Hi.
On 25 September 2011 18:10, Brent Yorgey
You must at least agree that it is short.
Not trying to start language wars here, but it is not terribly short for what it does. The following code does the same thing in C#, and isn't far longer. And it has more or less a one-to-one correspondence to the given Haskell code; open a file for reading, open a file for writing, read some number of bytes, apply the transformation, write it to the output file. Flushing the input/output buffers and closing the files are handled by the using construct, similar to withFile in the Haskell example. int chunksize = 4096; using (var r = new BinaryReader(File.OpenRead("infile"))) using (var w = new BinaryWriter(File.OpenWrite("outfile"))) for (var buffer = r.ReadBytes(chunksize); buffer.Length > 0; buffer = r.ReadBytes(chunksize)) w.Write(Array.ConvertAll(buffer, p => (byte) ~p)); I think the habit of using quite a few operators in Haskell does make the learning curve steeper. I am not trying to say that the C# code is *better. *Just that the Haskell code is not terribly short in this case and it can be a bit cryptic for a newbie. Best, Ozgur