rcj1@ihuxi.UUCP (01/25/84)
I have a Motorola MC68000 users manual that says 16-bit microprocessor on the cover. All the ads for Macintosh say it's a 32 bit micro... Which is it? Being more into hardware , I thought that having D0-D15 made it a 16 bit micro. Ray, ihnp4!ihuxi!rcj1
preece@uicsl.UUCP (01/28/84)
#R:ihuxi:-78000:uicsl:7000045:000:1112 uicsl!preece Jan 27 07:49:00 1984 I have a Motorola MC68000 users manual that says 16-bit microprocessor on the cover. All the ads for Macintosh say it's a 32 bit micro... Which is it? Being more into hardware , I thought that having D0-D15 made it a 16 bit micro. ---------- This is one of those unanswerable questions. You get to pick whether to classify machines by register width or by external buss width. Either answer may give confusing results. Consider the IBM System/360 family. As I recall, the buss width varied by model within the family from 8 to 64 bits (this is old knowledge and subject to the decay of my memory) and some operations dealt with up to 64 bits at a time, so you pick a size. It's generally thought of as a 32 bit family because most of the registers are 32 bits and that is referred to as the word size in the documentation. More modern processors have more kinds of operands, so there is no 'normal' word size. Clearly, though, the 68000 is an order of magnitude 'bigger' than the 8088, since in both data access width and register width it is twice as wide. scott preece ihnp4!uiucdcs!uicsl!preece
mzp@uicsg.UUCP (02/02/84)
#R:ihuxi:-78000:uicsg:7600006:000:442 uicsg!mzp Feb 1 13:37:00 1984 Another problem here is that even though the 68000 has 32-bit registers, it only has a 16-bit ALU. Therefore, it really can't be called a 32-bit machine in ANY sense of the word. The 16032, however, with a full 32-bit data path internally, is a 32-bit machine with a 16-bit bus. It's somewhat like comparing a Z80 and an 8088. Show me a real 16032/16081 micro and I'll buy that, but not this trash Apple stuff. Mark Papamarcos
rrm@unc-c.UUCP (02/07/84)
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.
-- rrm --
decvax/duke/mcnc/unc-c/rrmgranvold@tymix.UUCP (Tom Granvold) (02/07/84)
-
As far as I am concerned, Apple is playing by the same rules in calling
the 68000 as 32 bit machine. If IBM can call the 8088 a 16 bit machine, then
the 68000 is a 32 bit one.
Tom Granvold
Tymshare
Cupertino, Calif.
decvax!ucbvax!oliveb!tymix!granvoldcrandell@ut-sally.UUCP (Jim Crandell) (02/07/84)
> The trouble is > that if you dont play by the same rules as everyone else then you put > yourself at a marketing disadvantage. I don't know about that. It seems to have worked for Intel. -- Jim Crandell, C. S. Dept., The University of Texas at Austin {ihnp4,seismo,ctvax}!ut-sally!crandell
robertm@dartvax.UUCP (Robert P. Munafo) (02/09/84)
> Rating processors by their largest register is like selling chain by its > strongest link. Okay, fine - how about a benchmark? (is there a good one?) How long for each CPU to do a block move? How many bytes is the code to do it? Robert P. Munafo ...!{decvax,linus}|dartvax|robertm
guy@rlgvax.UUCP (Guy Harris) (02/09/84)
> 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
kds@intelca.UUCP (Ken Shoemaker) (02/09/84)
> As far as I am concerned, Apple is playing by the same rules in calling >the 68000 as 32 bit machine. If IBM can call the 8088 a 16 bit machine, then >the 68000 is a 32 bit one. close, but not quite, the 8088 has a 16 bit ALU, just like the 68000, so picking nits, an 8088 is closer to a 16 bit machine than a 68000 is to a 32 bit machine. But, in reality, people call things whatever they want to, regardless of reality. Personally, I have a 2**20 bit machine, since that is the amount of memory in my PC! (aren't numbers wonderful?) -- Ken Shoemaker, Intel, Santa Clara, Ca. {pur-ee,hplabs,ucbvax!amd70,ogcvax!omsvax}!intelca!kds
john@hp-pcd.UUCP (02/21/84)
Rating processors by their largest register is like selling chain by its strongest link. It looks good in the ads but hardly tells the customer anything about how the system will perform in real life. The trouble is that if you dont play by the same rules as everyone else then you put yourself at a marketing disadvantage. John Eaton !hplabs!hp-pcd!john