry in opengl as follows.


for(size_t i=0;i<indices.size();i+=4) {
   glm::vec3 p0 = positions[indices[i]];
   glm::vec3 p1 = positions[indices[i+1]];
   glm::vec3 p2 = positions[indices[i+2]];
   glm::vec3 p3 = positions[indices[i+3]];
			
   glBegin(GL_TRIANGLE_STRIP);	
	glVertex3f(p0.x,p0.y,p0.z);
	glVertex3f(p1.x,p1.y,p1.z);		
	glVertex3f(p3.x,p3.y,p3.z);
	glVertex3f(p2.x,p2.y,p2.z);		
	glVertex3f(p0.x,p0.y,p0.z);
	glVertex3f(p1.x,p1.y,p1.z);
   glEnd();
}

This renders fine. However, doing the same thing in the following geometry shader, some triangles are missing.


   //single strip
   gl_Position = p0; EmitVertex(); 
   gl_Position = p1; EmitVertex();  
   gl_Position = p3; EmitVertex(); 
   gl_Position = p2; EmitVertex(); 
   gl_Position = p0; EmitVertex(); 
   gl_Position = p1; EmitVertex();  EndPrimitive();

If I split the sequence into two strips (like this)


   //two strips
   gl_Position = p0; EmitVertex(); 
   gl_Position = p1; EmitVertex();  
   gl_Position = p3; EmitVertex(); 
   gl_Position = p2; EmitVertex(); EndPrimitive();
   gl_Position = p3; EmitVertex(); 
   gl_Position = p0; EmitVertex(); 
   gl_Position = p2; EmitVertex(); 
   gl_Position = p1; EmitVertex(); EndPrimitive();

it renders fine. Could anyone tell me why the single strip version does not work?

2

As always, after posting the question on opengl forum, I usually find the answer. I was outputting the line_strip rather than the triangle_strip primitive :slight_smile: Problem solved.