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到底什么是“对齐指针”？

发布于 2024-10-05 07:54:57 字数 37 浏览 0 评论 0原文

有人可以告诉我对齐指针实际上意味着什么吗？

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偏爱你一生 2024-10-12 07:54:57

这意味着所指向的地址可以被某个因子整除。

有时使用术语“自然对齐”，这通常意味着具有自然对齐的对象需要放置在可被对象大小整除的地址处。

对齐有时非常重要，因为许多与硬件相关的事物对这种对齐设置了限制。

例如，在经典的 SPARC 架构上（以及经典的 ARM，我认为），您无法从奇数地址读取大于一字节的整数。尝试这样做会立即因总线错误而停止您的程序。在 x86 架构上，CPU 硬件反而会处理该问题（通过根据需要对缓存和/或内存进行多次访问），尽管可能需要更长时间。 RISC:ier 架构通常不会为您执行此操作。

类似的事情也会影响填充，即在结构字段之间插入虚拟数据以保持对齐。像这样的结构：

struct example
{
  char initial;
  double coolness;
};

很可能最终在字段之间有 7 个字节的填充，以使 double 字段在可被其自身大小整除的偏移量上对齐（我假设为 8）。

当以二进制形式查看时，与 n 字节对齐的地址将其 log2(n) 最低有效位设置为零。例如，需要 32 字节对齐的对象将具有以（二进制）00000 结尾的正确对齐地址，因为 log2(32) 为 5。这也意味着可以通过清除所需的数字来强制地址对齐位。

It means that the address being pointed at is evenly divisible by some factor.

Sometimes the term "natural alignment" is used, which generally means that objects having natural alignment need to be placed at addresses that are evenly divisble by the object's size.

Alignment is somestimes very important, since many hardware-related things place restrictions on such alignment.

For instance, on the classic SPARC architecture (and also on classical ARM, I think), you can't read an integer larger than one byte from an odd address. Trying to do so will immediately halt your program with a bus error. On the x86 architecture, the CPU hardware instead handles the problem (by doing multiple accesses to cache and/or memory as needed), although it might take longer. RISC:ier architectures typically don't do this for you.

Things like these can also affect padding, i.e. the insertion of dummy data between e.g. structure fields in order to maintain alignment. A structure like this:

struct example
{
  char initial;
  double coolness;
};

would very likely end up having 7 bytes of padding between the fields, to make the double field align on an offset divisible by its own size (which I've assumed to be 8).

When viewed in binary, an address aligned to n bytes will have its log2(n) least-significant bits set to zero. For instance, an object that requires 32-byte alignment will have a properly-aligned address that ends with (binary) 00000, since log2(32) is 5. This also implies that an address can be forced into alignment by clearing the required number of bits.

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泪之魂 2024-10-12 07:54:57

为了补充 unwind 所解释的内容，这是我最近在作业中使用的一个 struct ：

struct infosale {               
    int   noseq;                
    char  salesman[30];         
    char  product[11];          
    int   count;                
};

您可能期望此struct的大小为(4+30+11+4=)49字节，但实际上是52 与 sizeof 相比。因为 noseq 是 4 字节 + salesman 是 32 字节（对齐）+ product是 12 字节（对齐），而 count 是 4 字节，因此是 52 字节。

To add to what unwind is explaining, here is a struct I have recently used in an assignment :

struct infosale {               
    int   noseq;                
    char  salesman[30];         
    char  product[11];          
    int   count;                
};

You may expect the size of this struct to be (4+30+11+4=) 49 bytes, but it is in fact 52 compared with sizeof. Because noseq is 4 bytes + salesman is 32 bytes (aligned) + product is 12 bytes (aligned) and count is 4 bytes, thus 52 bytes.