Question

我正在研究基于政策的设计，并创造了一个人为的例子。

#include <iostream>
#include <string>

template <typename T>
struct NewAllocator
{
    T* Allocate(){
        std::cout << "Allocated with new\n";
        return nullptr;
    }
};

template <typename T>
struct MallocAllocator
{
    T* Allocate(){
        std::cout << "Allocated with malloc\n";
        return nullptr;
    }
};

template <typename T>
struct CustomAllocator
{
    CustomAllocator(const char* name)
        :alloctr_name{name}
    {}

    T* Allocate(){
        std::cout << "Allocated with custom\n";
        return nullptr;
    }

    // Enriched behaviour
    void print_name()
    {
        std::cout << "Name of the allocator is: " << alloctr_name;
    }

    std::string alloctr_name;
};

template <typename T, template <typename Allocated> typename AllocationPolicy = NewAllocator>
struct my_container : public AllocationPolicy<T>
{
    my_container()
    :begin{AllocationPolicy<T>::Allocate()}
    {}

    // Host can use enriched behaviour from policies
    my_container(const char* alloctr_name)
    :AllocationPolicy<T>(alloctr_name)
    ,begin{AllocationPolicy<T>::Allocate()}
    {}

    // template only initialized if used
    void print_allocator_name()
    {
        AllocationPolicy<T>::print_name();
    }

    T *begin;
};

这是按预期工作但我很好奇。在my_container的第一个构造函数中，我调用策略类的Allocate方法将成员初始化为begin{AllocationPolicy<T>::Allocate()}，正如我所说的那样。但是，如果我尝试在同一行执行begin{Allocate()}，则会收到错误消息，指出Allocate是未声明的标识符。为什么会这样？这不是虚函数或任何东西。 Allocate是我基类的成员函数，我不应该在没有任何资格的情况下访问它吗？

调用基类成员函数而不限定它

0 个答案: