We Keep Coding

Hash a file as its being recived

End goal:
Users are uploading a large number of files in different sizes to my web site. And i dont want duplicate files on the disk.
The solution i have been using is a simple SH1 hash of the file when it is uploaded. With code like this:
public static string HashFile(string FileName)
{
using (FileStream stream = File.OpenRead(FileName))
{
SHA1Managed sha = new SHA1Managed();
byte[] checksum = sha.ComputeHash(stream);
string sendCheckSum = BitConverter.ToString(checksum).Replace("-",string.Empty);
return sendCheckSum;
}
}
This "works" fine for smaller files, but its a big pain when the file is 30gb. So i would like to hash the file as im reciving it from the client. I get the file from the client in "chunks" and size of the chunk is not always static.
Code that recives the file.
int chunk = context.Request["chunk"] != null ? int.Parse(context.Request["chunk"]) : 0;
int chunks = context.Request["chunks"] != null ? int.Parse(context.Request["chunks"]) : 0;
string fileName = context.Request["name"] != null ? context.Request["name"] : string.Empty;
HttpPostedFile fileUpload = context.Request.Files[0];
string fullFilePath = Path.Combine(SiteSettings.UploadTempFolder, fileName);
using (var fs = new FileStream(fullFilePath, chunk == 0 ? FileMode.Create : FileMode.Append))
{
var buffer = new byte[fileUpload.InputStream.Length];
fileUpload.InputStream.Read(buffer, 0, buffer.Length);
fs.Write(buffer, 0, buffer.Length);
**// Here i want the hash, when i have the file data in memory.**
}

You can always create your own stream :)
public class ActionStream : Stream
{
private readonly Stream _innerStream;
private readonly Action<byte[], int, int> _readAction;
public ActionStream(Stream innerStream, Action<byte[], int, int> readAction)
{
_innerStream = innerStream;
_readAction = readAction;
}
public override bool CanRead => true;
public override bool CanSeek => false;
public override bool CanWrite => false;
public override long Length => _innerStream.Length;
public override long Position
{
get { return _innerStream.Position; }
set { throw new NotSupportedException(); }
}
public override void Flush() { }
public override int Read(byte[] buffer, int offset, int count)
{
var bytesRead = _innerStream.Read(buffer, offset, count);
_readAction(buffer, offset, bytesRead);
return bytesRead;
}
public override long Seek(long offset, SeekOrigin origin)
{
throw new NotSupportedException();
}
protected override void Dispose(bool disposing)
{
if (disposing)
{
_innerStream.Dispose();
}
base.Dispose(disposing);
}
public override void SetLength(long value) { throw new NotSupportedException(); }
public override void Write(byte[] buffer, int offset, int count)
{
throw new NotSupportedException();
}
}
This allows you to bind together the two stream operations you're doing:
using (var fs = new FileStream(path, chunk == 0 ? FileMode.Create : FileMode.Append))
{
var as = new ActionStream(fileUpload.InputStream,
(buffer, offset, bytesRead) =>
{
fs.Write(buffer, offset, bytesRead);
});
var sha = new SHA1Managed();
var checksum = sha.ComputeHash(as);
}
This assumes that SHA1Manager reads through every single byte of the input stream in order - you should check that. I'm pretty sure that is how it works, though :)

This is a cut and paste from:
Compute a hash from a stream of unknown length in C#
MD5, like other hash functions, does not require two passes.
To start:
HashAlgorithm hasher = ..;
hasher.Initialize();
As each block of data arrives:
byte[] buffer = ..;
int bytesReceived = ..;
hasher.TransformBlock(buffer, 0, bytesReceived, null, 0);
To finish and retrieve the hash:
hasher.TransformFinalBlock(new byte[0], 0, 0);
byte[] hash = hasher.Hash;
This pattern works for any type derived from HashAlgorithm, including MD5CryptoServiceProvider and SHA1Managed.
HashAlgorithm also defines a method ComputeHash which takes a Stream object; however, this method will block the thread until the stream is consumed. Using the TransformBlock approach allows an "asynchronous hash" that is computed as data arrives without using up a thread.

Serial Ports - There is an error in XML document (5, 3870)

Here I have a class I've created for the purpose of sending text-files and images over a serial port:
public class SendItem
{
private byte[] bytes;
private string _fileName;
private string _extension;
private int bytesSin;
public string Extension
{
get { return _extension; }
set { _extension = value; }
}
public string FileName
{
get { return _fileName; }
set { _fileName = value; }
}
public byte[] Bytes
{
get { return bytes; }
set { bytes = value; }
}
public SendItem()
{
}
public void SendFile(SerialPort serialPort1)
{
if (serialPort1.IsOpen)
{
OpenFileDialog OFDialog = new OpenFileDialog();
OFDialog.Title = "Open File";
OFDialog.Filter = "Text Files (*.txt)" + "|*.txt|All files (*.*)|*.*";
OFDialog.InitialDirectory = #"C:\";
bool? userClickedOK = OFDialog.ShowDialog();
if (userClickedOK == true)
{
serialPort1.DiscardInBuffer();
string chosenFile = OFDialog.FileName;
string ext = Path.GetExtension(OFDialog.FileName);
_extension = ext;
byte[] data = File.ReadAllBytes(chosenFile);
SendItem newSendItem = new SendItem();
newSendItem._extension = ext;
newSendItem._fileName = System.IO.Path.GetFileNameWithoutExtension(chosenFile);
newSendItem.bytes = data;
byte[] view = newSendItem.Serialize();
string test = Convert.ToBase64String(data);
serialPort1.Write(newSendItem.Serialize(), 0, newSendItem.Serialize().Length);
//serialPort1.Write(data, 0, data.Length);
}
}
}
public override string ToString()
{
return string.Format("File: {0}{1}",_fileName, _extension);
}
}
There is quite a few redundant things in the SendFile method, but I ask that it would be ignored in favour of the following issue I keep on having.
Whenever I send a text file or just plain chat text from a textbox, the following block of code is executed without triggering the catch:
void _sPort_DataReceived(object sender, SerialDataReceivedEventArgs e)
{
SerialPort sp = (SerialPort)sender;
byte[] data = new byte[sp.BytesToRead];
string message = string.Empty;
sp.Read(data, 0, data.Length);
try
{
SendItem receivedObject = data.Deserialize<SendItem>();
File.WriteAllBytes(#"d:\" + receivedObject.FileName + receivedObject.Extension, receivedObject.Bytes);
message = "File has been recieved.";
sp.Write("File sent.");
}
catch (Exception exp)
{
errors = exp.Message;
message = Encoding.UTF8.GetString(data);
}
App.Current.Dispatcher.Invoke(new Action(() => _response.Add("Friend: " + message)));
}
The problem comes in when I try to send an image... It trigger the catch and gives the exception There is an error in XML document (5, 3870).
My serializer and deserializer I've written is as follow:
public static byte[] Serialize<T>(this T source)
{
XmlSerializer serializer = new XmlSerializer(typeof(T));
MemoryStream stream = new MemoryStream();
serializer.Serialize(stream, source);
byte[] buffer = stream.GetBuffer();
stream.Close();
return buffer;
}
public static T Deserialize<T>(this byte[] source)
{
XmlSerializer serializer = new XmlSerializer(typeof(T));
MemoryStream stream = new MemoryStream(source);
T result = (T)serializer.Deserialize(stream);
stream.Close();
return result;
}
Can anyone point out where I'm making my mistake? I've been debugging it for ages and I can't wrap my head around it.
**EDIT:
I am also including the serialised data for an image I tried to send, after looking at it, it appears my Byte might be too big for the serial port - is there any way to adjust this or allow the serial port to take all of the data sent over?
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

It may or may not be the whole problem, but this is definitely a problem:
byte[] buffer = stream.GetBuffer();
That's quite possibly returning more data than you want - because it's not limiting itself to the length of the stream. MemoryStream has the handy MemoryStream.ToArray method to simplify this:
public static byte[] Serialize<T>(this T source)
{
XmlSerializer serializer = new XmlSerializer(typeof(T));
MemoryStream stream = new MemoryStream();
serializer.Serialize(stream, source);
return stream.ToArray();
}
(There's no benefit from closing a MemoryStream that you've created like this - and if you did feel you needed to, you should use a using statement.)
Likewise your Deserialize method can be simplified to:
public static T Deserialize<T>(this byte[] source)
{
XmlSerializer serializer = new XmlSerializer(typeof(T));
return (T) serializer.Deserialize(new MemoryStream(source));
}
Given that you were effectively trying to deserialize "the XML document - and then probably a load of "blank" data, that may well be the issue.
EDIT: Now there's a second problem:
byte[] data = new byte[sp.BytesToRead];
string message = string.Empty;
sp.Read(data, 0, data.Length);
You're assuming that all of the data is ready to read in a single go. That won't be true if there's more data than the buffer size of the serial port. Instead, if you're trying to put multiple "messages" onto what is effectively a stream, you'll need to either indicate the end of a message (and keep reading until you find that) or write the length of the message before the message itself - then when reading, read that length and then make sure you read that many bytes.
This is a standard issue when you try to use a stream-oriented protocol as a message-oriented protocol.

You should use ToByteArray() instead of GetBuffer().

Why StringWriter.ToString return `System.Byte[]` and not the data?

UnZipFile method writes the data from inputStream to outputWriter.
Why sr.ToString() returns System.Byte[] and not the data?
using (var sr = new StringWriter())
{
UnZipFile(response.GetResponseStream(), sr);
var content = sr.ToString();
}
public static void UnZipFile(Stream inputStream, TextWriter outputWriter)
{
using (var zipStream = new ZipInputStream(inputStream))
{
ZipEntry currentEntry;
if ((currentEntry = zipStream.GetNextEntry()) != null)
{
var size = 2048;
var data = new byte[size];
while (true)
{
size = zipStream.Read(data, 0, size);
if (size > 0)
{
outputWriter.Write(data);
}
else
{
break;
}
}
}
}
}

The problem is on the line:
outputWriter.Write(data);
StringWriter.Write has no overload expecting a byte[]. Therefore, Write(Object) is called instead. And according to MSDN:
Writes the text representation of an object to the text string or stream by calling the ToString method on that object.
Calling ToString on a byte array returns System.byte[], explaining how you get that string in your StringWriter.

The reason is simple:
data is of type byte[]. There is no overload for byte[] on StringWriter so it uses the overload for object. And then calls ToString() on the boxed byte array which simply prints the type.
Your code is equivalent to this:
outputWriter.Write(data.ToString());

theateist,
Looking at the other answers here, I am going to have to agree that the reason for the "ToString()" returning System.Byte[] is because that is what you are putting into it, and everything put into the StringWriter calls it's own "ToString" method when doing so. (i.e. byte[].toString() = "System.byte[]"). In fact the whole idea is that the StringWriter is only ment for writing into a string "buffer" (StringBuilder), so in theory if your file was large enough(bigger than 2048), your output would be "System.Byte[]System.Byte[]" (etc.). Try this to deflate into a memory stream and then read from that stream, may be a better understanding of what you are looking at. (Code not tested, just example).
using (Stream ms = new MemoryStream())
{
UnZipFile(response.GetResponseStream(), ms);
string content;
ms.Position = 0;
using(StreamReader s = new StreamReader(ms))
{
content = s.ReadToEnd();
}
}
public static void UnZipFile(Stream inputStream, Stream outputWriter)
{
using (var zipStream = new ZipInputStream(inputStream))
{
ZipEntry currentEntry;
if ((currentEntry = zipStream.GetNextEntry()) != null)
{
int size = 2048;
byte[] data = new byte[size];
while (true)
{
size = zipStream.Read(data, 0, size);
if (size > 0)
{
outputWriter.Write(data);
}
else
{
break;
}
}
}
}
}
Another idea would actually be to using the endcoding to get the string
public string UnZipFile(Stream inputStream)
{
string tmp;
using(Stream zipStream = new ZipInputStream(inputStream))
{
ZipEntry currentEntry;
if(currentEntry = zipStream.GetNextEntry()) != null)
{
using(Stream ms = new MemoryStream())
{
int size = 2048;
byte[] data = new byte[size];
while(true)
{
if((size = zipStream.Read(data,0,size)) > 0)
ms.Write(data);
else
break;
}
tmp = Encoding.Default.GetString(ms.ToByteArray());
}
}
}
}
return tmp;
}
Or as one last idea, you could actually change your original code to have
outputWriter.Write(Encoding.Default.GetString(data));
Instead of
outputWriter.Write(data);
By the way, please avoid the var keyword in posts, maybe just my pet peev, but code is less readable when utilizing weak types.
StringWriter.Write:MSDN
StringWriter.ToString:MSDN

Loading Image from a string

Due to some datastorage limitations (noSQL) I need to store images as strings.
How can I serialize the image Bitmap to string and back.
Here is how I am doing it:
Uri testImageUri = new Uri("/DictionaryBasedVM;component/test.jpg", UriKind.Relative);
StreamResourceInfo sri = Application.GetResourceStream(testImageUri);
var stringData = GetString(sri.Stream);
ImageSource = stringData;
Where ImageControl is just a silverlight image control defined in xaml.
I am using the following utility functions:
//For testing
public static string GetString(Stream stream)
{
byte[] byteArray = ReadFully(stream);
return Encoding.Unicode.GetString(byteArray,0,byteArray.Length);
}
public static byte[] ReadFully(Stream input)
{
byte[] buffer = new byte[16 * 1024];
using (MemoryStream ms = new MemoryStream())
{
int read;
while ((read = input.Read(buffer, 0, buffer.Length)) > 0)
{
ms.Write(buffer, 0, read);
}
return ms.ToArray();
}
}
And the following property:
private string _ImageSource = "";
public string ImageSource
{
set
{
_ImageSource = value;
byte[] byteArray = Encoding.Unicode.GetBytes(value);
MemoryStream imageStream = new MemoryStream(byteArray);
BitmapImage imageSource = new BitmapImage();
imageSource.SetSource(imageStream);
ImageControl.Source = imageSource;
}
get
{
return _ImageSource;
}
}
I get the error : "Catastrophic failure (Exception from HRESULT: 0x8000FFFF (E_UNEXPECTED))" as shown:
Even if I don't store it as a string I am still curious why I can't do this.

Unicode is probably not the best encoding for this purpose. You'd be better of Base64 encoding the byte[] and storing that.

Base64 would add 30% size to the string. I don't think it's necessary here. You can safely "cast" bytes to chars and vice versa for this particular binary need, as long as you don't do anything with the string. Here is some code that seems to work:
// usage example
string encoded = FileToString("myimage.png");
Console.WriteLine(s.Length);
FileFromString(encoded, "copy.png");
public static void FileFromString(string input, string filePath)
{
if (input == null)
throw new ArgumentNullException("input");
if (filePath == null)
throw new ArgumentNullException("filePath");
using (FileStream stream = new FileStream(filePath, FileMode.OpenOrCreate, FileAccess.Write, FileShare.None))
{
byte[] buffer = FromString(input);
stream.Write(buffer, 0, buffer.Length);
}
}
public static byte[] FromString(string input)
{
if (input == null)
throw new ArgumentNullException("input");
char[] cbuffer = input.ToCharArray();
byte[] buffer = new byte[cbuffer.Length];
for (int i = 0; i < buffer.Length; i++)
{
buffer[i] = (byte)cbuffer[i];
}
return buffer;
}
public static string FileToString(string filePath)
{
if (filePath == null)
throw new ArgumentNullException("filePath");
using (FileStream stream = new FileStream(filePath, FileMode.Open, FileAccess.Read, FileShare.Write))
{
return ToString(stream);
}
}
public static string ToString(Stream input)
{
if (input == null)
throw new ArgumentNullException("input");
StringBuilder sb = new StringBuilder();
byte[] buffer = new byte[4096];
char[] cbuffer = new char[4096];
int read;
do
{
read = input.Read(buffer, 0, buffer.Length);
for (int i = 0; i < read; i++)
{
cbuffer[i] = (char)buffer[i];
}
sb.Append(new string(cbuffer, 0, read));
}
while (read > 0);
return sb.ToString();
}
However, it's possible the system where you store strings may not like strings that contains the 0 or other special numbers. In this case, base64 is still an option.

Have you tried using Convert.ToBase64String and Convert.FromBase64String methods instead? I'd guess the unicode GetString/GetBytes don't work, as your byte arrays don't line up with 'known characters'.

How to get a MemoryStream from a Stream in .NET?

I have the following constructor method which opens a MemoryStream from a file path:
MemoryStream _ms;
public MyClass(string filePath)
{
byte[] docBytes = File.ReadAllBytes(filePath);
_ms = new MemoryStream();
_ms.Write(docBytes, 0, docBytes.Length);
}
I need to change this to accept a Stream instead of a file path. Whats the easiest/most efficient way to get a MemoryStream from the Stream object?

In .NET 4, you can use Stream.CopyTo to copy a stream, instead of the home-brew methods listed in the other answers.
MemoryStream _ms;
public MyClass(Stream sourceStream)
_ms = new MemoryStream();
sourceStream.CopyTo(_ms);
}

Use this:
var memoryStream = new MemoryStream();
stream.CopyTo(memoryStream);
This will convert Stream to MemoryStream.

If you're modifying your class to accept a Stream instead of a filename, don't bother converting to a MemoryStream. Let the underlying Stream handle the operations:
public class MyClass
{
Stream _s;
public MyClass(Stream s) { _s = s; }
}
But if you really need a MemoryStream for internal operations, you'll have to copy the data out of the source Stream into the MemoryStream:
public MyClass(Stream stream)
{
_ms = new MemoryStream();
CopyStream(stream, _ms);
}
// Merged From linked CopyStream below and Jon Skeet's ReadFully example
public static void CopyStream(Stream input, Stream output)
{
byte[] buffer = new byte[16*1024];
int read;
while((read = input.Read (buffer, 0, buffer.Length)) > 0)
{
output.Write (buffer, 0, read);
}
}

You can simply do:
var ms = new MemoryStream(File.ReadAllBytes(filePath));
Stream position is 0 and ready to use.

You will have to read in all the data from the Stream object into a byte[] buffer and then pass that into the MemoryStream via its constructor. It may be better to be more specific about the type of stream object you are using. Stream is very generic and may not implement the Length attribute, which is rather useful when reading in data.
Here's some code for you:
public MyClass(Stream inputStream) {
byte[] inputBuffer = new byte[inputStream.Length];
inputStream.Read(inputBuffer, 0, inputBuffer.Length);
_ms = new MemoryStream(inputBuffer);
}
If the Stream object doesn't implement the Length attribute, you will have to implement something like this:
public MyClass(Stream inputStream) {
MemoryStream outputStream = new MemoryStream();
byte[] inputBuffer = new byte[65535];
int readAmount;
while((readAmount = inputStream.Read(inputBuffer, 0, inputBuffer.Length)) > 0)
outputStream.Write(inputBuffer, 0, readAmount);
_ms = outputStream;
}

I use this combination of extension methods:
public static Stream Copy(this Stream source)
{
if (source == null)
return null;
long originalPosition = -1;
if (source.CanSeek)
originalPosition = source.Position;
MemoryStream ms = new MemoryStream();
try
{
Copy(source, ms);
if (originalPosition > -1)
ms.Seek(originalPosition, SeekOrigin.Begin);
else
ms.Seek(0, SeekOrigin.Begin);
return ms;
}
catch
{
ms.Dispose();
throw;
}
}
public static void Copy(this Stream source, Stream target)
{
if (source == null)
throw new ArgumentNullException("source");
if (target == null)
throw new ArgumentNullException("target");
long originalSourcePosition = -1;
int count = 0;
byte[] buffer = new byte[0x1000];
if (source.CanSeek)
{
originalSourcePosition = source.Position;
source.Seek(0, SeekOrigin.Begin);
}
while ((count = source.Read(buffer, 0, buffer.Length)) > 0)
target.Write(buffer, 0, count);
if (originalSourcePosition > -1)
{
source.Seek(originalSourcePosition, SeekOrigin.Begin);
}
}

How do I copy the contents of one stream to another?
see that. accept a stream and copy to memory. you should not use .Length for just Stream because it is not necessarily implemented in every concrete Stream.

public static void Do(Stream in)
{
_ms = new MemoryStream();
byte[] buffer = new byte[65536];
while ((int read = input.Read(buffer, 0, buffer.Length))>=0)
_ms.Write (buffer, 0, read);
}

byte[] fileData = null;
using (var binaryReader = new BinaryReader(Request.Files[0].InputStream))
{
fileData = binaryReader.ReadBytes(Request.Files[0].ContentLength);
}