lua-users home
lua-l archive

[Date Prev][Date Next][Thread Prev][Thread Next] [Date Index] [Thread Index]




On Nov 16, 2018, at 2:04 PM, Sean Conner <sean@conman.org> wrote:

I was taught that a "word" is the number of bits the CPU can handle
natively (or without addtional work).  That's fine for most CPUs, but things
do get fuzzy.  For instance, the Intel 8088.  Internally it can do 16-bit
arithmatic so clearly it's "word" is 16-bits.  *BUT* it only has an 8-bit
data bus, so retreiving a 16-bit word requires more work.  So what is the
"word" size on an 8088?  8-bits or 16-bits?

Same here, but I always felt it was driven by the CPU architecture, not the bus width (after all, a modern x86 CPU has VERY wide busses, sometimes 256 bits, but no-one is claiming its a 256-bit architecture). To my mind the 8088 was a 16-bit CPU since most instructions (add/sub/xor/neg etc) naturally worked on 16-bit values in 16-bit registers (notwithstanding using AH/AL etc).

I think “word” wrt CPU architecture was never a precise term, and probably never will be.

-_Tim