Cache Line Alignment (Need clarification on article)

橙三吉。 提交于 2019-12-02 19:42:01

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.

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.

Imagine

#define CACHE_LINE_SIZE 32
sizeof(T) == 48

Now, consider how [[ align(CACHE_LINE_SIZE) ]], works. eg:

[[ align(32) ]] Foo foo;

This will force sizeof(Foo) == 32n for some n. ie align() will pad for you, if necessary, in order for things like Foo foo[10]; to have each foo[i] aligned as requested.

So, in our case, with sizeof(T) == 48, this means sizeof(cache_line_storage<T>) == 64.

So the alignment gives you the padding you were hoping for.

However, this is one 'error' in the template. Consider this case:

#define CACHE_LINE_SIZE 32
sizeof(T) == 32

Here we end up with char pad[1];. Which means sizeof(cache_line_storage<T>) == 64. Probably not what you want!

I think the template would need to be modified somewhat:

template <typename T, int padding>
struct pad_or_not
{
   T data;
   char pad[padding];
};

// specialize the 0 case
// As it is late, I am SURE I've got the specialization syntax wrong...
template <typename T, int>
struct pad_or_not<0>
{
   T data;
};

template<typename T>
struct cache_line_storage {
   [[ align(CACHE_LINE_SIZE) ]] pad_or_not<T, (sizeof(T) > CACHE_LINE_SIZE ? 0 : CACHE_LINE_SIZE - sizeof(T) ) > data;
};

or something like that.

"You can't have arrays of size 0, so 1 is required to make it compile" - GNU C does allow arrays dimensioned as zero. See also http://gcc.gnu.org/onlinedocs/gcc-4.1.2/gcc/Zero-Length.html

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