endianness

I'm writing a client and a server for a realtime offshore simulator, and, as I have to send a lot of data through a socket, I'm using binary data to maximize the ammount of data I can send. I already know about integers endianness, and how to use htonl and ntohl to circumvent endianness issues, but my application, as almost all simulation software, deals with a lot of floats. My question is: Is there some issue of endianness whean dealing with binary formats of floating point numbers? I know that all the machines where my code will run use IEEE implementation of floating points, but is there

I'm currently trying to create a C source code which properly handles I/O whatever the endianness of the target system. I've selected "little endian" as my I/O convention, which means that, for big endian CPU, I need to convert data while writing or reading. Conversion is not the issue. The problem I face is to detect endianness, preferably at compile time (since CPU do not change endianness in the middle of execution...). Up to now, I've been using this : #if __BYTE_ORDER__ == __ORDER_LITTLE_ENDIAN__ ... #else ... #endif It's documented as a GCC pre-defined macro, and Visual seems to

I just want to ask if my method is correct to convert from little endian to big endian, just to make sure if I understand the difference. I have a number which is stored in little-endian, here are the binary and hex representations of the number: ‭0001 0010 0011 0100 0101 0110 0111 1000‬ ‭12345678‬ In big-endian format I believe the bytes should be swapped, like this: 1000 0111 0110 0101 0100 0011 0010 0001 ‭87654321 Is this correct? Also, the code below attempts to do this but fails. Is there anything obviously wrong or can I optimize something? If the code is bad for this conversion can you

I have an array of Floats that need to be converted to a byte array and back to a float[]... can anyone help me do this correctly? I'm working with the bitConverter class and found myself stuck trying to append the results. The reason I'm doing this is so I can save runtime values into a IO Stream. The target storage is Azure Page blobs in case that matters. I don't care about what endian this is stored in, as long as it input matches the output. static byte[] ConvertFloatToByteArray(float[] floats) { byte[] ret = new byte[floats.Length * 4];// a single float is 4 bytes/32 bits for (int i = 0;

A real question that I've been asking myself lately is what design choices brought about x86 being a little endian architecture instead of a big endian architecture? Largely, for the same reason you start at the least significant digit (the right end) when you add—because carries propagate toward the more significant digits. Putting the least significant byte first allows the processor to get started on the add after having read only the first byte of an offset. After you've done enough assembly coding and debugging you may come to the conclusion that it's not little endian that's the strange

#include <stdio.h> int main() { float a = 1234.5f; printf("%d\n", a); return 0; } It displays a 0 !! How is that possible? What is the reasoning? I have deliberately put a %d in the printf statement to study the behaviour of printf . That's because %d expects an int but you've provided a float. Use %e / %f / %g to print the float. On why 0 is printed: The floating point number is converted to double before sending to printf . The number 1234.5 in double representation in little endian is 00 00 00 00 00 4A 93 40 A %d consumes a 32-bit integer, so a zero is printed. (As a test, you could printf(