缓存行对齐（需要澄清文章） [英] Cache Line Alignment (Need clarification on article)

问题描述

我最近遇到了我认为在我的应用程序中出现的假共享问题，我已经查看了 Sutter的文章如何对齐我的数据到缓存行。他建议使用以下C ++代码：

I've recently encountered what I think is a false-sharing problem in my application, and I've looked up Sutter's article on how to align my data to cache lines. He suggests the following C++ code:

// C++ (using C++0x alignment syntax)
template<typename T>
struct cache_line_storage {
   [[ align(CACHE_LINE_SIZE) ]] T data;
   char pad[ CACHE_LINE_SIZE > sizeof(T)
        ? CACHE_LINE_SIZE - sizeof(T)
        : 1 ];
};

我可以看到当 CACHE_LINE_SIZE> sizeof（T）是true - struct cache_line_storage 刚刚占用一个完整的缓存行内存。然而，当 sizeof（T）大于单个缓存行时，我认为我们应该通过 CACHE_LINE_SIZE - T％CACHE_LINE_SIZE 字节，以便生成的结构体的大小是缓存行大小的整数倍。我的理解有什么问题？为什么填充1个字节就足够了？

I can see how this would work when CACHE_LINE_SIZE > sizeof(T) is true -- the struct cache_line_storage just ends up taking up one full cache line of memory. However, when the sizeof(T) is larger than a single cache line, I would think that we should pad the data by CACHE_LINE_SIZE - T % CACHE_LINE_SIZE bytes, so that the resulting struct has a size that is an integral multiple of the cache line size. What is wrong with my understanding? Why does padding with 1 byte suffice?

推荐答案

你不能有大小为0的数组，编译。然而，规范的当前草案版本说这种填充是不必要的;编译器必须填充到结构的对齐方式。

You can't have arrays of size 0, so 1 is required to make it compile. However, the current draft version of the spec says that such padding is unecessary; the compiler must pad up to the struct's alignment.

请注意，如果 CACHE_LINE_SIZE 小于 alignof（T）。要解决这个问题，你应该使用 [[align（CACHE_LINE_SIZE），align（T）]] ，这将确保不会选择更小的对齐。

Note also that this code is ill-formed if CACHE_LINE_SIZE is smaller than alignof(T). To fix this, you should probably use [[align(CACHE_LINE_SIZE), align(T)]], which will ensure that a smaller alignment is never picked.

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