Read File in C#

Question

I found a C# source code that read from file which have 1000000 double number. The source code is below.

public void filereader()
{           
    using (BinaryReader b = new BinaryReader(File.Open("C:\\Users\\Hanieh\\Desktop\\nums.txt", FileMode.Open)))
    {
        int length = (int)b.BaseStream.Length;
        byte[] fileBytes = b.ReadBytes(length);

        for (int ii = 0; ii < fileBytes.Length - 32 ; ii++)
        {
            savg1[ii / 2] = (double)(BitConverter.ToInt16(fileBytes, ii) / 20000.0);// inja error index midee
            ii++;
        }
    }
}

when I run source code to read from text file I have an error that related to savg1 index that is out of bound. I debug step by step and result shows size of length= 24000000 but savg1=1000000. my question is here: how this source code work and how I can fix this problem.

Answer 1

I suggest something like this (File.ReadAllBytes and BitConverter.ToDouble):

  byte[] source = File.ReadAllBytes(@"C:\Users\Hanieh\Desktop\nums.txt");
  double[] data = new double[source.Length / sizeof(double)]; 

  for (int i = 0; i < data.Length; ++i)
    data[i] = BitConverter.ToDouble(source, i * sizeof(double));

Answer 2

I would solve it like:

double[] data;

using (BinaryReader b = new BinaryReader(File.Open("C:\\Users\\Hanieh\\Desktop\\nums.txt", FileMode.Open)))
{
    // create array/buffer for the doubles  (filelength / bytes per double)
    data = new double[b.BaseStream.Length / sizeof(double)];

    // read the data from the binarystream
    for (int i = 0; i < data.Length; i++)
        data[i] = b.ReadDouble();
}

MessageBox.Show("doubles read: " + data.Length.ToString());

Although, your file nums.txt implies that it is a textfile. You might not read it as a binary file.