Windows – safely assume that Windows installations will always be little-endian


I'm writing a userspace filesystem driver on Windows and endianness conversions are something I've been dealing with, as this particular filesystem always stores values in little-endian format and the driver is expected to convert them (if necessary) for the CPU it's running on. However, I find myself wondering if I even need to worry about endianness conversions, since as far as I can tell, desktop Windows only supports little-endian architectures (IA32, x86-84, etc.), and therefore, the on-disk little-endian values are perfectly fine sans conversion. Is this observation accurate, and if so, is it generally acceptable to make the assumption that Windows will always be running on little-endian hardware? Additionally, is it even possible (in 2011) to run Windows on a big-endian emulator or something, such that one could even test for endianness issues?

Edit: For additional clarity, the way my code currently works, I do an endianness check at startup time, and then every time I load a value off the disk, I run it through an inline function that uses an intrinsic to change endianness if the architecture is big-endian. The problem is, I don't know if I might have missed one or more of those places where I needed to do a conversion and the easiest way to see if I screwed up is to run the program on a big-endian architecture. So I'm interested in knowing (a) if it's even necessary to do these checks since Windows doesn't ordinarily run on little-endian platforms (today anyway), and (b) how I could possibly test my code, seeing as I can't think of a way to run Windows on a big-endian architecture, and manually reversing all the multibyte values on disk still involves a manual process that I might well screw up.

Best Solution

All versions of Windows that you'll see are little-endian, yes. The NT kernel actually runs on a big-endian architecture even today.

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