什么是错位的指针？ [英] What is a misaligned pointer ?

问题描述

据我所知，在下面的行，我们正在试图写入到一个无效的内存位置。但是，这实际上是一个不对齐指针也。有人可以解释什么是错位的指针又是怎样以下未对齐的指针？

  *（INT *）为0xffffffff = 0xbad;
 

解决方案

许多架构有一个概念叫的调整的，其中硬件的设计上是字大小的倍数的地址进行操作。例如，在32位处理器，对象可能被对准以32位边界（4字节），和在64位处理器，对象可能被排列为64位边界（8字节）。对准指针是一个指向一个地址的字的大小的倍数，和未对齐的指针是一个不是指向一个地址是这样的字大小的倍数。

在大多数架构，读取或写入未对齐的指针患有某种形式的处罚。在某些处理器，这样做会导致总线错误，这通常是立即终止程序。在其他国家，如X86，未对齐的读取和写入操作是合法的，但遭受性能损失是由于硬件的结构。

在你的code， 0xBAD = 2989 很可能不对齐，因为它不是一个多最常见的字长，并且你写指针也可能没有对齐。

希望这有助于！

I understand that in the following line we are attempting to write to an invalid memory location. But this is actually a misaligned pointer also. Can someone explain what is a misaligned pointer and how is the following misaligned pointer ?

*(int*)0xffffffff = 0xbad;

解决方案

Many architectures have a concept called alignment where the hardware is designed to operate on addresses that are multiples of the word size. For example, on a 32-bit processor, objects might be aligned to 32-bit boundaries (4 bytes), and on a 64-bit processor, objects might be aligned to 64-bit boundaries (8 bytes). An aligned pointer is one that points to an address that's a multiple of the word size, and an unaligned pointer is one that's not pointing to an address that's a multiple of the word size.

On most architectures, reading or writing unaligned pointers suffers some sort of penalty. On some processors, doing this causes a bus error, which usually terminates the program immediately. On others, such as x86, unaligned reads and writes are legal but suffer a performance penalty due to how the hardware is structured.

In your code, 0xBAD = 2989 is probably not aligned, since it's not a multiple of most common word sizes, and the pointer you write to is also probably not aligned.

Hope this helps!

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