blocks|key|1617331|text|我猜通常你会根据它的累加器/寄存器的大小来标记它。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1617332|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

I guess normally you label it by the size of it's accumulators/registers.

blocks|key|99635|text|关于CPU，我会说它是一个寄存器的宽度。一次只能对8+bits、16-bits、32-bits等进行操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|99636|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/8-bit|1|http://en.wikipedia.org/wiki/16-bit|2|http://en.wikipedia.org/wiki/32-bit^0|P|6|0|W|7|1|14|7|2|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]|$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|-4|5|6|7|Z|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]|L|$5|G|H|I|C|$J|M]]|N|$5|G|H|I|C|$J|O]]]]

With respect to a CPU, I'd say that it's the width of a register. You can do an operation on only <a href="http://en.wikipedia.org/wiki/8-bit" rel="nofollow noreferrer">8 bits</a>, <a href="http://en.wikipedia.org/wiki/16-bit" rel="nofollow noreferrer">16-bits</a>, <a href="http://en.wikipedia.org/wiki/32-bit" rel="nofollow noreferrer">32-bits</a>, etc. at a time.

blocks|key|315446|text|它的定义不是很好。广义上，正如xtofl所指出的，它是原子计算单元的大小(在早期的计算机中，它并不总是与“register”同义)。因此，PDP-10是36位机器，8080是8位，IBM360或英特尔80386是"32位“。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|315447|但也有例外。摩托罗拉68000和68010+CPU实现了32位寄存器集，但它是通过微码在主要是16位的内部架构上实现的。它们在当时的市场上通常被称为"16位“CPU。|315448|地址总线的大小几乎从来不是决定因素。例如，所有成功的"8位“CPU都实现了16位寻址(通常通过奇数破解来弥补单个地址寄存器c.f.的不足，6502的间接寻址模式或Z80的H/L寄存器)。正如您所提到的，80286使用段寄存器寻址使20条地址线工作(+8086将这一技巧扩展到24位物理地址)。另一方面，许多"32位“CPU有较小的地址总线，以节省不会在内存超过几兆字节的机器上使用的逻辑:+68000将寻址限制为24位，即使指针本身是32位。同样，现代64位CPU普遍实现少于64位的物理地址。|315449|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

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.

blocks|key|1505030|text|微处理器的位大小(8位、16位、32位)由硬件决定，特别是数据总线的宽度。Intel8086是一种16位处理器，因为它可以在数据总线上一次移动16位。Intel8088是一个8位处理器，即使它有相同的指令集。这类似于摩托罗拉68000和68008处理器。位大小不是由程序员的观点(寄存器宽度和地址范围)决定的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1505031|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

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).

blocks|key|1617417|text|我认为集成芯片的第一个数字是指处理器的类型。如果是IC8085，则表示它是一个8位处理器。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1617418|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

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.

blocks|key|1505137|text|据我所知，处理器的位宽是由内部数据处理电路一次接受多少位决定的。例如，如果ALU中的加法器、乘法器等接受16位操作数，那么CPU就是16位，如果它接受32位，那么它就是32位。数据总线或地址总线的宽度是多少并不重要。一般来说，累加器的位长将决定处理器的位长。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1505138|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

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.

blocks|key|99862|text|任何处理器都可以通过它的两个属性来指定|type|unstyled|depth|inlineStyleRanges|entityRanges|data|99863|99864|指令集体系结构&|ordered-list-item|99865|no.在单个时钟周期内可以处理的位数。|99866|99867|以英特尔的IA-32体系结构为例，也称为x86-32，其中x86表示体系结构，32表示32位处理器|99868|X-架构|99869|有许多架构|99870|x86+x86之前的版本|99871|-英特尔的IA-32体系结构，也称为x86-32+-x86-64+-+-with+AMD的AMD64和英特尔的64版本-摩托罗拉的6800和68000+a|99872|rchitectures+ARM7|99873|Y位处理器|99874|：简单-它是cpu/处理器在单个时钟周期内的数据处理能力。假设它是一个8位处理器，那么在一个时钟周期内，ALU只能对8位数据执行操作。(请注意，此操作可能是内部操作，如加/减以及将数据传输到其他IO设备)。|99875|classification+Based+on+Registers-|code-block|syntax|javascript|99876|除了ALU和CU之外，处理器还包含一些内存位置，称为寄存器。根据处理器的不同，寄存器通常可以存储8、16、32或64位。特定处理器的寄存器大小允许我们对处理器进行分类。寄存器大小为n位的处理器称为n位处理器，因此具有8位寄存器的处理器称为8位处理器。|99877|classification+Based+on+databus+width-|99878|由于alu在单个时钟周期中只能处理8位数据，因此数据总线宽度大于8位处理器将具有8位宽数据总线是没有意义的，因此数据总线宽度也可以是找出processor.for处理器的位处理能力的替代方式。|99879|关于问题：|99880|99881|“假设我们有一个32位的算术逻辑单元，即它一次可以取32位，我们的数据总线大小是16位，即它一次可以容纳16位的数据，那么ans将是什么。在这种情况下...？”|blockquote|99882|99883|99884|这种处理器例子将是Intel8088和Moto+68000|99885|使用总线宽度分类，Intel8088微处理器是一个8位处理器，因为它使用8位数据总线，尽管它的CPU寄存器实际上是16位寄存器。类似地，摩托罗拉68000被归类为16位处理器，即使它的CPU寄存器是32位寄存器。有时使用两种分类的组合，其中8088可被描述为8/16位处理器，而摩托罗拉68000可被描述为16/32位处理器。|99886|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1N|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|1O|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|1P|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|1Q|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|1R|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|1S|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|1T|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1U|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1V|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1W|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1X|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|10|7|20|8|@]|9|@]|A|$11|12]]|$1|13|3|14|5|6|7|21|8|@]|9|@]|A|$]]|$1|15|3|16|5|10|7|22|8|@]|9|@]|A|$11|12]]|$1|17|3|18|5|6|7|23|8|@]|9|@]|A|$]]|$1|19|3|1A|5|6|7|24|8|@]|9|@]|A|$]]|$1|1B|3|-4|5|6|7|25|8|@]|9|@]|A|$]]|$1|1C|3|1D|5|1E|7|26|8|@]|9|@]|A|$]]|$1|1F|3|-4|5|6|7|27|8|@]|9|@]|A|$]]|$1|1G|3|-4|5|6|7|28|8|@]|9|@]|A|$]]|$1|1H|3|1I|5|6|7|29|8|@]|9|@]|A|$]]|$1|1J|3|1K|5|6|7|2A|8|@]|9|@]|A|$]]|$1|1L|3|-4|5|6|7|2B|8|@]|9|@]|A|$]]]|1M|$]]

any processor can be designated by its' two attributes

<ol>
<li>instruction set architecture &amp;</li>
<li>no. of bits it can handle in single clock cycle.</li>
</ol>

take for example Intel's IA-32 architecture, also called x86-32 , here x86 indicates the architecture and 32 indicates 32-bit processor

<h2>X-Architecture</h2>

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

<h2>Y-bit processor</h2>

: 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)

<pre><code>classification Based on Registers-
</code></pre>

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.

<pre><code>classification Based on databus width-
</code></pre>

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 &amp; 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: 

<blockquote>
 "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...?"
</blockquote>

the example of such processor would be intel 8088 &amp; 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.

blocks|key|315577|text|微处理器的字长(8位、16位或32位)是执行单元中数据路径的大小。通常，这是累加器的大小。这是执行单元的大小。这很重要的一个例子是8088，它是一台运行在8位总线上的16位计算机。8085是8位的。8086/8088是16位的。80386是32位的。现代的英特尔处理器是64位的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|315578|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

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.

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?

How do we determine if a processor is 8-bit; 16-bit or 32-bit

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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它是由地址总线的大小决定的吗?如果是，那么8086是一个20位处理器吗？如果不是，那么分配8位、16位、32位这样的位数给处理器的标准是什么？

问如何确定处理器是8位、16位还是32位
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问如何确定处理器是8位、16位还是32位EN