C++ 中的模板参数类型分配器

Question

我有以下可用的玩具代码：

#include <cstdlib>
#include <vector>

template <typename T>
struct Dummy {
        using value_type = T;
        using size_type = size_t;
        using difference_type = ptrdiff_t;
        using pointer = value_type*;
        using const_pointer = const value_type*;
        using reference = value_type&;
        using const_reference = const value_type&;

        Dummy() = default;
        Dummy(const Dummy&) = default;
        
        pointer allocate(size_type n, const void* hint = 0) {
                (void)hint;
                return static_cast<pointer>(aligned_alloc(128, n * sizeof(T)));
        }

        void deallocate(void* p, size_type) {
                if (p)
                        free(p);
        }
};

int main()
{
        
        std::vector<int, Dummy<int, 16>> v;
        
        
        for (size_t i = 1; i < 5000; ++i)
                v.push_back(i);
        
        
        return 0;
}

现在我想对对齐进行模板化（而不是放入构造函数中）。我尝试向模板添加一个整数参数

template <typename T, size_t align>
struct Dummy {
        using value_type = T;
        using size_type = size_t;
        using difference_type = ptrdiff_t;
        using pointer = value_type*;
        using const_pointer = const value_type*;
        using reference = value_type&;
        using const_reference = const value_type&;

        Dummy() = default;
        Dummy(const Dummy&) = default;

        pointer allocate(size_type n, const void* hint = 0) {
                (void)hint;
                return static_cast<pointer>(aligned_alloc(align, n * sizeof(T)));
        }

        void deallocate(void* p, size_type) {
                if (p)
                        free(p);
        }
};

int main()
{
        
        std::vector<int, Dummy<int, 128>> v;
        
        
        for (size_t i = 1; i < 5000; ++i)
                v.push_back(i);
        
        
        return 0;
}

我在使用 g++ 和 clang++ 时遇到很多编译错误：例如

/usr/include/c++/10.2.0/bits/alloc_traits.h:78:11: error: no type named 'type' in 'struct std::__allocator_traits_base::__rebind<Dummy<int, 128>, int, void>'
   78 |     using __alloc_rebind
      |           ^~~~~~~~~~~~~~
In file included from /usr/include/c++/10.2.0/vector:67,
                 from aligned.cpp:4:
/usr/include/c++/10.2.0/bits/stl_vector.h: In instantiation of 'class std::vector<int, Dummy<int, 128> >':
aligned.cpp:62:43:   required from here
/usr/include/c++/10.2.0/bits/stl_vector.h:474:20: error: '_M_allocate' has not been declared in 'std::_Base<int, Dummy<int, 128> >'
  474 |       using _Base::_M_allocate;
      |                    ^~~~~~~~~~~
/usr/include/c++/10.2.0/bits/stl_vector.h:475:20: error: '_M_deallocate' has not been declared in 'std::_Base<int, Dummy<int, 128> >'
  475 |       using _Base::_M_deallocate;

我不明白这是怎么回事。我注意到，即使我不在 class 中使用 align，编译器也会以同样的方式抱怨。相反，如果我在模板中放置一个未使用的 typename ，一切都可以正常编译。你能帮帮我吗？

Answer 1

我想我找到了解决问题的方法，尽管我不确定我是否完全理解所有细节。

感谢：this and this我写了

template <typename T, size_t align>
struct Dummy {
        using value_type = T;
        using size_type = size_t;
        using difference_type = ptrdiff_t;
        using pointer = value_type*;
        using const_pointer = const value_type*;
        using reference = value_type&;
        using const_reference = const value_type&;

        template<class Other>
        struct rebind { using other =  Dummy<Other, align>; };



        
        Dummy() {};
        Dummy(const Dummy&) {};

        template <typename Other>
        Dummy(const Dummy<Other, align>&) {}
        
        pointer allocate(size_type n, const void* hint = 0) {
                (void)hint;
                return static_cast<pointer>(aligned_alloc(align, n * sizeof(T)));
        }

        void deallocate(void* p, size_type) {
                if (p)
                        free(p);
        }
};

编译并似乎具有正确的行为。

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