Convert Stream to Byte Array

Question

I created a simple extension to convert a Stream to Byte array:

public static Byte[] ToByte(this Stream value) 
{
  if (value == null)
    return null;

  if (value is MemoryStream)
  {
    return ((MemoryStream)value).ToArray();
  }
  else 
  {
     using (MemoryStream stream = new MemoryStream()) 
     {
          value.CopyTo(stream);
          return stream.ToArray();
      }
  }

}

This works fine the first time I use it ...

If I use it a second time on the same stream the resulting Byte[] is Byte[0].

I debugged and I think this happens because after the first conversion the Stream index goes to the end of the stream.

What is the correct way to fix this?

Keep in mind, stream is like a pointer to the memory. When you actually finish converting to the array, the pointer is at the end. Make sure to reset it if you do want to run it again. — Luo Sen
– Luo Sen, Commented Oct 8, 2014 at 22:38

BradleyDotNET · Accepted Answer · 2014-10-08 21:44:41Z

8

As you say, you are at the end of the stream after the first read. Thus, you need to reset the memory stream with:

stream.Seek(0, SeekOrigin.Begin);

Or

stream.Position = 0;

Before reading it again. Note that CanSeek on the stream must be true for either approach to work. MemoryStream is fine, some other streams will have this set to false.

answered Oct 8, 2014 at 21:44

BradleyDotNET

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3 Comments

AJ Richardson Over a year ago

This works as long as the stream is seekable (which it should be if it is a MemoryStream). If you are using a different type of stream that is not seekable, you will be out of luck unfortunately.

Miguel Moura Over a year ago

I also found the option stream.Position = 0; It seems to also work. Is it equivalent to your suggestion?

BradleyDotNET Over a year ago

@MDMoura Yes, it is equivalent, and also requires the stream to be seekable.

dee-see · Accepted Answer · 2014-10-08 21:46:49Z

1

set the stream index to the beginning.

stream.Seek(0, SeekOrigin.Begin);

edited Oct 8, 2014 at 21:46

dee-see

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answered Oct 8, 2014 at 21:46

mreiterer

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