On Tue, Aug 18, 1998 at 02:03:11PM +0200, Wojciech Zabolotny wrote: > Hi! > > Some of my programs have to work on big-endian and little-endian systems, > knowing what kind of system they are runing on (they are exchanging data > in binary format). To recognize the kind of the system I use the following > routine:
cool.... xfstt is lgpl'd...I dunno about the programs you use.... you mind if I use your code in it? (btw its a very nice aproach...its sort of how I suggested swapping the byte order when discussing the problem with the author once) I don't know if this code is NEEDED in fact... xfstt already detects if there is a byteorder DIFFERECE between the hosts but AFAIK (read: I didn't look THAT closely at it yet) I don't think it detects what type each end is... I think that wouldn't be needed tho... as long as a difference exists... all the swaps are the same...I think I can get away by just performing the right swaps if a global "endiandifference" flag is set without knowing what it is... alternarely...without a uninon... as for the swap... The authors concern was that on a big endian system network order and byte order are the same so hton* and ntoh* do nothing....so how do you do the swapps... I suggested a union first (like below) but...whats the "proper" way of doing it? Th ebest I can figure...you would set the value...and copy it to a temp variable... then bitwise AND the original against 11110000 and the copy against 00001111 then perform a left shift on one...a right shift on the other...and XOR them back together... (NB: the number of 111 and 000 may be wrong...I am writting this off the top of my head ;) ...its just to illistrate the idea ) is this sound right (I forget my real proposed code...I actually wrote the code in the e-mail to him)? -Steve -- /* -- Stephen Carpenter <[EMAIL PROTECTED]> --- <[EMAIL PROTECTED]>------------ */ E-mail "Bumper Stickers": "A FREE America or a Drug-Free America: You can't have both!" "honk if you Love Linux"
