.net reversing byte order?

Question

In the code below, why do X and Y take on different values than what I would think intuitively? If the bytes 0-7 are written to the buffer, shouldn't the resulting long have bytes in the same order? It's like it's reading the long values in reverse order.

x   0x0706050403020100  long
y   0x0706050403020100  long
z   0x0001020304050607  long

MemoryStream ms = new MemoryStream();
byte[] buffer = new byte[] { 0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07 };
ms.Write(buffer, 0, buffer.Length);
ms.Flush();
ms.Position = 0;

BinaryReader reader = new BinaryReader(ms);
long x = reader.ReadInt64();
long y = BitConverter.ToInt64(buffer, 0);
long z = BitConverter.ToInt64(buffer.Reverse<byte>().ToArray<byte>(), 0);

byte[] xbytes = BitConverter.GetBytes(x);
byte[] ybytes = BitConverter.GetBytes(y);
byte[] zbytes = BitConverter.GetBytes(z);

(I don't know what to tag this question, beyond just .net).

EDIT (Bounty started here)

BitConverter.IsLittleEndian 

is false. If my computer is big endian, why does this happen?

EDIT 2

• This is a Windows 7 64-bit machine
• Intel Core2 Quad Q9400 2.66GHz LGA 775 95W Quad-Core Processor Model BX80580Q9400
• SUPERMICRO MBD-C2SBX+-O LGA 775 Intel X48 ATX Intel Motherboard

The results of this code (in response to Jason's comment):

byte[] buffer = new byte[] { 0x00, 0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07 }; 
long y = BitConverter.ToInt64(buffer, 1);
Console.WriteLine(BitConverter.IsLittleEndian);
Console.WriteLine(y);

Result:

False    
506097522914230528

Answer 1

You're on a little endian machine, where integers are stored least-significant byte first.

Answer 2

BinaryReader assumes Little Endian order: http://msdn.microsoft.com/en-us/library/system.io.binaryreader.readint64.aspx

Answer 3

If you care about the endian-ness of your bytes, Jon Skeet wrote a class to allow you to choose the endian-order when you do the conversion.

See http://stackoverflow.com/questions/217980/c-little-endian-or-big-endian

Answer 4

BinaryReader.ReadInt64 is little endian by design. From the documentation:

BinaryReader reads this data type in little-endian format.

In fact, we can inspect the source for BinaryReader.ReadInt64 using Reflector.

public virtual long ReadInt64() {
    this.FillBuffer(8);
    uint num = (uint) (((this.m_buffer[0] |
              (this.m_buffer[1] << 0x08)) |
              (this.m_buffer[2] << 0x10)) |
              (this.m_buffer[3] << 0x18));
    uint num2 = (uint) (((this.m_buffer[4] |
               (this.m_buffer[5] << 0x08)) |
               (this.m_buffer[6] << 0x10)) |
               (this.m_buffer[7] << 0x18));
    return (long) ((num2 << 0x20) | num);
}

Showing that BinaryReader.ReadInt64 reads as little endian independent of the underlying machine architecture.

Now, BitConverter.ToInt64 is suppose to respect the endianness of your underlying machine. In Reflector we can see

public static unsafe long ToInt64(byte[] value, int startIndex) {
    // argument checking elided
    fixed (byte* numRef = &(value[startIndex])) {
        if ((startIndex % 8) == 0) {
            return *(((long*) numRef));
        }
        if (IsLittleEndian) {
            int num = (numRef[0] << 0x00) |
                      (numRef[1] << 0x08) |
                      (numRef[2] << 0x10) |
                      (numRef[3] << 0x18);
            int num2 = (numRef[4] << 0x00) |
                       (numRef[5] << 0x08) |
                       (numRef[6] << 0x10) |
                       (numRef[7] << 0x18);
            return (((long) ((ulong) num)) | (num2 << 0x20));
        }
        int num3 = (numRef[0] << 0x18) |
                   (numRef[1] << 0x10) |
                   (numRef[2] << 0x08) |
                   (numRef[3] << 0x00);
        int num4 = (numRef[4] << 0x18) |
                   (numRef[5] << 0x10) |
                   (numRef[6] << 0x08) |
                   (numRef[7] << 0x00);
        return (((long) ((ulong) num4)) | (num3 << 0x20));
}

So what we see here is that if startIndex is congruent to zero modulo eight that a direct cast is done from eight bytes starting at address numRef. This case is handled specially because of alignment issues. The line of code

return *(((long *) numRef));

translates directly to

    ldloc.0      ;pushes local 0 on stack, this is numRef
    conv.i       ;pop top of stack, convert to native int, push onto stack
    ldind.i8     ;pop address off stack, indirect load from address as long
    ret          ;return to caller, return value is top of stack

So we see that in this case the key is the ldind.i8 instruction. The CLI is agnostic about the endianness of the underlying machine. It lets the JIT compiler handle that issue. On a little-endian machine, ldind.i8 will load higher addresses into more significant bits and on a big-endian machine ldind.i8 will load higher addresses into less significant bytes. Therefore, in this case, endianness is handled properly.

In the other case, you can see that there is an explicit check of the static property BitConverter.IsLittleEndian. In the case of little endian the buffer is interpreted as little endian (so that memory { 0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07 } is interpreted as the long 0x0706050403020100) and in case of big endian the buffer is interpreted as big endian (so that memory { 0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07 } is interpreted as the long 0x0001020304050607). So, for BitConverter it all comes down to the endianness of the underyling machine. I note that you're on an Intel chip on Windows 7 x64. Intel chips are little endian. I note that in Reflector, the static constructor for BitConverter is defined as the following:

static BitConverter() {
    IsLittleEndian = true;
}

This is on my Windows Vista x64 machine. (It could differ on, say, .NET CF on an XBox 360.) There is no reason for Windows 7 x64 to be any different. Consequently, are you sure that BitConverter.IsLittleEndian is false? It should be true and therefore the behavior that you are seeing is correct.