Question

Why these two pieces of codes, running on same conditions, does not have difference in amount of running time?

Code 1

static long time = 0;

static int n = 200;

static float[] vq = new float[200 * 200 * 200 * 3];
static int[] iq = new int[200 * 200 * 200];

static FloatBuffer verts = BufferUtils.createFloatBuffer(vq.length);
static IntBuffer ind = BufferUtils.createIntBuffer(iq.length);

static void draw() {

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            for (int k = 0; k < n; k++) {
                int index = (i * n * n) + (j * n) + k;
                vq[3 * index + 0] = i;
                vq[3 * index + 1] = j;
                vq[3 * index + 2] = k;

                iq[index] = index;
            }
        }
    }

    verts.put(vq);
    ind.put(iq);

    verts.flip();
    ind.flip();

    GL11.glEnableClientState(GL11.GL_VERTEX_ARRAY);

    GL11.glVertexPointer(3, 0, verts);

    GL11.glDrawElements(GL11.GL_QUADS, ind);

    GL11.glDisableClientState(GL11.GL_VERTEX_ARRAY);

    long newTime = System.currentTimeMillis();
    System.out.println(newTime - time);
    time = newTime;
}

Code 2

static int n = 200;

static long time = 0;

static void draw() {

    GL11.glBegin(GL11.GL_QUADS);

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            for (int k = 0; k < n; k++) {
                GL11.glVertex3f(i, j, k);
            }
        }
    }

    GL11.glEnd();

    long newTime = System.currentTimeMillis();
    System.out.println(newTime - time);
    time = newTime;
}

The reason I think Code 2 should be slower than Code1, is that it has near 8 millions native calls, but Code 1 has just a few, and they do the same thing in result.

What is the reason, and how could I improve performance of my code?

Answer 1

Code 1 includes copying your values into an array and then copying the array into native memory – if your n stays the same you can avoid this overhead by only doing it once.