Question

我有这样的层次结构：

#include <boost/shared_ptr.hpp>
//base cache
template <typename KEY, typename VAL>  class Cache;
//derived from base to implement LRU
template <typename KEY, typename VAL>  class LRU_Cache :public Cache<KEY,VAL>{};
//specialize the above class in order to accept shared_ptr only
template <typename K, typename VAL>
class LRU_Cache<K, boost::shared_ptr<VAL> >{
public:
    LRU_Cache(size_t capacity)
    {
        assert(capacity != 0);
    }
//....
};

//so far  so good

//extend the LRU
template<typename K, typename V>
class Extended_LRU_Cache : public LRU_Cache<K,V >
{
public:
    Extended_LRU_Cache(size_t capacity)
    :LRU_Cache(capacity)// <-- this is where the error comes from
    {
        assert(capacity != 0);
    }
};

错误说：

template-1.cpp: In constructor ‘Extended_LRU_Cache<K, V>::Extended_LRU_Cache(size_t)’:
template-1.cpp:18:38: error: class ‘Extended_LRU_Cache<K, V>’ does not have any field named ‘LRU_Cache’
  Extended_LRU_Cache(size_t capacity):LRU_Cache(capacity)

你可以帮我找到缺失的部分吗？

感谢

Answer 1

调用基础构造函数时，需要父级的完整定义（即使用模板参数）：

Extended_LRU_Cache(size_t capacity)
:LRU_Cache<K,V>(capacity)
{
    assert(capacity != 0);
}

Answer 2

您需要指定基本类型的模板参数：

Extended_LRU_Cache(size_t capacity)
:LRU_Cache<K, V>(capacity)
{
    assert(capacity != 0);
}

因为您没有提供它们，所以编译器不会建立您尝试调用基础构造函数的连接，因为您不能从名为LRU_Cache的类型派生，所以它查找名称为LRU_Cache的字段以进行初始化。现在应该很清楚为什么会收到这个pariticular错误消息。

模板参数是必需的，因为您可以使用不同的模板参数从相同的模板化类型派生两次，然后基础构造函数调用将是不明确的：

template <typename T>
class Base
{
public:
    Base() { }
};

template <typename T, typename U>
class Derived : Base<T>, Base<U>
{
public:
    // error: class ‘Derived<T, U>’ does not have any field named ‘Base’
    // Derived() : Base() { }
    //             ^
    // Which Base constructor are we calling here?

    // But this works:
    Derived() : Base<T>(), Base<U>() { }
};

不能从派生类调用Base模板化类的构造函数

2 个答案: