Elegant way the find the Vertices of a Cube

Question

Nearly every OpenGL tutorial lets you implement drawing a cube. Therefore the vertices of the cube are needed. In the example code I saw a long list defining every vertex. But I would like to compute the vertices of a cube rather that using a overlong list of precomputed coordinates.

A cube is made of eight vertices and twelve triangles. Vertices are defined by x, y, and z. Triangles are defined each by the indexes of three vertices.

Is there an elegant way to compute the vertices and the element indexes of a cube?

Answer 1

I solved this problem with this piece code (C#):

public CubeShape(Coord3 startPos, int size) {
    int l = size / 2;
    verts = new Coord3[8];
    for (int i = 0; i < 8; i++) {
        verts[i] = new Coord3(
            (i & 4) != 0 ? l : -l,
            (i & 2) != 0 ? l : -l,
            (i & 1) != 0 ? l : -l) + startPos;
    }

    tris = new Tris[12];
    int vertCount = 0;
    void AddVert(int one, int two, int three) =>
        tris[vertCount++] = new Tris(verts[one], verts[two], verts[three]);
        
    for (int i = 0; i < 3; i++) {
        int v1 = 1 << i;
        int v2 = v1 == 4 ? 1 : v1 << 1;
        AddVert(0, v1, v2);
        AddVert(v1 + v2, v2, v1);
        AddVert(7, 7 - v2, 7 - v1);
        AddVert(7 - (v1 + v2), 7 - v1, 7 - v2);
    }
}

If you want to understand more of what is going on, you can check out the github page I wrote that explains it.