模板化 ctor 上的编译器错误(这是编译器错误吗?)

Compiler error on templated ctor (is this a compiler bug?)

当我尝试在 Visual Studio 2019 年编译来自 this post 的代码(见下文)时,我在模板化复制构造函数的行收到以下编译器错误:

Error   C2439   'PreAllocator<_Newfirst>::memory_ptr': member could not be initialized
Error   C2440   'initializing': cannot convert from 'T *const ' to 'T *'
Error   C2248   'PreAllocator<int>::memory_size': cannot access private member declared in class 'PreAllocator<int>'    
Error   C2248   'PreAllocator<int>::memory_ptr': cannot access private member declared in class 'PreAllocator<int>' 

这是编译器错误还是我遗漏了什么?


template <typename T>
class PreAllocator
{
private:
    T* memory_ptr;
    std::size_t memory_size;

public:
    typedef std::size_t     size_type;
    typedef T* pointer;
    typedef T               value_type;

    PreAllocator(T* memory_ptr, std::size_t memory_size) : memory_ptr(memory_ptr), memory_size(memory_size) {}

    PreAllocator(const PreAllocator& other) throw() : memory_ptr(other.memory_ptr), memory_size(other.memory_size) {};

    template<typename U>
    PreAllocator(const PreAllocator<U>& other) throw() : memory_ptr(other.memory_ptr), memory_size(other.memory_size) {};

    template<typename U>
    PreAllocator& operator = (const PreAllocator<U>& other) { return *this; }
    PreAllocator<T>& operator = (const PreAllocator& other) { return *this; }
    ~PreAllocator() {}

    pointer allocate(size_type n, const void* hint = 0) { return memory_ptr; }
    void deallocate(T* ptr, size_type n) {}

    size_type max_size() const { return memory_size; }
};

int main()
{
    int my_arr[100] = { 0 };
    std::vector<int, PreAllocator<int>> my_vec(0, PreAllocator<int>(&my_arr[0], 100));
}
template<typename U>
PreAllocator(const PreAllocator<U>& other) throw() : memory_ptr(other.memory_ptr), memory_size(other.memory_size) {};

此构造函数尝试从 U* 类型的 other.memory_ptr 初始化 T* 类型的 memory_ptr,这与 T* 不同(因为否则一个副本构造函数将被调用)。 std::vector 实现显然使用内部类型来实例化分配器以分配存储,并且指向该类型的指针不能隐式转换为您期望的 int*

您必须显式转换指针,或在内部存储 void*,然后您将在 allocate 成员函数中将其转换为 T*

template <typename T>
class PreAllocator
{
private:
    void* memory_ptr;

    ...

public:
    template<typename U>
    PreAllocator(const PreAllocator<U>& other) throw() :
        memory_ptr(other.memory_ptr), memory_size(other.memory_size) {}

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

    ...
};

此外,在同一个构造函数中,您正在访问 PreAllocator<U>memory_ptrmemory_size 成员,这是与 PreAllocator<T> 不同且不相关的类型(同样,因为 TU 是不同的类型)。这些成员是私有的,因为 class,这给了你访问错误。

您可以通过使这些成员成为 public 或为他们添加 public 访问器来缓解这种情况,或者通过如下声明使 PreAllocator 朋友的所有专业化:

template <typename T>
class PreAllocator
{
    template< typename U >
    friend class PreAllocator;

    ...
};