I have to write a binary file.
The data is supposed to be written in Little Endian format, which is a type of byte order.
Does anyone know how to do this?
Glossary:
byte order:
byte order sets the endian form of the resulting data. Byte order, or endian form, indicates whether integers are represented in memory from most-significant byte to least-significant byte or vice versa.little-endian—The least-significant byte occupies the lowest memory address. Used on Windows and Linux.
Yes, I know how to do that. [That is the question you asked! :-)]
Do you have any thoughts on the matter?
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Mats
Ok, I'll be kind and give a hint: Try finding out what byte-order your machine is...
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Mats
Your reference saysThe part that says "Used in Windows and Linux" is true - just like "dogs have a tail" - doesn't mean that all animals with a tail are dogs, right?little-endian—The least-significant byte occupies the lowest memory address. Used on Windows and Linux.
I don't know about Windows, but Linux can and does definitely run on big-endian machines, no doubt at all about that.
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Mats
first you need to determine the byte ordering of the machine your program is running under.
a simple way to determine this would be:
if it's big endian, simply reverse the bytes of each 'native' data type.Code:int is_big_endian_machine(void) { static unsigned tst = 1; static int stat = (*(unsigned char*)&tst) != 1; return stat; }
the easiest way to do that is to cast the address of the data to an unsigned char* and then work with the data as an array of bytes.
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
Isn't it also possible to do it portably without caring about a particular implementation's underlying multibyte storage? That is, where you truncate to a byte some integral value to get the least significant byte to put to a file, and then right-shift the value by CHAR_BIT; doing so for each byte in the object.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
The only problem I can see is if you wanted to move a file written from one type of machine to the other. Guess that's why you can't move files from Macs to PCs.Originally Posted by chico1st
You want to try and convert a file from one to the other? Unless you know the specific data types that make up the file, that would be impossible.
Yet more thigh-slapping hilarity.
Yeah, I kind of wondered about that.
little/big endian is processor related, not OS
At least to a part, hence my comment:But at the same time, there are several processors I know of (Mips, 29K) that can actually set the byte-ordering by changing a bit in a control-register. I think ARM can do this too. In this case, the OS can decide what byte-ordering to use.The part that says "Used in Windows and Linux" is true - just like "dogs have a tail" - doesn't mean that all animals with a tail are dogs, right?
I don't know about Windows, but Linux can and does definitely run on big-endian machines, no doubt at all about that.
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Mats
Don't Macs use Motorola processors and PC's use Intel processors? I never mentioned anything about OS's...
What's so funny?
If I made a binary file on a PC writing the words 10, 20, 30, and 40, and you tried to read that on a Mac, you're telling me you would get 10, 20, 30 and 40?
Last edited by MacNilly; 08-22-2007 at 05:47 PM.
Given that my Mac has an Intel chip, I would have to say... yes.
Properly written software takes endianness into account. Just because a file is "binary" doesn't mean it can't be manipulated on processors of differing endianness.
so to switch the byte order my code would look like this:
or something to that extent?Code:buffer1 = (unsigned char *)OldNumber; bit[0] = find the first bit; bit[1] = find the second bit etc... bit[8] = ... sprintf(Buffer2, "%c%c%c%c%c%c%c%c", bit[0], bit[1],...); NewNumber = (unsigned short)Buffer2;
