How many bits does a WORD contain in 32/64 bit OS respectively?

本小妞迷上赌 提交于 2019-12-18 11:00:15

问题


Anyone has a definite answer?

Someone says that on 32 bit OS a WORD means 16bit,true?


回答1:


The concept of a "word" has several meanings. There's 3 meanings embedded in the question.

  • The generic term "processor word", in context of CPU architectures
  • The "bit size" of software/OS, vs the "bit size" of hardware
  • The all-caps term WORD, meaning a 16 bit value - This is a part of the Windows "Win32" C language API

When describing the Win32 WORD type definition, this also comes up:

  • The Intel/AMD instruction set concept of a "Word", "Doubleword", and "Quadword"

The generic term "processor word", in context of CPU architectures

In common/generic usage, a "processor word" refers to the size of a processor register. It can also refer to the size of CPU instruction, or the size of a pointer (depending on which exact CPU architecture). In simple cases, a 32 bit processor will have a 32 bit "word" size (and pointer size). A 64 bit processor will have a 64 bit "word" size (and pointer size).

There is a wikipedia article on this "processor word" concept, which details all the generic uses of the term, and the sizes for several current and historical CPU architectures.

"Bit size" of software/OS vs the "bit size" of hardware

A "64 bit" CPU and a "64 bit" OS are necessary in order to run "64 bit" software. This much is probably obvious.

"64 bit software" uses 64 bit instructions (e.g. adding 64 bit numbers together, or copying 64 bits of data from a processor register to RAM at the same time). It also can use a 64 bit pointer size. This means that instead of only being able to use a maximum of 4 Gigabytes of RAM (like "32 bit software"), it can theoretically use about 17 Billion Gigabytes of RAM (16 Exabytes).

A "64 bit" x64/x86 CPU can also run "32 bit" (or even "16 bit") software. It can do this without any changes to the code, and without having to rebuild the software. This is because all the old CPU instructions still exist on new CPUs, and they are backwards compatible.

These concepts aren't strictly the same as the generic concept of a "processor word", but are closely related.

Note: This concept starts getting slightly more complicated when you talk about older and more specialized processors (especially older video game systems), but the question wasn't really about those so I won't go into detail. Those tend to be talked about as "64 bit" or "8 bit" systems, but the truth is a bit more complicated than that. See the "processor word" wiki article I linked above, or an article about the specific system in question.

The question's specific context - WORD, in all-caps

The capitalization and the specific sizes in the question (16 bit for WORD, on a 32 bit OS) imply something different than the generic term "processor word".

In legacy Windows programming (the Win32 API), there is a macro defined called WORD, the size of which is 16 bits. This made sense when processors were 16 bit. However, even when you compile code that contains this macro for a 32 bit or 64 bit target, it will still be 16 bits. A DWORD in the Win32 API is 32 bits, and a QWORD is 64 bits.

This is because Microsoft really tries very hard in their Win32 API to support backwards compatibility without having to do any changes to code. For the most part you can compile the Win32 samples from the Windows 95 era without changes, and they'll still work exactly the same way today.

Microsoft very likely inherited this naming scheme from Intel (and possibly AMD) documentation.

The Intel/AMD instruction set concept of a "Word", "Doubleword", etc

In Intel docs, a "Word" (Win32 WORD) is 16 bits. A "Doubleword" (Win32 DWORD) is 32 bits. A "Quadword" (Win32 QWORD) is 64 bits. The related assembly instruction names also reflect this naming scheme (e.g. MMX Add Packed Integers PADD instructions: PADDW, PADDD, PADDQ).

For some examples, you can check this wikipedia article on the x86 instruction set, or the Intel software development manuals.

This naming scheme doesn't necessarily make sense in terms of the general concept of a "processor word", since these concepts only address a part of a register. However they do make sense in terms of creating a stable programming interface for x86 programs. This is a big part of why you can use "32 bit" (and 16 bit) programs on top of a "64 bit" OS.




回答2:


One Word is the size of 16 bit DWord(double word) is double the size of word that is 32 bit when used in programing but...

The name Word in OS is the number shown nexto the OS so if it's saying 64 bit one word (adres for storege) is 64 bit's in this case

So it depends from what angle your looking at it from programing or the OS number




回答3:


There is no definitive answer.

Traditionally the term "word" refers to the size of the processor's registers and main data path. By that definition a "word" would be 32 bit on your 32-bit system and 64-bit on your 64-bit system.

However when processor families were extended to add wider registers/operating modes the manufacturers and users of those processors sometimes continued to use "word" to refer to the word size of the original processor. The same can happen when software is ported from one processor family to another.

Intel x86 documentation uses the term "word"¹ to refer to a 16 bit quantity. This usage bleeds over into software environments that were originally developed for x86 such as the windows API and Borland-style pascal.

On the other hand arm documentation uses the term "word"² to refer to a 32-bit quantity.

¹ see section 4.1 in https://software.intel.com/sites/default/files/managed/a4/60/253665-sdm-vol-1.pdf

² see section A1.4 in https://static.docs.arm.com/ddi0487/db/DDI0487D_b_armv8_arm.pdf



来源:https://stackoverflow.com/questions/5295903/how-many-bits-does-a-word-contain-in-32-64-bit-os-respectively

标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!