Is C# Endian sensitive?

Question

Is C# ever Endian sensitive, for example, will code such as this:

int a = 1234567;
short b = *(short*)&i;

always assign the same value to b. If so, what value will it be?

If not, what good ways are there to deal with endianness if code with pointers in?

Answer 1

C# doesn't define the endianness. In reality, yes it will probably always be little-endian (IIRC even on IA64, but I haven't checked), but you should ideally check BitConverter.IsLittleEndian if endianness is important - or just use bit-shifting etc rather than direct memory access.

To quote a few lines from protobuf-net (a build not yet committed):

WriteInt64(*(long*)&value);
if (!BitConverter.IsLittleEndian)
{   // not fully tested, but this *should* work
    Reverse(ioBuffer, ioIndex - 8, 8);
}

i.e. it checks the endianness and does a flip if necessary.

Answer 2

Yes, I believe that code is endian-sensitive. The value of b will be the least-significant bytes on a little-endian processor, and the most-significant bytes on a big-endian processor. To make this simpler to see, let's switch to hex:

using System;

class Test
{
    unsafe static void Main()
    {
        int a = 0x12345678;
        short b = *(short*)&a;
        Console.WriteLine(b.ToString("x"));
    }
}

On my x86 box, that prints "5678" showing that the least-significant bytes were at the "start" of the vaue of a. If you run the same code on a processor running in big-endian mode (probably under Mono) I'd expect it to print "1234".