对于具有两个模板化变量的模板类，可以使一个var引用另一个var？ [英] for a templated class with two templated variable, can one var be made to reference the other var?

问题描述

我正在寻找一种方法来引用A的t从真正的类型B和C.在下面的代码中，你看到我的第一个倾向是尝试并初始化它。其他尝试我尝试使用完美的转发，继承和添加更多的模板参数到B和C.有人可以建议一个路径前进？有什么新的方法可能有帮助吗？我是关闭还是不可能？

I am looking for a way to reference A's t from within real types B and C. In the code below you see that my first inclination is to try and initialize it. Other attempts I have tried were using perfect forwarding, inheritance and adding more templates parameters to B and C. Can someone suggest a path forward? Are there new constructs on the way that may help? Am I close or is this impossible?

struct D {};
struct E {};

template< typename U1 > 
struct B 
{
  B() : u1(???)
  U1& u1; // how to reference A's t variable?
};

template< typename U2 > 
struct C 
{ 
  C() : u2(???)
  U2& u2; // how to reference A's t variable?
};

template< typename T, typename U >
struct A
{
  T t;
  U u;
};

int main()
{
  A< D, B< D > > a1;

  A< E, C< E > > a2;

  return 0;
}

推荐答案

是模板模板参数。

这是他的代码一次审查：

I think what the OP wants is a template template parameter.
Here is his code once reviewed:

#include<memory>

struct D {};
struct E {};

template< typename U > 
struct B 
{
    B(std::shared_ptr<U> v) : u{v} {}
    std::shared_ptr<U> u;
};

template< typename U > 
struct C 
{ 
    C(std::shared_ptr<U> v) : u{v} {}
    std::shared_ptr<U> u;
};

template< typename T, template<typename> typename U >
struct A
{
    A(): t{std::make_shared<T>()}, u{t} {}
    std::shared_ptr<T> t;
    U<T> u;
};

int main()
{
    A< D, B > a1;
    A< E, C > a2;
    return 0;
}

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