No, and it would be wrong for it to do so, because you've given it a very explicit set of instructions about what to do: "Give me the value of the byte of memory at the start of x."

To do what you're asking, you'd do something like this:

  unsigned char c = (unsigned char)x;

That will give you the low order byte of x. But to do that, on a big endian system, when you've told it to get you the byte at the base address of x, is simply wrong behavior. At least in C. I can't speak to higher level languages since I don't work in them.

To expand slightly on Syonyk said: the compiler cannot do it, because the object is stored between addresses c and c + sizeof(unsigned int). You can use this information to, for example, copy the object to another place with memcpy, and that of course wouldn't work if c wasn't pointing to the "leftmost" byte in the memory.

Unless, I suppose, sizeof was negative :).

If you wanted to return 04 on big-endian architectures, you can use a binary mask - (int &0xFF).

Since this compiles to FF 00 00 00 in big-endian and 00 00 00 FF in little-endian, it would work on both platforms.

If you’re reading a file in binary format from disk, though, you always have to know whether the byte you are reading is little-endian or big-endian on disk.