c++ - 为什么 Visual Studio 无法在模板类中选择正确的构造函数？

Question

当我实例化模板类时，Visual Studio 看不到正确的构造函数。我哪里做错了？

我已经尝试过明确/删除复制/移动构造函数。没有帮助。

#include <set>

using namespace std;

template <class T, template<class> class ConnectionType>
struct node
{
    T value;
    node(const T& value) : value(value) {}

    set<ConnectionType<T>> connections;
};

template <class T>
struct connection
{
    node<T, connection>* n;

    connection(node<T, connection>* n) :
        n(n) {}

    bool operator<(const connection& b) const
    {
        return n < b.n;
    }
};

int main()
{
    node<int, connection> a(0);
    connection<int> c(&a); // ERROR HERE

    return 0;
}

错误:

error C2664:  'connection<T>::connection(connection<T> &&)': cannot convert argument 1 from 'node<int, connection> *' to 'node<T, connection<T>> *'

Answer 1

好像是VS的bug。 VS 似乎正在处理 injected class name connection作为相当于 connection<T> 的类型名称, 但它应该被视为类模板本身的模板名称，即 connection在 node<T, connection>* n;和 connection(node<T, connection>* n) ，因为 node 的第二个模板参数是模板模板参数。

(强调我的)

In the following cases, the injected-class-name is treated as a template-name of the class template itself:

• it is followed by <
• it is used as a template argument that corresponds to a template template parameter
• it is the final identifier in the elaborated class specifier of a friend class template declaration.

Otherwise, it is treated as a type-name, and is equivalent to the template-name followed by the template-parameters of the class template enclosed in <>.

template <template <class, class> class> struct A;

template<class T1, class T2>
struct X {
    X<T1, T2>* p; // OK, X is treated as a template-name
    using a = A<X>; // OK, X is treated as a template-name
    template<class U1, class U2>
    friend class X; // OK, X is treated as a template-name
    X* q; // OK, X is treated as a type-name, equivalent to X<T1, T2>
};

PS:您的代码可以很好地编译 clang .

PS: 处理为connection<T>在 bool operator<(const connection& b) const .