endianness

问题 I receive a datapacket containing a byte array and I have to get some integer values from it. Here is a part of the documentation. Can someone help me please? This comes in a 4-byte array. Year from 1990 to 2052 (6 bit), Month from 1 to 12 (4 bit), Day from 1 to 31 (5 bit), Hour from 0 to 23 (5 bit), Minute from 0 to 59 (6 bit), Second from 0 to 59 (6 bit) Default value: 1 January 2000, 12:00:00 The format of the message is in little endian. 回答1: What you need is some bitwise operations.

问题 I'm using DataOutputStream to write to a file, however I want to change the endian of the data. This is how i'm writing the byte data to the file (it outputs in Little endian by default) public void generateBinObjFile(String outputFile) try { // Create file DataOutputStream stream = new DataOutputStream( new FileOutputStream(outputFile)); stream.writeShort(this.quantize(this.xComponents.get(index), //<-- Short is written in little Endian this.min_x, this.max_x) - 32768); } // catch statements

问题 I don't know much python but from what I can tell from the documentation the code: str = "AAAA" str += pack("<I", 0) would append the result of the pack function to str, which would be the integer value of 0 in little-endian style. My question is what the C equivalent of this would be. Would it just be: char str[20] = "AAAA"; strcat(str, "\x00"); ?... 回答1: strcat() stops at the first NUL, so no. char str[20] = "AAAA"; int val = 0; int nval = htole32(val); memcpy(str + 4, (char*)&nval, 4); 来源：

问题 I have a radiograph .img file without the header file. However, the researchers who have published the file have given this information about it High resolution (2048 x 2048 matrix size, 0.175mm pixel size) Wide density range (12bit, 4096 gray scale) Universal image format (no header, big-endian raw data) Using this information, I tried fread command in Matlab to read the image into Matlab. fid = fopen('image.img','r','B'); oneSlice = fread(fid, [2048 2048], '*uint8','B'); imshow(oneSlice)

问题 Let L and B be two machines. L order its bits from LSB (Least Significant Bit) to MSB (Most Significant Bit) while B order from MSB to LSB. Or, in other words, L uses Little Endian while B uses Big Endian bit - not to be confused with byte - ordering. Problem 1 SOLVED : We are writing the following code which we want to be portable: #include <stdio.h> int main() { unsigned char a = 1; a <<= 1; printf("a = %d\n", (int) a); return 0; } on L , it will print 2, but what happens on B ? Will it

问题 Let L and B be two machines. L order its bits from LSB (Least Significant Bit) to MSB (Most Significant Bit) while B order from MSB to LSB. Or, in other words, L uses Little Endian while B uses Big Endian bit - not to be confused with byte - ordering. Problem 1 SOLVED : We are writing the following code which we want to be portable: #include <stdio.h> int main() { unsigned char a = 1; a <<= 1; printf("a = %d\n", (int) a); return 0; } on L , it will print 2, but what happens on B ? Will it

问题 I'm currently working on a big project that involve pictures. One of the big issues I'm having is with the endianness of the picture (jpeg to be clearer). I always though that in our modern world we didn't have to botter about this subject, but now I'm not sure. What I do: I do an HTTP request to an IP camera, the camera return me an array of byte. I parse these byte into an object Image in .NET using Image.FromStream. I take my Image object and do a Save to a physical file on the hard disk.

问题 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); 回答1: BitConverter.ToInt32 isn't very fast in the first place. I'd simply

问题 How to convert the 32 bit integer to network byte order. What is the right way to do that? [1024].pack("N") OR [1,0,2,4].pack("N") Thanks 回答1: To start, look at the output of each: >> [1024].pack("N") => "\000\000\004\000" >> [1,0,2,4].pack("N") => "\000\000\000\001" Note what the second is missing: >> [1,0,2,4].pack("NNNN") => "\000\000\000\001\000\000\000\000\000\000\000\002\000\000\000\004" 来源： https://stackoverflow.com/questions/12571715/how-to-convert-32-bit-integer-to-network-byte-order

问题 to convert a byte array from another machine which is big-endian, we can use: long long convert(unsigned char data[]) { long long res; res = 0; for( int i=0;i < DATA_SIZE; ++i) res = (res << 8) + data[i]; return res; } if another machine is little-endian, we can use long long convert(unsigned char data[]) { long long res; res = 0; for( int i=DATA_SIZE-1;i >=0 ; --i) res = (res << 8) + data[i]; return res; } why do we need the above functions? shouldn't we use hton at sender and ntoh when