Returning a stream from File.OpenRead()

Solution 1:

You forgot to seek:

str.CopyTo(data);
data.Seek(0, SeekOrigin.Begin); // <-- missing line
byte[] buf = new byte[data.Length];
data.Read(buf, 0, buf.Length);

Solution 2:

Options:

• Use data.Seek as suggested by ken2k

• Use the somewhat simpler Position property:

  data.Position = 0;
  

• Use the ToArray call in MemoryStream to make your life simpler to start with:

  byte[] buf = data.ToArray();
  

The third option would be my preferred approach.

Note that you should have a using statement to close the file stream automatically (and optionally for the MemoryStream), and I'd add a using directive for System.IO to make your code cleaner:

byte[] buf;
using (MemoryStream data = new MemoryStream())
{
    using (Stream file = TestStream())
    {
        file.CopyTo(data);
        buf = data.ToArray();
    }
}

// Use buf

You might also want to create an extension method on Stream to do this for you in one place, e.g.

public static byte[] CopyToArray(this Stream input)
{
    using (MemoryStream memoryStream = new MemoryStream())
    {
        input.CopyTo(memoryStream);
        return memoryStream.ToArray();
    }
}

Note that this doesn't close the input stream.