iPhone: Converting C# code to Objective-C

Question

Can you guys help me converting this C# code to Objective-C? I don't have a clue about C#/Visual Studio!

public static class BezierSpline
{

public static void GetCurveControlPoints(Point[] knots,
        out Point[] firstControlPoints, out Point[] secondControlPoints)
    {
        int n = knots.Length - 1;

        // Calculate first Bezier control points
        // Right hand side vector
        double[] rhs = new double[n];

        // Set right hand side X values
        for (int i = 1; i < n - 1; ++i)
            rhs[i] = 4 * knots[i].X + 2 * knots[i + 1].X;
        rhs[0] = knots[0].X + 2 * knots[1].X;
        rhs[n - 1] = (8 * knots[n - 1].X + knots[n].X) / 2.0;
        // Get first control points X-values
        double[] x = GetFirstControlPoints(rhs);

        // Set right hand side Y values
        for (int i = 1; i < n - 1; ++i)
            rhs[i] = 4 * knots[i].Y + 2 * knots[i + 1].Y;
        rhs[0] = knots[0].Y + 2 * knots[1].Y;
        rhs[n - 1] = (8 * knots[n - 1].Y + knots[n].Y) / 2.0;
        // Get first control points Y-values
        double[] y = GetFirstControlPoints(rhs);

        // Fill output arrays.
        firstControlPoints = new Point[n];
        secondControlPoints = new Point[n];
        for (int i = 0; i < n; ++i)
        {
            // First control point
            firstControlPoints[i] = new Point(x[i], y[i]);
            // Second control point
            if (i < n - 1)
                secondControlPoints[i] = new Point(2 * knots
                    [i + 1].X - x[i + 1], 2 *
                    knots[i + 1].Y - y[i + 1]);
            else
                secondControlPoints[i] = new Point((knots
                    [n].X + x[n - 1]) / 2,
                    (knots[n].Y + y[n - 1]) / 2);
        }
    }

    private static double[] GetFirstControlPoints(double[] rhs)
    {
        int n = rhs.Length;
        double[] x = new double[n]; // Solution vector.
        double[] tmp = new double[n]; // Temp workspace.

        double b = 2.0;
        x[0] = rhs[0] / b;
        for (int i = 1; i < n; i++) // Decomposition and forward substitution.
        {
            tmp[i] = 1 / b;
            b = (i < n - 1 ? 4.0 : 3.5) - tmp[i];
            x[i] = (rhs[i] - x[i - 1]) / b;
        }
        for (int i = 1; i < n; i++)
            x[n - i - 1] -= tmp[n - i] * x[n - i]; // Backsubstitution.

        return x;
    }
}

thanks.

Answer 1

double[] tmp = new double[n];

tmp is an array of length n. Each value is not initialized explicitly, but it is implicitly set to the default value of the double type, which is 0. So tmp is an n length array of zeros. {0,0,0,0,0, ... 0}