endianness

Due to poor documentation and lack of experience with Netty, i faced with little problem. I have no clue how can i set a default ByteOrder. I need a Little-Endian set by default. I'll be glad, if someone will give me some hints about this. You could use Bootstrap.setOption() to do this. serverBootstrap.setOption("child.bufferFactory", new HeapChannelBufferFactory(ByteOrder.LITTLE_ENDIAN)); ... or ... clientBootstrap.setOption("bufferFactory", new HeapChannelBufferFactory(ByteOrder.LITTLE_ENDIAN)); 来源： https://stackoverflow.com/questions/9758537/netty-and-byteorder

I'm working on a PowerPC machine with in-core crypto. I'm having trouble porting AES key expansion from big endian to little endian using built-ins. Big endian works, but little endian does not. The algorithm below is the snippet presented in an IBM blog article . I think I have the issue isolated to line 2 below: typedef __vector unsigned char uint8x16_p8; uint8x64_p8 r0 = {0}; r3 = vec_perm(r1, r1, r5); /* line 1 */ r6 = vec_sld(r0, r1, 12); /* line 2 */ r3 = vcipherlast(r3, r4); /* line 3 */ r1 = vec_xor(r1, r6); /* line 4 */ r6 = vec_sld(r0, r6, 12); /* line 5 */ r1 = vec_xor(r1, r6); /*

I'm trying to write a byteswap routine for a C++ program running on Win XP. I'm compiling with Visual Studio 2008. This is what I've come up with: int byteswap(int v) // This is good { return _byteswap_ulong(v); } double byteswap(double v) // This doesn't work for some values { union { // This trick is first used in Quake2 source I believe :D __int64 i; double d; } conv; conv.d = v; conv.i = _byteswap_uint64(conv.i); return conv.d; } And a function to test: void testit() { double a, b, c; CString str; for (a = -100; a < 100; a += 0.01) { b = byteswap(a); c = byteswap(b); if (a != c) { str

I'm working with Node.JS. Node's buffers support little-endian UCS-2, but not big-endian, which I need. How would I do so? According to wikipedia, UCS-2 should always be big-endian so it's odd that node only supports little endian. You might consider filing a bug. That said, switching endian-ness is fairly straight-forward since it's just a matter of byte order. So just swap bytes around to go back and forth between little and big endian, like so: function swapBytes(buffer) { var l = buffer.length; if (l & 0x01) { throw new Error('Buffer length must be even'); } for (var i = 0; i < l; i += 2)

I currently have a Java ByteBuffer that already has the data in Big Endian format. I then want to write to a binary file as Little Endian. Here's the code which just writes the file still in Big Endian: public void writeBinFile(String fileName, boolean append) throws FileNotFoundException, IOException { FileOutputStream outStream = null; try { outStream = new FileOutputStream(fileName, append); FileChannel out = outStream.getChannel(); byteBuff.position(byteBuff.capacity()); byteBuff.flip(); byteBuff.order(ByteOrder.LITTLE_ENDIAN); out.write(byteBuff); } finally { if (outStream != null) {

I cant tell, does boost asio handle endian? Asio does convert things like port into network order. The conversion functions are not exposed as part of the official interface and are hidden in the detail namespace instead (e.g. boost::asio::detail::socket_ops::host_to_network_short ). boost::asio::socket_'s do not perform any byte order conversion. 来源： https://stackoverflow.com/questions/524453/boost-asio-and-endian

I've been looking for the answer for how to use BSWAP for lower 32-bit sub-register of 64-bit register. For example, 0x0123456789abcdef is inside RAX register, and I want to change it to 0x01234567efcdab89 with a single instruction (because of performance). So I tried following inline function: #define BSWAP(T) { \ __asm__ __volatile__ ( \ "bswap %k0" \ : "=q" (T) \ : "q" (T)); \ } And the result was 0x00000000efcdab89 . I don't understand why the compiler acts like this. Does anybody know the efficient solution? Ah, yes, I understand the problem now: the x86-64 processors implicitly zero