endianness | 易学教程

Efficient way to read big endian data in C#

阅读更多关于 Efficient way to read big endian data in C#

I use the following code to read BigEndian information using BinaryReader but I'm not sure if it is the efficient way of doing it. Is there any better solution? Here is my code: // some code to initialize the stream value // set the length value to the Int32 size BinaryReader reader =new BinaryReader(stream); byte[] bytes = reader.ReadBytes(length); Array.Reverse(bytes); int result = System.BitConverter.ToInt32(temp, 0); CodesInChaos BitConverter.ToInt32 isn't very fast in the first place. I'd simply use public static int ToInt32BigEndian(byte[] buf, int i) { return (buf[i]<<24) | (buf[i+1]<

Is there a way to enforce specific endianness for a C or C++ struct?

阅读更多关于 Is there a way to enforce specific endianness for a C or C++ struct?

问题 I've seen a few questions and answers regarding to the endianness of structs, but they were about detecting the endianness of a system, or converting data between the two different endianness. What I would like to now, however, if there is a way to enforce specific endianness of a given struct . Are there some good compiler directives or other simple solutions besides rewriting the whole thing out of a lot of macros manipulating on bitfields? A general solution would be nice, but I would be

ASCII strings and endianness

阅读更多关于 ASCII strings and endianness

问题 An intern who works with me showed me an exam he had taken in computer science about endianness issues. There was a question that showed an ASCII string "My-Pizza", and the student had to show how that string would be represented in memory on a little endian computer. Of course, this sounds like a trick question because ASCII strings are not affected by endian issues. But shockingly, the intern claims his professor insists that the string would be represented as: P-yM azzi I know this can't

Big Endian and Little Endian

阅读更多关于 Big Endian and Little Endian

问题 Given is the snap shot of memory of a byte-addressable computer. What would be loaded into register $16 after execution of instruction lw $16, 24($17) if machine is big endian and when Little Endian. Register $17 contains 200 . Now according to me, four bytes would be copied from the memory (224-227) irrespective of Little Endian or Big Endian,then if the machine is Big Endian then they will be copied to the register as they are. If the machine is Little Endian then will be reversed and then

Why does BinaryReader.ReadUInt32() reverse the bit pattern?

阅读更多关于 Why does BinaryReader.ReadUInt32() reverse the bit pattern?

问题 I am trying to read a binary file with the BinaryReader class, and I need to read it in as blocks of UInt32, and then do some bit shifting etc. afterwords. But, for some reason bit order is reversed when I use the ReadUInt32 method. If I for example have a file where the first four bytes looks like this in hex, 0x12345678 , they end up like this after being read by ReadUInt32: 0x78563412 . If I use the ReadBytes(4) method, I get the expected array: [0x00000000] 0x12 byte [0x00000001] 0x34

Byte order with a large array of characters in C

阅读更多关于 Byte order with a large array of characters in C

问题 Hey guys, question from a C/Networking newbie... I'm doing some socket programming in C and trying to wrestle with byte order problems. My request (send) is fine but when I receive data my bytes are all out of order. I start with something like this... char * aResponse= (char *)malloc(512); int total = recv(sock, aResponse, 511, 0); When dealing with this response, each 16bit word seems to have it's bytes reversed (I'm using UDP). I tried to fix that by doing something like this... unsigned

Is it true that endianness only affects the memory layout of numbers,but not string?

阅读更多关于 Is it true that endianness only affects the memory layout of numbers,but not string?

问题 Is it true that whether the architecture is big or little endian ,only the memory layout of numbers differ,that of the string is the same. 回答1: If you have a simple 8-bit character representation (e.g. extended ASCII), then no, endianness does not affect the layout, because each character is one byte. If you have a multi-byte representation, such as UTF-16, then yes, endianness is still important (see e.g. http://en.wikipedia.org/wiki/UTF-16#Byte_order_encoding_schemes). 回答2: For strings of 1

Endianness inside CPU registers

阅读更多关于 Endianness inside CPU registers

I need help understanding endianness inside CPU registers of x86 processors. I wrote this small assembly program: section .data section .bss section .text global _start _start: nop mov eax, 0x78FF5ABC mov ebx,'WXYZ' nop ; GDB breakpoint here. mov eax, 1 mov ebx, 0 int 0x80 I ran this program in GDB with a breakpoint on line number 10 (commented in the source above). At this breakpoint, info registers shows the value of eax=0x78ff5abc and ebx=0x5a595857 . Since the ASCII codes for W, X, Y, Z are 57, 58, 59, 5A respectively; and intel is little endian, 0x5a595857 seems like the correct byte

Calculating CoreMIDI Pitch Bend Values For iOS?

阅读更多关于 Calculating CoreMIDI Pitch Bend Values For iOS?

问题 I need to hand assemble 14bit MIDI Pitch Bend values from raw UInt16 values in iOS. I'm wondering if anybody out there has had a chance to come up with an elegant solution? Here's where I'm at - I'll get a chance to test this probably later today, but if I hear back before then, great: First, some MIDI preliminaries for anybody curious. MIDI Pitch Bend is broken up into one Status Byte followed by two Data Bytes (it's a 14bit controller), these two Data Bytes are associated with their Status

How to write endian agnostic C/C++ code?

阅读更多关于 How to write endian agnostic C/C++ code?

I did some googling and couldn't find any good article on this question. What should I watch out for when implementing an app that I want to be endian-agnostic? Pubby This might be a good article for you to read: The byte order fallacy The byte order of the computer doesn't matter much at all except to compiler writers and the like, who fuss over allocation of bytes of memory mapped to register pieces. Chances are you're not a compiler writer, so the computer's byte order shouldn't matter to you one bit. Notice the phrase "computer's byte order". What does matter is the byte order of a