Reversing byte order in .NET

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().ToArray(), 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.) ``` BitConverter.IsLittleEndian ``` is false. If my computer is big endian, why does this happen? -…

Accepted Answer

`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.

Reversing byte order in .NET

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