> Because:
>    goto $       ; this is a continuous loop, hence
>    goto $-1    ; jumps to btfss   port,anotherbit
>    goto $-2    ; jumps to goto    action
>    goto $-3    ; jumps to btfsc   port,bit
>
> So there!   ;-)

If the character following a "$" is in the set [0-9A-Fa-f] then the dollar
sign and all succeeding alphanumerics form a hex number (if some of the
succeeding alphanumerics are in the range [G-Zg-z] then the code contains
a syntax error).  It's somewhat harder to allow "%" for binary and "&" for
octal (they aren't nearly as universal anyway) but those can be done without
ambiguity: if either of those characters is encountered at a place where a
digit could legally appear, it's a radix specifier; otherwise it's either
an operator or a syntax error.