endianness

Why does std::bitset expose bits in little-endian fashion?

痴心易碎 提交于 2019-11-27 04:51:26
问题 When I use std::bitset<N>::bitset( unsigned long long ) this constructs a bitset and when I access it via the operator[] , the bits seems to be ordered in the little-endian fashion. Example: std::bitset<4> b(3ULL); std::cout << b[0] << b[1] << b[2] << b[3]; prints 1100 instead of 0011 i.e. the ending (or LSB) is at the little (lower) address, index 0. Looking up the standard, it says initializing the first M bit positions to the corresponding bit values in val Programmers naturally think of

Is mars MIPS simulator Big or Little Endian

Deadly 提交于 2019-11-27 04:48:36
问题 I have to determine if the mars simulator is big or little endian as homework, this seems pretty straightforward at first, but I am having some issues. First I tried storing 4 bytes in memory with .byte 0, 0, 0, 1, in memory this appears as 0x01000000, so, in reverse order, which seems to indicate that the simulator is little endian, however, when I load the 4 bytes as an integer to a register, what appears in the register is 0x01000000 again, as I understand if it was little endian what

Marshalling a big-endian byte collection into a struct in order to pull out values

泄露秘密 提交于 2019-11-27 04:40:46
There is an insightful question about reading a C/C++ data structure in C# from a byte array , but I cannot get the code to work for my collection of big-endian (network byte order) bytes. (EDIT: Note that my real struct has more than just one field.) Is there a way to marshal the bytes into a big-endian version of the structure and then pull out the values in the endianness of the framework (that of the host, which is usually little-endian)? (Note, reversing the array of bytes will not work - each value's bytes must be reversed, which does not give you the same collection as reversing all of

Little Endian vs Big Endian?

China☆狼群 提交于 2019-11-27 04:37:19
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? 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 a few exceptions) address bytes at the individual address level. Taking the -12 example: Little endian, in

Type casting char pointer to integer pointer

给你一囗甜甜゛ 提交于 2019-11-27 04:32:07
问题 So I saw a few example on how the endianness of an architecture could be found. Let's say we have an integer pointer that points to an int data type. And let's say the int value is 0x010A0B12. In a little endian architecture, the least significant byte, i.e, 12, will be stored in the lowest memory address, right? So the lowest byte in a 4-byte integer will be 12. Now, on to the check. If we declare a char pointer p, and type cast the integer pointer to a char * and store it in p, and print

C# - Binary reader in Big Endian?

北慕城南 提交于 2019-11-27 04:03:39
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 a mimic class of Binary reader that returns a reversed array of bytes? Is there something I can change

Converting float values from big endian to little endian

非 Y 不嫁゛ 提交于 2019-11-27 03:52:05
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? simply reverse the four bytes works float ReverseFloat( const float inFloat ) { float retVal; char *floatToConvert = ( char* ) & inFloat; char *returnFloat = ( char* ) & retVal; // swap the bytes into a temporary buffer returnFloat[0] = floatToConvert[3];

create AudioClip from byte[]

只愿长相守 提交于 2019-11-27 03:29:50
问题 I have problem. I use sqlite to store sounds. I get sound from it in byte[]. Then convert byte[] to float[]: private float[] ConvertByteToFloat(byte[] array) { float[] floatArr = new float[array.Length / 4]; for (int i = 0; i < floatArr.Length; i++) { if (BitConverter.IsLittleEndian) Array.Reverse(array, i * 4, 4); floatArr[i] = BitConverter.ToSingle(array, i * 4); } return floatArr; } float[] f = ConvertByteToFloat(bytes); Then create AudioClip: AudioClip audioClip = AudioClip.Create(

Converting byte array values in little endian order to short values

孤街醉人 提交于 2019-11-27 03:15:35
问题 I have a byte array where the data in the array is actually short data. The bytes are ordered in little endian: 3, 1, -48, 0, -15, 0, 36, 1 Which when converted to short values results in: 259, 208, 241, 292 Is there a simple way in Java to convert the byte values to their corresponding short values? I can write a loop that just takes every high byte and shift it by 8 bits and OR it with its low byte, but that has a performance hit. 回答1: With java.nio.ByteBuffer you may specify the endianness

Why is network-byte-order defined to be big-endian? [closed]

落爺英雄遲暮 提交于 2019-11-27 03:02:53
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? 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 octet on the right. The reference they make is to On Holy Wars and a Plea for Peace Cohen, D. Computer The