le 6/06/03 2:26, Jeff Ishaq � [EMAIL PROTECTED] a �crit�:
> 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
>
ARM processors need 4 byte algnment for any value larger than 1 byte.
So
struct {
> UInt8 byVal1;
> UInt8 byVal2;
> UInt8 byVal3;
> }
Is 3 bytes
struct {
> UInt8 byVal1;
> UInt8 byVal2;
> UInt8 byVal3;
> UInt16 wVal4;
> UInt8 byVal5;
> }
Requires padding after byVal3, and is 7 bytes
You need to tell the compiler to align on 4 bytes and/or do the padding by
hand (which is safer)
-------------------------------
Eric VERGNAUD - JLynx Software
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