> Yeah, "was the 8088 an 8-bit or a 16-bit processor?" doesn't have a straightforward answer..."Bitness" is usually held to refer to the size of a processor's main register set, so the size of data units it is capable of processing internally most of the time, ignoring special cases. By that common definition the 8088 is definitely a 16-bit CPU. Especially as it was a modification of the 16-bit 8086 design and fully code compatible with it, the modification being that 8 bit data bus which slowed down talking to the outside world.
Similarly the 386SX is a 32-bit unit. They took the original 386 and modified it in much the same way, giving it a 16-bit data bus to allow it to be used on much cheaper motherboard designs being the main defining change. The 386DX actually came first technically because it was essentially the original 386 lines renamed once the SX variants came along.
The Z80 could perform a limited set of 16 bit operations such as partially 16-bit multiply (8-bit inputs to a 16-bit output) but was definitely an 8-bit unit overall as the majority of it's inwards were (and the majority of the 8088s innards were 16-bit, and the 386's (original/DX, or SX) were 32. If we count instruction outputs and not general purpose register size then the 6502 family would be 9-bit not 8 as most of its arithmetic instructions output 9 bits (the main 8, plus overflow in the flags register).
The 486 would not be called 80-bit because of its floating point unit having 80-bit registers for intermediate operations, and early-ish Pentiums were still 32-bit when 64-bit chewing MMX instructions were added (in fact the very first Pentiums had 64-bit data busses so they could pump data into the on-die cache twice as fast to try keep up with the demands of the fast pipelined internals but where still considered 32-bit as said internals were).
There are some CPU/GPU/other-PU designs where the distinction is rather muddy, but for Intel's main lines and those inheriting from them I'd say their bitness is pretty well defined.