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2013-07-22T11:09:27.957
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2017-05-23T10:25:15.457
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I agree with Henk Holtermans Response. You'll have to do the split yourself. What you can do however is to not calculate the complete hash with a single ComputeHash call, but to do it in chunks of Bytes with <code>TransformBlock</code> calls. <a href="https://stackoverflow.com/questions/2124468/possible-to-calculate-md5-or-other-hash-with-buffered-reads">See here for an example</a>. With this you can yourself instantiate a buffer of a size and give it as Parameters to the consequent parallel TransformBlock Calls. Edit: here is some code that does the Job <pre><code> static void Hash2Md5inParallel() { string strFullPath = YourFilePathGoesHere; byte[] Buffer = new Byte[1000]; //Instantiate Buffer to copy bytes. byte[] DumpBuffer = new Byte[1000]; //Send output to bin. System.Security.Cryptography.MD5 md5_1 = new System.Security.Cryptography.MD5CryptoServiceProvider(); System.Security.Cryptography.MD5 md5_2 = new System.Security.Cryptography.MD5CryptoServiceProvider(); System.IO.FileStream file = new System.IO.FileStream(strFullPath, FileMode.Open); file.Seek(1000, SeekOrigin.Begin); //skip some chars if need be int BytesToHash = 0; do { BytesToHash = file.Read(Buffer, 0, 1000); md5_1.TransformBlock(Buffer, 0, BytesToHash, DumpBuffer, 0); //enter some code to skip some bytes for the other hash if you like... md5_2.TransformBlock(Buffer, 0, BytesToHash, DumpBuffer, 0); } while (BytesToHash > 0); //Repeat until no more bytes. //call TransformFinalBlock to finish hashing - empty block is enough md5_1.TransformFinalBlock(new byte[0], 0, 0); md5_2.TransformFinalBlock(new byte[0], 0, 0); //Get both Hashs. byte[] hash1 = md5_1.Hash; byte[] hash2 = md5_2.Hash; } </code></pre>
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1. POsplit a FileStream for multiple consumers in C#
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