Question

#include <iostream>
#include <string>

using namespace std;

struct Uid {
  typedef int type;  
};

struct Name {
  typedef string type;
};


struct Age {
  typedef int type;
};


template <class T1, class T2, class T3>
class People {
private:
  typename T1::type val1;
  typename T2::type val2;
  typename T3::type val3;
  //add a get function here
  }
};


int main() {
  People<Uid, Name, Age> people;
  people.get<Uid>(); //make this validate
}

这是我的代码，我想在类中添加一个get函数，使函数调用get in main validate。我尝试在类中添加一个tempalte get及其特化版本，但它是一个无效的方法，编译器说：非命名空间范围'class People'中的显式特化。有人说这种方法适用于vs，但它打破了标准。

Answer 1

您需要一个模板化的get（）成员函数可以使用的辅助类。辅助类可以在命名空间范围内。

#include <iostream>
#include <string>

using std::string;
using std::cout;

struct Uid {
  typedef int type;
};

struct Name {
  typedef string type;
};

struct Age {
  typedef int type;
};

// Helper class that can be specialized to get different members of People.
template <class P, class U> struct PeopleGet;


template <class T1, class T2, class T3>
class People {
public:
  People(
    typename T1::type const& val1,
    typename T2::type const& val2,
    typename T3::type const& val3
  )
  : val1(val1),
    val2(val2),
    val3(val3)
  {
  }

  template <class U> typename U::type get()
  {
    return PeopleGet<People<T1,T2,T3>,U>::get(*this);
  }
private:
  typename T1::type val1;
  typename T2::type val2;
  typename T3::type val3;

  template <class P,class U> friend class PeopleGet;
};


template <class T1,class T2,class T3>
struct PeopleGet<People<T1,T2,T3>,T1> {
  static typename T1::type get(const People<T1,T2,T3> &people)  { return people.val1; }
};

template <class T1,class T2,class T3>
struct PeopleGet<People<T1,T2,T3>,T2> {
  static typename T2::type get(const People<T1,T2,T3> &people)  { return people.val2; }
};

template <class T1,class T2,class T3>
struct PeopleGet<People<T1,T2,T3>,T3> {
  static typename T3::type get(const People<T1,T2,T3> &people)  { return people.val3; }
};


int main()
{
  People<Uid, Name, Age> people(5,"name",47);
  cout << people.get<Uid>() << "\n";
  cout << people.get<Name>() << "\n";
  cout << people.get<Age>() << "\n";
  return 0;
}

如何在类中使用成员函数专业化？

1 个答案: