使用减少的模板参数数量推导模板类

我有一个 wrapper 需要a的课程 template<typename> class 作为一个参数。但是,我有一个带有额外参数的类,例如 template<typename T, typename P> Special ,我想将其放入包装器类的参数中。

我唯一能想到的解决方案如下:

// test.cpp

// Base class
class Base {
public:
  int data_base = 1;
};

// A derive class
template<typename P> 
class Derived : public Base
{
  P data_child;
};

// A wrapper to accept any type similar to Derived<P> as the base class
template<template<typename> class BT, typename P>
class Wrapper : public BT<P>
{
  int do_something;
};

// However, a special derived class need an extra template parameter
template<typename T, typename P>
class Special : public Derived<P>
{
  T special;
};

// Try to fit Special into the type of BT of Wrapper
template<typename T>
class SpecialBridge
{
public:
  template<typename P>
  using BT = Special<T, P>;
};

// actually define the type to use (Error!)
template<typename T, typename P>
using Actual = Wrapper<SpecialBridge<T>::BT, P>;


int main() {
  Actual<int, int> obj;
  return obj.data_base;
}

然而,c++17和c++20都声称相同。编译器无法推断 SpecialBridge<T>::BT

$ g++ -std=c++17 test.cpp -o test
test.cpp:40:47: error: type/value mismatch at argument 1 in template parameter list for ‘template<template<class> class BT, class P> class Wrapper’
   40 | using Actual = Wrapper<SpecialBridge<T>::BT, P>;
      |                                               ^
test.cpp:40:47: note:   expected a class template, got ‘SpecialBridge<T>::BT’
test.cpp: In function ‘int main()’:
test.cpp:44:3: error: ‘Actual’ was not declared in this scope
   44 |   Actual<int, int> obj;
      |   ^~~~~~
test.cpp:44:10: error: expected primary-expression before ‘int’
   44 |   Actual<int, int> obj;
      |          ^~~
test.cpp:45:10: error: ‘obj’ was not declared in this scope
   45 |   return obj.data_base;
      |          ^~~

我有什么办法可以把这件事做好吗。很抱歉,我描述这个问题的方式不太标准。示例代码或多或少是可以自我解释的。