Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1a 12/4/83; site rlgvax.UUCP Path: utzoo!watmath!clyde!floyd!harpo!seismo!rlgvax!guy From: guy@rlgvax.UUCP (Guy Harris) Newsgroups: net.micro Subject: Re: Mac: 16 or 32 bit? Message-ID: <1689@rlgvax.UUCP> Date: Wed, 8-Feb-84 23:39:35 EST Article-I.D.: rlgvax.1689 Posted: Wed Feb 8 23:39:35 1984 Date-Received: Fri, 10-Feb-84 03:39:57 EST References: <1239@unc-c.UUCP> Organization: CCI Office Systems Group, Reston, VA Lines: 39 > The M68000 is indeed a 16-bit microprocessor. I would > assume that the Macintosh is called a 32-bit > machine because of its data path. It could be > more correctly identified as a 16/32 machine, > similar to the IBM PC designation of 8/16 (8-bit > M6800 microprocessor / 16-bit data path). The > differences are usually invisible to the user. The MC68000 has, if I correctly remember what I've heard stated on net.micro before, a 32-bit data path for (some) address arithmetic and a 16-bit data path for other arithmetic. (The instruction timings can give that impression.) Its registers are 32 bits wide, as are its virtual and physical addresses (well, 24 bits really, but it's the thought that counts). Its external data path, however, is 16 bits wide. The Intel 8088 (not MC6800) in the IBM PC has a 16-bit data path for arithmetic and an 8-bit data path, and its registers are 16 bits wide; its physical addresses are 20 bits wide, and its virtual addresses are 16 or 20 bits wide depending on how you look at it. It is only an 8-bit micro in terms of its path to the outside world; by that measure, the MC68000 is a 16-bit micro and the MC68008 is an 8-bit micro, even though any program that runs on the MC68000 will run on the MC68008 without change (modulo any funny instructions; my officemate's copy of the M68000 manual isn't here at the moment) and the MC68008 has the same 24-bit linear address space as the MC68000 has. The question of which width is important depends on what you're interested in. The width of the external and internal data paths has a bearing on the speed of the processor, but you can have a 64-bit linear address space on a machine with a 1-bit data path (if that's your idea of a good time). The compactness of code and the complexity of handling more than 64K code and 64K data, and thus the speed of the processor as well as the convenience of programming, depend on the size of the virtual address space of the processor rather than the size of its data paths, and the amount of physical memory you can attach to the machine depends on the size of the physical address space of the machine. Guy Harris {seismo,ihnp4,allegra}!rlgvax!guy