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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:
Quote:
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.
Quote:
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?
--
Mats
Ok, I'll be kind and give a hint: Try finding out what byte-order your machine is...
--
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?Quote:
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.
--
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.
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.
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.Quote:
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.
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;
Yet more thigh-slapping hilarity.Quote:
Originally Posted by MacNilly
Yeah, I kind of wondered about that.Quote:
Originally Posted by brewbuck
little/big endian is processor related, not OS
At least to a part, hence my comment:Quote:
Originally Posted by KIBO
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.Quote:
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.
--
Mats
>> or something to that extent?
consider the 4 byte hexidecimal number 0xbebafeca. on a big endian machine this would be stored in contigiuos memory as [be][ba][fe][ca], whereas on a little endian machine the format would be [ca][fe][ba][be]. viewed as an array of unsigned char, to reverse the byte order you would need simply to swap the first element of the array with the last and the second with the third.
oh, phew i thought i had to switch every single bit.
so if i had 'unsigned shorts' (the data type), would i literally take the number lets say 16, which is 00000000 00010000 in binary, little endian format would be 00010000 00000000, which in integer is 4096. Then write that number to the binary file i have to write?
ok i googled the topic and i found a very simple code to swap these endians .. but it is in C++ therefore i dont know whats going on.
Can you help me to convert this to C?
its from here.. and there is a little story to go with the articalCode:unsigned short ByteSwap4 (unsigned short nValue)
{
return (((nValue>> 8)) | (nValue << 8));
}
http://www.codeproject.com/cpp/endianness.asp
ok there is a function in:
#include <winsock2.h>
called:
ntohs(DataPoint); <datapoint is a short... ntohs works for shorts
but when i try to use it I get this
here is the MSDN for it:Code:Linking...
TestTile.obj : error LNK2001: unresolved external symbol __imp__ntohs@4
Debug/TestTile.exe : fatal error LNK1120: 1 unresolved externals
Error executing link.exe.
TestTile.exe - 2 error(s), 0 warning(s)
http://msdn.microsoft.com/library/de...e_ordering.asp
which means i am missing a library... does anyone know anything about this?
OR
does anyone know how that function i posted earlier does?
if anyone here has VS 2005.. can you look inside winsock2.h to see if there are any library includes.
You need to link with wsock32.libQuote:
Originally Posted by chico1st
wow you rock man.. thanks :D
that worked perfectly
is there any way to check the endian-ness of a variable?
Yes. Something like this will work:Quote:
Originally Posted by chico1st
--Code:#define BIG_ENDIAN 1
#define LITTLE_ENDIAN 2
int endian_ness()
{
int x = 0x01000002;
char *p = (char *)&x;
if (x == 0x01)
return BIG_ENDIAN;
else if (x == 0x02)
return LITTLE_ENDIAN;
else {
printf("Strange integer format or bad compiler\n");
exit(1);
}
}
Mats
It's worth noting that ALL variables on a particular system has the same endianness. [By the way, the word endian is related to the book "Gulliver's travels", where the two islands were at war because they wouldn't agree on which end of the egg one should start eating, the "thick" (big) end, or the "thin" (little) end - forming the warring factions of Big Endians and Little Endians - and of course, computer architects would discuss which is better at length!]
--
Mats
>> is there any way to check the endian-ness of a variable?
endian-ness is simply the storage convention used by a CPU (within it's registers) to manipulate numbers. it has no effect on the numbers themselves.
Don't Macs use Motorola processors and PC's use Intel processors? I never mentioned anything about OS's...Quote:
Originally Posted by KIBO
What's so funny?Quote:
Originally Posted by brewbuck
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?
No, it's the EXTERNAL storage format in the memory that is defined by the endian-ness. The internal storage within the processor is completely invisible to an external observer - for all we know, the bits can be stored in any order in a register internally in the machine, as long as it "appears" that bit 31 is the most significant bit (and this doesn't matter if it's a BE or LE machine) and bit 0 is the least significant, and all bits in between come in the appropriate order. [That's for a 32-bit word - but it applies analogously to any size data that the processor will process as one unit].Quote:
Originally Posted by Sebastiani
When the data is stored in memory, however, the data will be stored in a particular order: Big endian stores the most significant byte first, Little Endian stores the least significant byte first.
--
Mats
IIRC, Macs used to use PowerPCs, and they relatively recently switched to Intels.Quote:
Originally Posted by MacNilly
Given that my Mac has an Intel chip, I would have to say... yes.Quote:
Originally Posted by MacNilly
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.