endianness

问题 I have an integer which represents a frame length. I would like to know how I can convert the integer to an unsigned 32 bit (4 bytes) big endian byte array in Java 回答1: A big endian byte sequence is simply 'big numbers first'. But of course, converted into binary. So it's shockingly easy with almost any 'hex' conversion - that's the default output. It depends rather which language you're intending to use, but sprintf is pretty common. The format string to do this is %X so in perl you'd have

问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: Why is x86 little endian? I personally, and I bet many will agree to this, find that big-endian byte ordering notation is easier to read and understand. I suppose big-endian byte ordering is just as easy for the CPU to process as little-endian byte ordering. In the article The Dark Corners of x86, David Chisnall states that "[...] features that looked like a good idea at the time were gradually added until we

问题 Do I have to worry about endianness in this case (integers MUST be 0-127): int a = 120; int b = 100; int c = 50; char theBytes[] = {a, b, c}; I think that, since each integer sits in its own byte, I don't have to worry about Endianess in passing the byte array between systems. This has also worked out empirically. Am I missing something? 回答1: Endianness only affects the ordering of bytes within an individual value. Individual bytes are not subject to endian issues, and arrays are always

问题 I think I’m a kind of at a loss here. I trying such a simple thing, I can’t believe that there is nothing build-in Qt (using Qt 5.6.2). I try to convert the data inside a QByteArray from big endian to little endian. Always starts with the same test QByteArray like this. QByteArray value; value.append(0x01); value.append(0x02); value.append(0x03); qDebug() << "Original value is: " << value.toHex(); // “010203” like expected What I need is the little endian, which means the output should be

问题 Is there a portable and safe way to interpret the bit-pattern made by a boost::uint16_t as a boost::int16_t ? I have a uint16_t , which I know represents a signed 16-bit integer encoded as little-endian. I need to do some signed arithmetic on this value, so is there anyway to convince the compiler that it already is a signed value? If I a not mistaken, a static_cast<int16_t> would convert the value, perhaps changing its bit-pattern. 回答1: If you are looking for something different than a cast,

问题 I've done a little bit of reading on endianness and its role in C, but nothing has really managed to clarify this for me. I'm just starting out with C and I saw this example: #include <stdio.h> int main(void) { int x = 017; int y = 12; int diff = x - y; printf("diff is %d\n", diff); return 0; } and it asks what will print. I compiled and ran the example and got that diff is 3, so x is 15. I sort of see why this is, but would really appreciate if somebody really clarified it for me. [1] I've

问题 Is the x86 edition able to run old apps? Does it take care of all endianness stuff? Please don't send me to Google :) I honestly tried and was not able to find a solid answer with some references. 回答1: No, you can't. You must use an emulator like QuickTransit to do that You can find more information on Google with the appropriate keywords Running Solaris Sparc Apps on X86 Solaris Runing Solaris SPARC software on x86-64 Transitive Translates SPARC Solaris Apps on Windows Solaris SPARC to x86

问题 I'm trying to convert a 2D-array composed by integers to a big endian binary file using Python by this way: import struct; fh=open('file.bin','wb') for i in range(width): for j in range(height): fh.write(struct.pack('>i2',data[i,j])) fh.close() when I open it with numpy: a=np.fromfile('file.bin',dtype='>i2') The result is an array with zeros between original data: [266, 0, 267, 0, 268, 0, 272, 0, 268, 0, 264, 0, 266, 0, 264, 0, 263, 0, 263, 0, 267, 0, 263, 0, 265, 0, 266, 0, 266, 0, 267, 0,

问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: When does Endianness become a factor? reading this tuto on endianess, i fall on this example where endianess does matter. It is about writting a char* filled with 1 and 0. it can then be converted to a short, and results depends on endianess, little or big. Here is the example, quoted. unsigned char endian[2] = {1, 0}; short x; x = *(short *) endian; What would be the value of x? Let's look at what this code is

问题 I have the following hex string: 00000000000008a3a41b85b8b29ad444def299fee21793cd8b9e567eab02cd81 but I want it to look like this: 81cd02ab7e569e8bcd9317e2fe99f2de44d49ab2b8851ba4a308000000000000 (Big endian) I think I have to reverse and swap the string, but something like this doesn't give me right result: String hex = "00000000000008a3a41b85b8b29ad444def299fee21793cd8b9e567eab02cd81"; hex = new StringBuilder(hex).reverse().toString(); Result: