Thread: Endianness programme help

Hello Everyone,

How can i write the c code to check whether my machine is little indian or big indian????
and how to convert one from other?? Please explain me the logic????????/

Code:

int x = 1;
if (*(char *)&x)
    // Little endian
else
    // Big endian

It works by reading the lowest byte (char) of the variable x. If it's non zero, then the machine is little endian (first byte in memory is the 0x01 byte of 0x00000001).

--
Mats

Originally Posted by tabstop

I'm sure Mats will edit the & into his example here shortly.

As to converting, you just reassemble the bytes in the opposite order (and you can peel off the bytes using pointers as above). Generally you only do this if you need to communicate between machines, and if so, then using something like htons and friends would be the Right Thing to do.

Yes, I have fixed the missing & - too much typing, not enough thinking, as per usual!

htons() and such functions are definitely the best solution. Many processors have clever instructions that do (or at least help) convert byte-order, so writing your own function to read/write the bytes is not a good idea.

--
Mats

We are assuming there are only two kinds of machines... lsb -> msb and the reverse. But what about those whose 16-bit words may be in one order yet the bytes are in another. Or similarly with 32-bit or 64-bit sub chunks may not necessarily be bytes in simple ascending or descending major order.

I find it very unlikely that a machine would use different endianness for different data sizes. And having something else like 2301 ordering seems even less likely.

If I were you, I'd write most of my code as portably as possible, and have a few wrapper functions to convert between types which just use e.g. htons() etc. Then if you actually encounter such a bizarre machine as you have described, you can modify these wrapper functions as necessary.

Well, I stand corrected.

But I still think that this is a very platform-specific problem which can be solved by writing platform-specific code for your program. I don't think (but watch me be wrong again :P) that your code would necessarily need to deal with all these kinds of endianness in one compiled binary program. I could see you needing to deal with the "local" endianness as well as a "generic", platform-independent endianness for reading binary data files or transferring data over a network, for example; but does your one program need to be able to read PDP-endianness and big-endianness data without being re-compiled?

I agree with you... the program should not need to be compiled/running any endienness natively. The original problem was to determine the endienness and be able to convert one to another.

The reason I brought up other oddball endienness is to contend matsp's solution. His is making the assumption there are only two possible orderings.

I would suggest determining endienness thus:

Code:

union {
    long l;
    unsigned char c[4]; } un; 

un.l = 0x03020100l;
printf("Endienness of this machine (lsb to msb) is: %d %d %d %d\n", un.c[0], un.c[1], un.c[2], un.c[3]);

Which gives:

Endienness of this machine (lsb to msb) is: 0 1 2 3

on my Intel based Dell.

The second part of the originally stated problem has not as yet been addressed. How to convert one endienness to another.

I would suggest that once the transformation vector is found, to loop through the bytes of the source 32-bit integer and transposing them using it as guide. This would also require the knowledge of the source endienness as well to find the complete transpose permutation.