endianness

问题 I just want to ask if my method is correct to convert from little endian to big endian, just to make sure if I understand the difference. I have a number which is stored in little-endian, here are the binary and hex representations of the number: ‭0001 0010 0011 0100 0101 0110 0111 1000‬ ‭12345678‬ In big-endian format I believe the bytes should be swapped, like this: 1000 0111 0110 0101 0100 0011 0010 0001 ‭87654321 Is this correct? Also, the code below attempts to do this but fails. Is

问题 i am making application in c#.In that application i have byte array containing hex values. Here i am getting data as a big endian but i want it as a little endian. Here i am using Bitconverter.toInt32 method for converting that value to integer. But my problem is that before converting value,i have to copy that 4 byte data into temporary array from source byte array and then reverse that temporary byte array. I cant reverse source array because it contains other data also. Because of that my

问题 I'm using WebGL to render a binary encoded mesh file. The binary file is written out in big-endian format (I can verify this by opening the file in a hex editor, or viewing the network traffic using fiddler). When I try to read the binary response using a Float32Array or Int32Array, the binary is interpreted as little-endian and my values are wrong: // Interpret first 32bits in buffer as an int var wrongValue = new Int32Array(binaryArrayBuffer)[0]; I can't find any references to the default

问题 I'm working on an implementation of the memcache protocol which, at some points, uses 64 bits integer values. These values must be stored in "network byte order". I wish there was some uint64_t htonll(uint64_t value) function to do the change, but unfortunately, if it exist, I couldn't find it. So I have 1 or 2 questions: Is there any portable (Windows, Linux, AIX) standard function to do this ? If there is no such function, how would you implement it ? I have in mind a basic implementation

问题 I'm working on an implementation of the memcache protocol which, at some points, uses 64 bits integer values. These values must be stored in "network byte order". I wish there was some uint64_t htonll(uint64_t value) function to do the change, but unfortunately, if it exist, I couldn't find it. So I have 1 or 2 questions: Is there any portable (Windows, Linux, AIX) standard function to do this ? If there is no such function, how would you implement it ? I have in mind a basic implementation

问题 In the documentation of hardware that allows us to control it via UDP/IP, I found the following fragment: In this communication protocol, DWORD is a 4 bytes data, WORD is a 2 bytes data, BYTE is a single byte data. The storage format is little endian, namely 4 bytes (32bits) data is stored as: d7-d0, d15-d8, d23-d16, d31-d24; double bytes (16bits) data is stored as: d7-d0 , d15-d8. I am wondering how this translates to C#? Do I have to convert stuff before sending it over? For example, if I

问题 I am receiving big endian data over UDP and converting it to little endian. The source says the integers are signed but when I swap the bytes of the signed ints (specifically 16-bit) I get unrealistic values. When I swap them as unsigned ints I get what I expect. I suppose the source documentation could be incorrect and is actually sending unsigned 16-bit ints. But why would that matter? The values are all supposed to be positive and well under 16-bit INT_MAX so overflow should not be an

问题 Is C endian-neutral? Ok, another way of asking this question. I am currently translating a lot of code from C to Matlab on the same platform (PC). Do I need to care about endianess? Both are endian-neutral languages but C (not so sure), Matlab (pretty sure). By the same token I am also translating C to Python. So my question, has anybody in his experience, (translating from C to another endian-neutral language) met an unexpected problem with big/little endianness. Obviously we are only

问题 I am working on a file whose endian is different from my desktop and I need to convert it, but When I visualized the vtk it worked. So are vtkreaders of vtk files endian independent 回答1: ParaView can read VTK files written using either big- or little-endian byte ordering and will try to work out which format has been used when writing a given file. From the VTK File format documentation: Binary Files. Binary files in VTK are portable across different computer systems as long as you observe

问题 In C11, support for portable wide char types char16_t and char32_t are added for UTF-16 and UTF-32 respectively. However, in the technical report, there is no mention of endianness for these two types. For example, the following snippet in gcc-4.8.4 on my x86_64 computer when compiled with -std=c11 : #include <stdio.h> #include <uchar.h> char16_t utf16_str[] = u"十六"; // U+5341 U+516D unsigned char *chars = (unsigned char *) utf16_str; printf("Bytes: %X %X %X %X\n", chars[0], chars[1], chars[2