Why my 2 bytes (0x00 0x40) that should be 64 produce 16384 instead?

Question

Please see this file using Hex Editor:

I am reading the 2 bytes this way:

BinaryReader binaryReader;
int wlen = binaryReader.ReadUInt16();

When I read this 2 bytes my BinaryReader.BaseStream.Position is 14 but wlen = 16384 and this should be 64. What am I doing wrong?

I'll give you a hint... 0x4000 == 16384, you need to swap the bytes. — Ron Beyer, Nov 23 '15 at 16:46
Show the *real* code. This fragment just generates a `NullReferenceException`. — Ondrej Tucny, Nov 23 '15 at 16:46
What @Ron said. Before your start parsing arbitrary bytes, learn about endianness. — CodeCaster, Nov 23 '15 at 16:47

score 3 · Answer 1 · edited May 23 '17 at 12:03

Your data file appears to be produced by some big-endian writer, while BinaryReader reads data assuming the little-endian representation.

If you have control over the format of the file, changing the writer to produce little endian representation would let you avoid making changes to your C# program.

There are several ways to read big-endian data in .NET, too. You could use BitConverter, but reversing the bytes manually is probably the most performant one:

public static short ToInt16Be(byte[] buf, int pos) {
    return (short)(buf[pos]<<8 | buf[pos+1]);
}

score 0 · Answer 2 · answered Nov 23 '15 at 16:48

0

binaryReader.ReadUInt16() reads using little endian.

See: https://msdn.microsoft.com/en-us/library/system.io.binaryreader.readuint16(v=vs.110).aspx

answered Nov 23 '15 at 16:48

fjardon

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Why my 2 bytes (0x00 0x40) that should be 64 produce 16384 instead?

2 Answers2