在32位架构的读取短裤（例如） [英] Reading shorts in 32-bits architectures (for example)
问题描述
首先，对不起，我的英语水平。
First of all, sorry for my English.
我知道架构是非常复杂的，有的情况下一个宽s prectrum，但一种常见的概括是，如果一个计算机体系结构具有32位的话，装置的寄存器，存储器访问和公共汽车用的32位字的工作长（但我觉得有很多的变种在当前的体系结构）。
I know architectures are very complex and there's a broad sprectrum of situations, but a common generalization is if a computer architecture has 32-bits words, means registers, memory accesses and buses work with words of 32-bits long (but I think there's a lot of variants in current architectures).
好吧，让我们假设这是规则，我们的架构是一个小端之一，86。在这种情况下，如果我们要读短整型（2字节长），内存读取，然后在4字节字包含了短。让我们假设包含词W是 0xf1342ea0 ，内存：
Ok, let's suppose this is the rule and our architecture is a little-endian one, as x86. In such a case, if we want to read a short int (2-bytes long), the memory reads then the 4-bytes word which contains our short. Let's suppose the containing word W is 0xf1342ea0, in memory:
{a0, 2e, 34, f1} // a0 is the byte in the lowest address.
和我们的半字H被在W的最高部分，然后，H是0xf134。据我所知，处理器接收，从内存，一个字与短转变：
and our half-word H is in the highest part of W, then, H is 0xf134. I understand the processor receives, from the memory, a word with the short shifted:
{34, f1, 00, 00}
由于0x0000f134等于0xf134。
since 0x0000f134 equals 0xf134.
通过这个画面在脑海中，由于处理器是4字节长，它是这样neccesary通过各种手段包换，为什么一定要2个字节的数据在2字节字边界对齐？
With this picture in mind, since the processor is 4-bytes long and it is thus neccesary by all means a shifting, why must 2-bytes data to be aligned in 2-bytes word boundaries?
在换句话说：
为什么令人鼓舞建议不要读短中字0xf134：
Why is encouragingly recommended not to read the short 0xf134 in the word:
{ff, 34, f1, 0a}
修改：EX $ P $的其他方式pssing同样的疑问是：为什么对齐的定义是
EDIT: Other way of expressing the same doubt is: why the definition of alignment is
A object of size N and address d is aligned if d is divisible by N.
而不是
A object of size N and direction d is aligned respect to an architecture
of B bytes if d is divisible by B, or ⌊d/B⌋ == ⌊(d+N)/B⌋ if N < B.
注意：属性⌊d/B⌋==⌊（D + N）/B⌋意味着对象属于对齐的字
NOTE: The property ⌊d/B⌋ == ⌊(d+N)/B⌋ implies the object belongs to an aligned word.
推荐答案
如果内存{FF，34，F1，0A}，那么它不是一个x86处理器的一个问题。然而，如果存储器是{FF，FF，FF，34} {F1，氨基酸，氨基酸，氨基酸}，处理器必须执行两个总线周期来检索短的值。 （还要注意有一些根本不支持未对齐访问的基于RISC的处理器。）
If the memory is { ff, 34, f1, 0a }, then it's not a problem for an x86 processor. However, if the memory is { ff, ff, ff, 34 } {f1, aa, aa, aa }, the processor must perform two bus cycles to retrieve the value of the short. (Also note there are some RISC-based processors that do not support misaligned accesses at all.)
这篇关于在32位架构的读取短裤（例如）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
