endianness

问题 I haven't found a specific question here on SO, if this is a duplicate please point it out to me and I'll delete this. So really, does endianness have anything to do with bit order? This seems to imply the answer is NO, while other sources (I fail to find one right now, but surely I've read some time ago some articles) imply that endianness is the order of bytes and bits . To be more specific: in a Big Endian architecture, where MSB is first, inside any byte, is also MSb first? Conversly, on

问题 I haven't found a specific question here on SO, if this is a duplicate please point it out to me and I'll delete this. So really, does endianness have anything to do with bit order? This seems to imply the answer is NO, while other sources (I fail to find one right now, but surely I've read some time ago some articles) imply that endianness is the order of bytes and bits . To be more specific: in a Big Endian architecture, where MSB is first, inside any byte, is also MSb first? Conversly, on

问题 Many implementations of htonl() or ntohl() test for the endianness of the platform first and then return a function which is either a no-op or a byte-swap. I once read a page on the web about a few tricks to handle to/from big/little-endian conversions, without any preconceived knowledge of the hardware configuration. Just taking endianness for what it is : a representation of integers in memory. But I could not find it again, so I wrote this : typedef union { uint8_t b[4]; uint32_t i; }

问题 I wrote a simple program in c which calls a function called while_loop with arguments 4,3,2. The function is just basically a while loop, I don't think it's really that relevant to my question since it's more of a generic question. I was told to run objdump -d, so I did. I have multiple questions so here it goes: I understand that in the leftmost column there are addresses and they increment according to the number of bytes in front. What I don't understand very well is the second column. Is

问题 I wrote a simple program in c which calls a function called while_loop with arguments 4,3,2. The function is just basically a while loop, I don't think it's really that relevant to my question since it's more of a generic question. I was told to run objdump -d, so I did. I have multiple questions so here it goes: I understand that in the leftmost column there are addresses and they increment according to the number of bytes in front. What I don't understand very well is the second column. Is

问题 What is the simplest way to work with big-endian values in RISC-V at the assembly language level? That is, how to load a big-endian value from memory into a register, work with the register value in native-endian (little-endian), then store it back into memory in big-endian. 16, 32 and 64 bit values are used in many network protocols and file formats. I couldn't find a byte-swap instruction (equivalent to BSWAP on x86 or REV on ARM) in the manual, nor anything about big-endian loads and

问题 Literally confused about htonl(). In so many links I found that code to do htonl is : #define HTONL(n) (((((unsigned long)(n) & 0xFF)) << 24) | \ ((((unsigned long)(n) & 0xFF00)) << 8) | \ ((((unsigned long)(n) & 0xFF0000)) >> 8) | \ ((((unsigned long)(n) & 0xFF000000)) >> 24)) If the same code is ran on both the machines, it is going to swap the byte orders. Example : uint32_t a = 0x1; On Little Endian: Addr value 100 1 101 0 102 0 103 0 After htonl(a) Addr value 100 0 101 0 102 0 103 1 ====

问题 If I write the following section .data align 4 X: db 1 Y: dw 5 Z: db 0x11 section .text add dword [X], 0xAA000101 I'm trying to understand the differences between the big endian and the little endian representations, and I don't understand what will be the value of each variable for each representation? Will they be the same? 回答1: Have a look at these pictures: This is the list of endiannesses for all architectures/instruction sets 回答2: In a big-endian configuration, the most significant byte

问题 I have following byte slice which from which i need to extract bits and place them in a []int as i intend to fetch individual bit values later. I am having a hard time figuring out how to do that. below is my code data := []byte{3 255}//binary representation is for 3 and 255 is 00000011 11111111 what i need is a slice of bits -- > [0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1] What i tried I tried converting byte slice to Uint16 with BigEndian and then tried to use strconv.FormatUint but that fails with

问题 We are using Java ByteBuffer for socket communication with a C++ server. We know Java is Big-endian and Socket communication is also Big-endian. So whenever the byte stream received and put into a ByteBuffer by Java, we call getInt() to get the value. No problem, no conversion. But if somehow we specifically set the ByteBuffer byte order to Little-endian (my co-worker actually did this), will the Java automatically convert the Big-endian into the Little-endian when the data is put into the