endianness

Endianness conversion in ARM

随声附和 提交于 2020-01-09 10:04:28
问题 How do I convert big endian to little endian in ARM? 回答1: Are you talking about ARM's endian modes, or reading something written by some other big endian processor, etc? Normally converting to/from big/little endian you swap the bytes around. So 0xABCD is 0xCDAB when viewed as a 16 bit number 0x12345678 is 0x78563412 when viewed as a 32 bit number. ARM cores armv5 and older (ARM7, ARM9, etc) have an endian mode known as BE-32, meaning big endian word invariant. armv6 and newer (mpcore, cortex

Endianness conversion in ARM

你。 提交于 2020-01-09 10:03:12
问题 How do I convert big endian to little endian in ARM? 回答1: Are you talking about ARM's endian modes, or reading something written by some other big endian processor, etc? Normally converting to/from big/little endian you swap the bytes around. So 0xABCD is 0xCDAB when viewed as a 16 bit number 0x12345678 is 0x78563412 when viewed as a 32 bit number. ARM cores armv5 and older (ARM7, ARM9, etc) have an endian mode known as BE-32, meaning big endian word invariant. armv6 and newer (mpcore, cortex

How do I convert a value from host byte order to little endian?

痴心易碎 提交于 2020-01-09 05:18:25
问题 I need to convert a short value from the host byte order to little endian. If the target was big endian, I could use the htons() function, but alas - it's not. I guess I could do: swap(htons(val)) But this could potentially cause the bytes to be swapped twice, rendering the result correct but giving me a performance penalty which is not alright in my case. 回答1: Something like the following: unsigned short swaps( unsigned short val) { return ((val & 0xff) << 8) | ((val & 0xff00) >> 8); } /*

How to extract individual fields from byte array (which is in BIG-ENDIAN) in C++

扶醉桌前 提交于 2020-01-06 08:08:18
问题 I am tring to read couple of bytes from byteData as mentioned below in my C++ code. The actual value within byteData is a binary blob byte array in BIG-ENDIAN byte order format. So I cannot simply just "cast" the byte array into a String.. byteData byte array is composed of these three things - First is `schemaId` which is of two bytes (short datatype in Java) Second is `lastModifiedDate` which is of eight bytes (long datatype in Java) Third is the length of actual `byteArray` within

Assembly language - Why are characters stored in register as little endian?

岁酱吖の 提交于 2020-01-06 05:30:18
问题 I am new to assembly language. I am trying the below code and as you can see the below code. bits 64 global _start section .text _start: mov rcx, 1234567890 xor rcx, rcx mov rcx, 'wxyz' mov rax, 60 mov rdi, 0 syscall I would like to know why digits are stored as Big endian in register and characters are stored in registers as Little-endian Below screenshots are from the debugger. I thought only in the memory, data is stored as Little endian. But I don't understand why the characters are

Get int from bytes in Java, encoded first in C#

女生的网名这么多〃 提交于 2020-01-04 13:37:12
问题 I am having a problem converting an int from a byte array first encoded in C#. I first convert it into big-endian because Java works in big-edian rather than small. The following code encodes the into into bytes Console.WriteLine("A new data client has connected!"); byte[] welcomeMessage = ASCIIEncoding.ASCII.GetBytes("Welcome young chap! Please let me know about anything you need!"); welcomeMessage = Byte.add(BitConverter.GetBytes(System.Net.IPAddress.HostToNetworkOrder(20)), welcomeMessage)

Get int from bytes in Java, encoded first in C#

别来无恙 提交于 2020-01-04 13:37:12
问题 I am having a problem converting an int from a byte array first encoded in C#. I first convert it into big-endian because Java works in big-edian rather than small. The following code encodes the into into bytes Console.WriteLine("A new data client has connected!"); byte[] welcomeMessage = ASCIIEncoding.ASCII.GetBytes("Welcome young chap! Please let me know about anything you need!"); welcomeMessage = Byte.add(BitConverter.GetBytes(System.Net.IPAddress.HostToNetworkOrder(20)), welcomeMessage)

Java, Search for a long in a binary file input, 8 byte aligned, big endian

断了今生、忘了曾经 提交于 2020-01-04 03:18:09
问题 public static void main(String[] args) { File inFile = null; if (0 < args.length) { inFile = new File(args[0]); } BufferedInputStream bStream = null; try { int read; bStream = new BufferedInputStream(new FileInputStream(inFile)); while ((read = bStream.read()) > 0) { getMarker(read, bStream); System.out.println(read); } } catch (IOException e) { e.printStackTrace(); } finally { try { if (bStream != null)bStream.close(); } catch (IOException ex) { ex.printStackTrace(); } } } private static

BitConverter VS ToString for Hex

≯℡__Kan透↙ 提交于 2020-01-03 17:16:13
问题 Just wondering if someone could explain why the two following lines of code return "different" results? What causes the reversed values? Is this something to do with endianness? int.MaxValue.ToString("X") //Result: 7FFFFFFF BitConverter.ToString(BitConverter.GetBytes(int.MaxValue)) //Result: FF-FF-FF-7F 回答1: int.MaxValue.ToString("X") outputs 7FFFFFFF , that is, the number 2147483647 as a whole . On the other hand, BitConverter.GetBytes returns an array of bytes representing 2147483647 in

Endianness in programming languages

风流意气都作罢 提交于 2020-01-03 09:08:44
问题 Well, the "Endianness" theme was always a little bit confusing to me, but i have never faced any problems which required me to even think about the default behaviour of binary writers/readers that i used. I am writing a PNG decoder in c# right now. PNG file format specification states that all numbers are stored in a big endian notation (which i find very natural). However, i was very surprised when i noticed, that .NET's BinaryReader/Writer works with a little endian notation. What confused