endianness

问题 I\'m trying to improve my understanding of the STFS file format by using a program to read all the different bits of information. Using a website with a reference of which offsets contain what information, I wrote some code that has a binary reader go through the file and place the values in the correct variables. The problem is that all the data is SUPPOSED to be Big Endian, and everything the binary reader read is Little Endian. So, what\'s the best way to go about fixing this? Can I create

问题 Is it possible to convert float s from big to little endian? I have a big endian value from a PowerPC platform that I am sendING via TCP to a Windows process (little endian). This value is a float , but when I memcpy the value into a Win32 float type and then call _byteswap_ulong on that value, I always get 0.0000? What am I doing wrong? 回答1: simply reverse the four bytes works float ReverseFloat( const float inFloat ) { float retVal; char *floatToConvert = ( char* ) & inFloat; char

问题 A real question that I\'ve been asking myself lately is what design choices brought about x86 being a little endian architecture instead of a big endian architecture? 回答1: Largely, for the same reason you start at the least significant digit (the right end) when you add—because carries propagate toward the more significant digits. Putting the least significant byte first allows the processor to get started on the add after having read only the first byte of an offset. After you've done enough

问题 As written in the heading, my question is, why does TCP/IP use big endian encoding when transmitting data and not the alternative little-endian scheme? 回答1: RFC1700 stated it must be so . (and defined network byte order as big-endian). The convention in the documentation of Internet Protocols is to express numbers in decimal and to picture data in "big-endian" order [COHEN]. That is, fields are described left to right, with the most significant octet on the left and the least significant

问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: C Macro definition to determine big endian or little endian machine? int main() { int x = 1; char *y = (char*)&x; printf(\"%c\\n\",*y+48); } If it\'s little endian it will print 1. If it\'s big endian it will print 0. Is that correct? Or will setting a char* to int x always point to the least significant bit, regardless of endianness? 回答1: In short, yes. Suppose we are on a 32-bit machine. If it is little endian

问题 I\'m having troubles wrapping my head on the two. I understand how to represent something in big endian. For example -12 is 1111 1111 1111 0100 But why is the little endian representation 1111 0100 1111 1111 instead of 0100 1111 1111 1111? 回答1: Endianness is about byte address order . Little endian means the lower significant bytes get the lower addresses. Big endian means the other way around. So it's about the bytes (8-bit chunks) not nibbles (4-bit chunks). Most computers we use (there are

问题 I\'m trying to build a 32-bit float out of its 4 composite bytes. Is there a better (or more portable) way to do this than with the following method? #include <iostream> typedef unsigned char uchar; float bytesToFloat(uchar b0, uchar b1, uchar b2, uchar b3) { float output; *((uchar*)(&output) + 3) = b0; *((uchar*)(&output) + 2) = b1; *((uchar*)(&output) + 1) = b2; *((uchar*)(&output) + 0) = b3; return output; } int main() { std::cout << bytesToFloat(0x3e, 0xaa, 0xaa, 0xab) << std::endl; // 1

问题 How to check whether a system is big endian or little endian? 回答1: In C, C++ int n = 1; // little endian if true if(*(char *)&n == 1) {...} See also: Perl version 回答2: In Python: from sys import byteorder print(byteorder) # will print 'little' if little endian 回答3: Another C code using union union { int i; char c[sizeof(int)]; } x; x.i = 1; if(x.c[0] == 1) printf("little-endian\n"); else printf("big-endian\n"); It is same logic that belwood used. 回答4: If you are using .NET: Check the value of

问题 I am using BinaryWriter class to write a binary file to disk. When I invoke the Write method, passing an unsigned short value, it writes it in little-endian format. For example: bw.Write(0xA000); writes the value in the binary file as 0x00 0xA0. Is there a way to make BInaryWriter use Big Endian? If not, is it possible to create a new class, inheriting BinaryWriter and overload the Write function to make it write big endian? 回答1: You can use my EndianBinaryWriter in MiscUtil. That lets you

问题 What endianness does Java use in its virtual machine? I remember reading somewhere that it depends on the physical machine it\'s running on, and then other places I have read that it is always, I believe, big endian. Which is correct? 回答1: Multibyte data in the class files are stored big-endian. From The Java Virtual Machine Specification, Java SE 7 Edition, Chapter 4: The class File Format: A class file consists of a stream of 8-bit bytes. All 16-bit, 32-bit, and 64-bit quantities are