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.NET Forum / Languages / C# / October 2007BitConverter.ToInt16 doesn't work correct
Hello. I have a problem with getting short value from 2 byte array. I have this code. There are 2 short values in bytes. byte[] cast = { 18, 152, 00, 80 }; Int32 port = BitConverter.ToInt16(cast, 0); // -26606 - wrong! Int32 port2 = BitConverter.ToInt16(cast, 2); // 20480 - wrong! //port and port2 get not correct values //and I found solution in google short result1 = (short)(((short)cast[1]) | (short)(cast[0] * 256)); //4760 - correct! short result2 = (short)(((short)cast[3]) | (short)(cast[2] * 256)); //80 - correct! why it happends? How make BitConverter work correct? > Hello. I have a problem with getting short value from 2 byte array. > I have this code. There are 2 short values in bytes. > > byte[] cast = { 18, 152, 00, 80 }; > Int32 port = BitConverter.ToInt16(cast, 0); // -26606 - wrong! -26606 is the correct result. Your 16-bit number is 0x9812, which when interpreted as a signed 16-bit integer is in fact -26606. > Int32 port2 = BitConverter.ToInt16(cast, 2); // 20480 - wrong! Likewise, 20480 is the correct result. Your 16-bit number is 0x5000, which when interpreted as a signed 16-bit integer is in fact 20480. > //port and port2 get not correct values It appears to me that you are confused about the native, little-endian number format used by the Intel hardware running your code, and the "network order" used in TCP/IP, which is big-endian. > //and I found solution in google > short result1 = (short)(((short)cast[1]) | (short)(cast[0] * 256)); //4760 - > correct! > short result2 = (short)(((short)cast[3]) | (short)(cast[2] * 256)); //80 - > correct! Those are only the correct values if your data is in big-endian format. If your data is in big-endian format, it is a mistake to ask BitConverter, which operates only on little-endian data, to convert it. > why it happends? How make BitConverter work correct? Don't lie to it about the data you're passing it. If you want it to convert data that you have that starts in big-endian format, you need to convert that data to little-endian before you pass it to BitConverter by swapping the bytes so that they are in the correct order. Pete > Don't lie to it about the data you're passing it. If you want it to > convert data that you have that starts in big-endian format, you need to > convert that data to little-endian before you pass it to BitConverter How to make this convert to little-endian? >> Don't lie to it about the data you're passing it. If you want it to >> convert data that you have that starts in big-endian format, you need to >> convert that data to little-endian before you pass it to BitConverter > > How to make this convert to little-endian? It was in my post. You trimmed out the procedure I described in the above quote. Beyond that, if you are going to be using BitConverter, you should probably learn more about actual data formats, and in this case especially about byte-order and how to deal with differences in that. And in particular, you seem to be doing network address work here, where the byte-ordering for the format is very important, since many network APIs use big-endian even on computers that are natively little-endian. .NET translates for you, when you give it plain little-endian integers, but it's quite common to see network addresses stored as big-endian. Basically, you should probably take a step back and gain better familiarity with the byte-ordering issue before you attempt to write code that is dependent on byte-ordering. Pete Here is a function I wrote that is designed for a BinaryReader: byte[] ReadBytes(BinaryReader reader, int length, bool littleEndian) { byte[] bytes = new byte[length]; if (littleEndian) return reader.ReadBytes(length); else { for (int i = length - 1; i > -1; i--) bytes[i] = reader.ReadByte(); return bytes; } } Simply put, when the byte order is Big-Endian, you reverse the order of the bytes.  SignatureHTH, Kevin Spencer Microsoft MVP DSI PrintManager, Miradyne Component Libraries: http://www.miradyne.net >>> Don't lie to it about the data you're passing it. If you want it to >>> convert data that you have that starts in big-endian format, you need to [quoted text clipped - 19 lines] > > Pete Note that this will only work if the binary data is a single value (however many bytes that is for the type in question); if it is a true stream of successive values you could be in a pickle. Minor aside; personally I would have called reader.ReadBytes() to fill the buffer in either case, and then (if littleEndian) reversed the buffer locally. Horses for courses... Marc > Here is a function I wrote that is designed for a BinaryReader: > [quoted text clipped - 13 lines] > Simply put, when the byte order is Big-Endian, you reverse the order of > the bytes. Doesn't this reverse the order of the entire array? Michael from inspection, it appears to - hence my reply; valid in some circumstances, but by no means all. For the paranoid: if you are going to start reversing things to manually compensate for CPU endian-ness (for BitConverter usage), you might also want to double-check that you need to! BitConverter.IsLittleEndian will tell you how the CPU works. Itanium leans the other way... so to speak. Marc > from inspection, it appears to - hence my reply; valid in some > circumstances, but by no means all. [quoted text clipped - 3 lines] > want to double-check that you need to! BitConverter.IsLittleEndian will > tell you how the CPU works. Itanium leans the other way... so to speak. Another option would be to convert the bytes back to short and then do something like if(BitConverter.IsLittleEndian) //or should that be not? { res = ((res & 0xff00) >> 8) || ((res & 0xff) << 8) } If res wasn't unsigned then might need to convert first. This might work too: res = (res >> 8) || (res << 8) Michael That should probably be a bitwise "or" (|), not a logical "or" (||). And again, it only helps with a few cases (single, simple values like short). IMO, the better solution is to leave the byte-stream alone, and use EndianBitConverter. This will work for multiple sequenced values, and all data types. Marc Hi Michael, Yes, it does reverse the order of an entire array. I should have mentioned that this is part of a set of tools for parsing files, the most general of the tools. I created it for working with GeoTiff images, which can be either Big-Endian or Little-Endian. Once the Endian-ness of the file is determined, the class actually sets a global variable, rather than passing a boolean value to any function. Notice the BinaryReader instance parameter. This is due to the fact that it is parsing a file. The actual tool set has multiple methods that employ this method to read the data in the file a "chunk" at a time. The data type of each "chunk" is known, and passed to this method as the length parameter. There are ReadShort, ReadLong, ReadInt32, ReadUInt32, ReadInt16, etc, methods that employ this method. I can understand your concern, as it is only the byte order within each individual "chunk" of data that is reversed, and reading the entire file into such a method would not work. I hope this clarifies your question. SignatureHTH, Kevin Spencer Microsoft MVP DSI PrintManager, Miradyne Component Libraries: http://www.miradyne.net >> Here is a function I wrote that is designed for a BinaryReader: >> [quoted text clipped - 17 lines] > > Michael Note that Jon Skeet has an EndianBitConverter (below) which may be of use: http://www.yoda.arachsys.com/csharp/miscutil/ Marc > Hello. I have a problem with getting short value from 2 byte array. > I have this code. There are 2 short values in bytes. [quoted text clipped - 12 lines] > > why it happends? How make BitConverter work correct? You could just do cast[0] * 256 + cast[1]. You will need to work out how to handle values of 128 or above for the first byte. Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. Simply choose a title, complete the application form and submit it. Read more ...
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