>Andy & Barry:
>I understand what you're saying. I have looked at the movwf instruction in
>the mid-range manual and I completely understand how the bits are used. But
>it looks to me as if the compiler is mapping the byte to bank-0 by the
>"^0x80" add-on and that ain't particularly desirable. Or am I missing
>something?
>        Sam....

Well it *is* compiling it as though it were bank 0.  But that's
just the compiler.  When the PIC is actually running, that
instruction may hit one of the other banks, depending on
which bank bits in the status register have been previously
set.  That's a separate operation, and since it can appear
anywhere in the program, the compiler can't know which
bank bits are set at any given time.    Indeend,  you could
write a program where the same instruction hit a different
bank each time (but I wouldn't post the code here! :)

Barry

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