endianness

问题 I am not sure why htons and ntohs both exist in the standard library. They do exactly the same thing -- unless I'm confused somehow! The same goes for htonl and ntohl . 回答1: They make for self-documenting code that tells the reader whether the data is in host- or network- order. 回答2: This is in case a machine has some sort of unusual endianness besides big-endian or little-endian that isn't one or more simple byte swaps. For example, if the value 0x0A0B0C0D was represented internally as 0B 0C

问题 I am not sure why htons and ntohs both exist in the standard library. They do exactly the same thing -- unless I'm confused somehow! The same goes for htonl and ntohl . 回答1: They make for self-documenting code that tells the reader whether the data is in host- or network- order. 回答2: This is in case a machine has some sort of unusual endianness besides big-endian or little-endian that isn't one or more simple byte swaps. For example, if the value 0x0A0B0C0D was represented internally as 0B 0C

问题 I thought that endianness is not supposed to affect structs of size at most 1 byte. But here's the code on my little endian machine: #include <iostream> #include <bitset> #include <cstdint> #include <cstring> using namespace std; static_assert(sizeof(uint8_t) == 1, "Wrong uint8_t size"); struct Pieces { uint8_t flag : 1; uint8_t value : 7; }; static_assert(sizeof(Pieces) == 1, "Something went wrong with Pieces size"); int main() { uint8_t value = 0b10001111; Pieces pieces; std::memcpy(&pieces

问题 I thought that endianness is not supposed to affect structs of size at most 1 byte. But here's the code on my little endian machine: #include <iostream> #include <bitset> #include <cstdint> #include <cstring> using namespace std; static_assert(sizeof(uint8_t) == 1, "Wrong uint8_t size"); struct Pieces { uint8_t flag : 1; uint8_t value : 7; }; static_assert(sizeof(Pieces) == 1, "Something went wrong with Pieces size"); int main() { uint8_t value = 0b10001111; Pieces pieces; std::memcpy(&pieces

问题 Hello i have a small dout in little endian and big endian i know this question has asked n no of times but i could not figure out some below points lets take int i=10 it is store in binary as 00000000 00000000 00000000 00001010 in stack section as below:- 00000000 |00000000 |00000000 |00001010 // In case of little endian MSB-------------------------------------------LSB Big endian 00001010 |00000000 |00000000 |00000000 // In case of in big endian MSB------------------------------------------

问题 This is the source code I have: section .data msg: db "pppaaa" len: equ $ section .text global main main: mov edx,len mov ecx,msg mov ebx,1 mov eax,4 int 0x80 And when I debug this code I will see: (gdb) info register ecx ecx 0x804a010 134520848 (gdb) x 0x804a010 0x804a010 <msg>: 0x61707070 (gdb) x 0x804a014 0x804a014: 0x00006161 "70" here represents the character 'p' and "61" the character 'a' obviously. What I am confused about is, why is the data in location 0x804a010 is 0x61707070 (appp)

问题 This is the source code I have: section .data msg: db "pppaaa" len: equ $ section .text global main main: mov edx,len mov ecx,msg mov ebx,1 mov eax,4 int 0x80 And when I debug this code I will see: (gdb) info register ecx ecx 0x804a010 134520848 (gdb) x 0x804a010 0x804a010 <msg>: 0x61707070 (gdb) x 0x804a014 0x804a014: 0x00006161 "70" here represents the character 'p' and "61" the character 'a' obviously. What I am confused about is, why is the data in location 0x804a010 is 0x61707070 (appp)

问题 This is the source code I have: section .data msg: db "pppaaa" len: equ $ section .text global main main: mov edx,len mov ecx,msg mov ebx,1 mov eax,4 int 0x80 And when I debug this code I will see: (gdb) info register ecx ecx 0x804a010 134520848 (gdb) x 0x804a010 0x804a010 <msg>: 0x61707070 (gdb) x 0x804a014 0x804a014: 0x00006161 "70" here represents the character 'p' and "61" the character 'a' obviously. What I am confused about is, why is the data in location 0x804a010 is 0x61707070 (appp)

问题 I have a school assignment that requires me to convert a word from little endian to big endian in three different ways. One of them is by using multiplication and division. I know that shifting to the left multiplies the number by 2 but i still cant figure out how to utilise that. Here is me doing it using rotate. Can someone help me step on this and do it with division and multiplication? .data .text .globl main main: li $t0,0x11223344 #number to be converted in t0 rol $t1,$t0,8 li $t2

问题 I am trying to write an Integer from C# and read it from Java. An integer is 4 bytes in both languages. However when I write it from C#, integer 1 is written in the following bytes 1000. Meaning the first byte is 1 and the rest is 0. But in Java the same thing is written as 0001. Meaing the first 3 bytes are 0 and the last is 1. Is there a simple way of reading and writing among these languages instead of manually reversing every 4 byte? The code for Java ByteBuffer buffer = ByteBuffer