Luit van Drongelen wrote: > Thanks for that one Borut. I just knew there was something 14-bit wide > in 12(C/F)/16F devices which is 16-bit in 18F chips :) so the > '8-bit'-ness of the chip is actually the width of the busses inside > the real processor? >
As I already wrote, there is no clean cut. Internal buses in the same CPU can have different widths. For example PIC 16F* has 14-bit internal instruction bus and 8-bit internal data bus and 13-bit address bus (if I'm not wrong). It has 8 level stack, so we could say that it has 3-bit stack bus. CPUs with von Newman architecture has (usually)2 buses: data and address bus. Most common situations are: - address width = 2 * data width or - address width = data width Borut > Luit > > On 3/31/07, *Borut Razem* <[EMAIL PROTECTED] > <mailto:[EMAIL PROTECTED]>> wrote: > > Luit van Drongelen wrote: > > If I recall correctly, the 18Fxxx(x) chips are 16-bit chips, hence > > PIC16 architecture > > the 16Fxx(x) chips are 14 bit, supported by the PIC architecture > port > > > Actually 18F* has 16-bit wide instructions while 16F* has 14-bit wide > instructions. > > I think that there is no precise definition of what 8, 16, ... bit CPU > means: the address space, the widest chunk of date memory which can be > accessed with one instruction, the width of accumulators, ... I would > say that both 16F* and 18F* families are 8 bit, which is also > Microchip's statement. > > > Borut > ------------------------------------------------------------------------- Take Surveys. Earn Cash. Influence the Future of IT Join SourceForge.net's Techsay panel and you'll get the chance to share your opinions on IT & business topics through brief surveys-and earn cash http://www.techsay.com/default.php?page=join.php&p=sourceforge&CID=DEVDEV _______________________________________________ Sdcc-user mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/sdcc-user
