模板參數不取代

Question

我試圖做一個通用的代碼，會導致編譯錯誤，如果B不是D的祖先我想出什麼樣的主意：

template<typename B, typename D> 
struct assert_base_of { 
    enum {value = sizeof(B::D)}; 
} 

它不起作用。當我稱之爲'像這樣：

assert_base_of<A2,A1>::value; 

我得到以下錯誤（G ++ 4.8.2）：

main.cpp:16:22: error: ‘D’ is not a member of ‘A2’ 

它看起來像模板參數d沒有得到取代A1。任何人都可以解釋並提出解決方案嗎？

Answer 1

繼承不會將派生類封裝到基類的作用域，因此使用作用域解析運算符是沒有意義的。正確的選擇（也可與多重繼承）是濫用重載決策規則：

#include <iostream> 
#include <type_traits> 

template<typename Base, typename Derived, 
     bool = std::is_same<Base, Derived>::value> 
struct is_base_of_impl 
{ 

    typedef typename std::remove_cv<Base>::type  no_cv_base;  
    typedef typename std::remove_cv<Derived>::type no_cv_derived; 


    template<typename _Up> 
    static std::true_type test(no_cv_derived&, _Up); 
    static std::false_type test(no_cv_base&, int); 

    //Black Magic 
    struct converter 
    { 
    operator no_cv_derived&(); 
    operator no_cv_base&() const; 
    }; 

    static const bool value = decltype(test(converter(), 0))::value; 
}; 

template<typename Base, typename Derived> 
struct is_base_of_impl<Base, Derived, true> 
{ 
    static const bool value = std::is_same<Base, Derived>::value; 
}; 

template<typename Base, typename Derived> 
struct is_base_of 
: public std::integral_constant<bool, 
       is_base_of_impl<Base, Derived>::value> 
{ }; 


struct A {}; 
struct B1 : A {}; 
struct B2 : A {}; 
struct C : B1, B2 {}; 

int main() 
{ 
    std::cout << is_base_of<A, B1>::value << "\n"; 
    std::cout << is_base_of<B1, C>::value << "\n"; 
    std::cout << is_base_of<A, C>::value << "\n"; 
    return 0; 
} 

欲瞭解更多信息，看看到這些鏈接：

How does `is_base_of` work?

https://groups.google.com/d/msg/comp.lang.c++.moderated/xv4VlXq2omE/--WAroYkW2QJ