Question

使用SFINAE，可以访问可变参数类模板的各个元素。当基类从可变参数类模板继承，然后派生类再次从基类和可变参数类模板（使用不同的模板参数）继承时，我的问题就出现了。存在关于应该遵循哪个继承链的含糊不清。有没有办法解决歧义？

例如：

// compile with flag: -std=c++11
#include <type_traits>

struct A { int x; };
struct B { int x; };
struct C { int x; };
struct D { int x; };

template <class ... Params> class Parameter { };

template <class Param, class ... Tail>
class Parameter<Param, Tail ...> : public Param, public Parameter<Tail ...>
{
  public:
    //! Get a parameter
    template <class Param2>
    typename std::enable_if<std::is_same<Param, Param2>::value, int>::type
    getParam() const
    { return Param::x; }

    //! Get a parameter. Delegate false template matches down the Tail... inheritance line (see SFINAE)
    template <class Param2>
    typename std::enable_if<! std::is_same<Param, Param2>::value, int>::type
    getParam() const
    { return Parameter<Tail ...>::template getParam<Param2>(); }
};

class Base : public Parameter<A, B>
{ };

class Derived : public Base, public Parameter<C, D>
{ };

int main(int const argc, char const * argv[])
{
  Base base;
  int a = base.getParam<A>(); // ok
  int b = base.getParam<B>(); // ok

  Derived derived;
  int c0 = derived.getParam<C>(); // error: request for member ‘getParam’ is ambiguous
  int c1 = derived.Derived::getParam<C>();  // error: request for member ‘getParam’ is ambiguous
  int c2 = derived.Parameter<C, D>::getParam<C>(); // ok but syntax overly complex, especially if many params

  int a0 = derived.getParam<A>(); // error: request for member ‘getParam’ is ambiguous
  int a1 = derived.Base::getParam<A>(); // ok, could be acceptable if also worked on Derived
  int a2 = derived.Parameter<A, B>::getParam<A>(); // ok but syntax overly complex and confusing

  return 0;
}

我意识到几个主要的设计变化可以解决这个问题：1）摆脱可变参数类模板，而不是一次从每个参数派生2）使参数成员变量。但我的问题是我是否可以保留可变参数类模板，但避免歧义。

Answer 1

我会删除不匹配的getParam以避免歧义，如下所示：Demo

template <class ... Params> class Parameter;

template <class Param>
class Parameter<Param> : public Param
{
public:
    //! Get a parameter
    template <class Param2>
    typename std::enable_if<std::is_same<Param, Param2>::value, int>::type
    getParam() const
    { return Param::x; }
};

template <class Param, class ... Tail>
class Parameter<Param, Tail ...> : public Parameter<Param>, Parameter<Tail...>
{
public:
    using Parameter<Param>::getParam;
    using Parameter<Tail...>::getParam;
};

class Base : public Parameter<A, B> {};

class Derived : public Base, public Parameter<C, D>
{
public:
    using Base::getParam;
    using Parameter<C, D>::getParam;
};

当基类和派生类都具有可变继承链时，避免模糊的SFINAE访问

1 个答案: