Rules for ARM <-> 68K

[Jeff Ishaq]

I need to read raw data from a OS preference resource.  On OS4 machines, 
it's simply a matter of locking down the returned resource handle and 
casting it to the appropriate OS-defined structure pointer.  However, on 
OS5, ARM alignment and byte ordering take effect, so it becomes more difficult.

I have learned so far:

1) You must byte-swap 16-bit and 32-bit values
2) ARM enforces DWord alignment
I understand and can do #1, no problem.  I can't wrap my head around "DWord 
alignment", however.  Does this mean that any value that is smaller than a 
32 bits ALWAYS becomes padded with enough zeros to make it 32 bits?  So 
this structure:

struct {
UInt8  byVal1;
UInt8  byVal2;
UInt8  byVal3;
}
would occupy 32 + 32 + 32 bits?

Or does the compiler just pad an extra 8 bits to the structure?  Exactly 
where do these extra bits go?  What if we had:

struct {
UInt8 byVal1;
UInt8 byVal2;
UInt8 byVal3;
UInt16 wVal4;
UInt8 byVal5;
}
Is there some online reference manual or FAQ somewhere that might lay down 
the law for this?  I would very much appreciate it!

Thanks,
-Jeff
--
For information on using the Palm Developer Forums, or to unsubscribe, please see 
http://www.palmos.com/dev/support/forums/