endianness

问题 Suppose I have unsigned long long x = 0x0123456789ABCDEF . Which of the following is correct? (I can verify only the first one): On a 32-bit little-endian processor, it will appear in memory as 67 45 23 01 EF CD AB 89 . On a 64-bit little-endian processor, it will appear in memory as EF CD AB 89 67 45 23 01 . On a 32-bit big-endian processor, it will appear in memory as 01 23 45 67 89 AB CD EF . On a 64-bit big-endian processor, it will appear in memory as 01 23 45 67 89 AB CD EF . 回答1: The

问题 My task is to convert a data file from big endian to little endian & vice versa using C. I have been looking online for about 3 hours now for other examples and reading my text book, however I am so stuck on how to even start this function. So far I have the order of events correct (1 through 4) but inside my convert_and_save function do I have to create a char array using → char buffer[4]; ? Can someone please help me? even if you just give me clues on what to look up, I would greatly

问题 Is the endianness of format params guaranteed in RIFF WAV files? I have heard conflicting answers to this including references to a RIFX file format. 回答1: Yes. If the file starts with RIFF, then it's little endian. If it starts with FFIR or RIFX, then it's probably not. Generally, supporting the WAV format means supporting RIFF files, although adding RIFX support should not prove difficult. The AES31 specification for BWF (Broadcast Wave Format) references this specification for RIFF: http:/

问题 I'm reading an ESRI Shapefile, and to my dismay it uses big endian and little endian at different points (see, for instance, the table at page 4, plus the tables from page 5 to 8). So I created two functions in C++, one for each endianness. uint32_t readBig(ifstream& f) { uint32_t num; uint8_t buf[4]; f.read((char*)buf,4); num = buf[3] | buf[2]<<8 | buf[1]<<16 | buf[0]<<24; return num; } uint32_t readLittle(ifstream& f) { uint32_t num; f.read(reinterpret_cast<char *>(&num),4); //f.read((char*

问题 Is it possible to compile some Linux Kernel and run it over QEMU, emulating some Big Endian ARM processor? If QEMU is not capable of that, I'd love to hear about other system emulators than can. My basic goal is to run and debug dedicated Big Endian ELFs in as much as possible native environment. Every close solution or idea would help! 回答1: QEMU has support for big-endian ARM CPUs, but it does not currently have support for emulation of any specific machines (boards) which have big-endian

问题 Ok, if i've got a binary file encoded either in little endian or big endian under .NET, what is the best way to read / write to it? In the .NET framework i've only managed to found BinaryWritters / BinaryReaders which use little endian as default, so my approach was implement my own BinaryReader / BinaryWritter for reading / writting data in big endian, but I wonder if there is a better aproach. 回答1: I like this one: Miscellaneous Utility Library 来源： https://stackoverflow.com/questions/80784

问题 I'm writing a program on Linux in C to analyze core files produced from an embedded system. The core files might be little endian (ARM) or big endian (MIPS), and the program to analyze them might be running on a little endian host (x86) or big endian (PowerPC). By looking at the headers I know whether the core is LE or BE. I'd rather my program not need to know whether the host it runs on is little or big endian, I'd like to use an API to handle it for me. If there is no better option, I

问题 I'm writing a program on Linux in C to analyze core files produced from an embedded system. The core files might be little endian (ARM) or big endian (MIPS), and the program to analyze them might be running on a little endian host (x86) or big endian (PowerPC). By looking at the headers I know whether the core is LE or BE. I'd rather my program not need to know whether the host it runs on is little or big endian, I'd like to use an API to handle it for me. If there is no better option, I

问题 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;

问题 How do I convert big endian to little endian in ARM? 回答1: Are you talking about ARM's endian modes, or reading something written by some other big endian processor, etc? Normally converting to/from big/little endian you swap the bytes around. So 0xABCD is 0xCDAB when viewed as a 16 bit number 0x12345678 is 0x78563412 when viewed as a 32 bit number. ARM cores armv5 and older (ARM7, ARM9, etc) have an endian mode known as BE-32, meaning big endian word invariant. armv6 and newer (mpcore, cortex