问题
Is it determined by size of the address buss; if yes then was 8086 a 20-bit processor? If no what is criteria for assigning a bit number like 8-bit, 16-bit, 32-bit to processor?
回答1:
It's not well defined. Broadly, as xtofl points out, it's the size of the atomic unit of computation (in early computers, this wasn't always synonymous with "register"). So the PDP-10 was a 36 bit machine, a 8080 was 8 bit, and a IBM 360 or Intel 80386 is "32 bits".
But there are exceptions. The Motorola 68000 and 68010 CPUs implemented a 32 bit register set, but did it via microcode on top of a mostly 16 bit internal architecture. They were usually marketed as "16 bit" CPUs at the time.
The size of the address bus is almost never the defining factor. All successful "8 bit" CPUs implemented 16 bit addressing, for example (often via odd hacks to make up for the lack of a single address register, c.f. 6502's indirect addressing modes or the Z80's H/L registers). And the 8086, as you mention, used its segment register addressing to get 20 address lines to work (the 80286 extended this trick to 24 bits of physical address). And in the other direction, many "32 bit" CPUs had smaller address buses to save logic that wouldn't be used on a machine that would never have more than a few megabytes of memory: the 68000 limited addressing to 24 bits, even though the pointers themselves were 32. Likewise modern 64 bit CPUs universally implement less than 64 bits of physical address.
回答2:
I guess normally you label it by the size of it's accumulators/registers.
回答3:
With respect to a CPU, I'd say that it's the width of a register. You can do an operation on only 8 bits, 16-bits, 32-bits, etc. at a time.
回答4:
As far as i know the bit width of the processor is determined by how many bits the internal data processing circuits accept at once. Like if the adders, multipliers etc in the ALU accept 16 bit operands then the CPU is 16 bit, and if it accepts 32 bits then it is 32 bit. It does not matter what is the width of the data bus or the address bus. In general the bit length of the Accumulator would determine the bit length of the processor.
回答5:
The bit size (8-bit, 16-bit, 32-bit) of a microprocecessor is determined by the hardware, specifically the width of the data bus. The Intel 8086 is a 16-bit processor because it can move 16 bits at a time over the data bus. The Intel 8088 is an 8-bit processor even though it has an identical instruction set. This is similar to the Motorola 68000 and 68008 processors. The bit size is not determined by the programmer's view (the register width and the address range).
回答6:
I think the first number of Integrated chip refers to the type of the processor. If it is IC 8085 means it is a 8 bit processor.
回答7:
any processor can be designated by its' two attributes
- instruction set architecture &
- no. of bits it can handle in single clock cycle.
take for example Intel's IA-32 architecture, also called x86-32 , here x86 indicates the architecture and 32 indicates 32-bit processor
X-Architecture
there are a number of architectures
Pre-x86 x86
-Intel's IA-32 architecture, also called x86-32 -x86-64 - -with AMD's AMD64 and Intel's Intel 64 version of it - Motorola's 6800 and 68000 a
rchitectures ARM7
Y-bit processor
: simply- its the data handling capability of cpu/processor in a single clock cycle. suppose it is an 8 bit processor then in a single clock cycle, the ALU can perform operation on 8 bit data only.(note that this operation may be an internal operation like add/sub as well as transferring data to other IO device)
classification Based on Registers-
Processor in addition to ALU and CU contains some memory locations as well, called as registers. depending on the processor, a register may typically store 8, 16, 32 or 64 bits. The register size of a particular processor allows us to classify the processor. Processors with a register size of n-bits are called n-bit processors, so that processors with 8-bit registers are called 8-bit processors.
classification Based on databus width-
since the alu can handle only 8 bit data in a single clock cycle it won't make sense to have data bus width more than that & 8 bit processor will have 8 bit wide databus, hence databus width can also be an alternate way to find out the bit processing capability of processor.for a processor with n bit databus means that the CPU can transfer n-bits to another device in a single operation.
for the question:
"suppose we have a 32 bit ALU i.e. it can take 32 bits at a time and our data bus size is 16 bit i.e. it can hold 16 bit of data at a time thn wht will be the ans. In this case...?"
the example of such processor would be intel 8088 & Moto 68000
Using bus width classification, the Intel 8088 microprocessor is an 8-bit processor since it uses an 8-bit data bus, although its CPU registers are in fact 16-bit registers. Similarly the Motorola 68000 is classified as a 16-bit processor, even though its CPU registers are 32-bit registers. Sometimes a combination of the two classifications is used where the 8088 might be described as 8/16-bit processor and the Motorola 68000 as a 16/32-bit processor.
回答8:
The word size(8-bits, 16-bits or 32-bits) of a microprocessor is the size of the data path in the execution unit. Typically, this is the size of the accumulator. This is the execution unit size. An example where this matters is the 8088, which is a 16 bit computer running on an 8 bit bus. The 8085 is 8-bits. The 8086/8088 is 16-bits. The 80386 is 32-bits. Mordern Intel Processors are 64-bits.
来源：https://stackoverflow.com/questions/1434606/how-do-we-determine-if-a-processor-is-8-bit-16-bit-or-32-bit