Class as a non type template parameter of the template

Template parameter extension

We mentioned earlier ： We cannot use a custom class type as a non type template parameter , But let's think about “ Can you put the static of the class / Non static members are used as non type template parameters ”？

① Take the non static member of the class type as the non type template parameter

#include <iostream>  
using namespace std;  
  
class Student  
{  
public:  
    int age;  
public:  
    Student() :age(0) {};  
    void ShowInf();  
};  
  
void Student::ShowInf()  
{  
    cout << this->age << endl;  
}  
  
template <typename T,typename F>  
void Test(T& obj, F Func)  
{  
    (obj.*Func)();  
}  
  
int main()  
{  
    Student obj;  
    Test(obj, &Student::ShowInf);  
}  

Be careful ： A normal member function cannot exist without a class object , So we have to pass in a class object in the function , Let this class object call ordinary member functions of the class .

② Take the static member function of the class type as the non type parameter of the template

#include <iostream>  
using namespace std;  
  
class Student  
{  
public:  
    static int age;  
public:  
    static void ShowInf();  
};  
  
int Student::age = 0;  
  
void Student::ShowInf()  
{  
    cout << Student::age << endl;  
}  
  
template <typename F>  
void Test(F Func)  
{  
    Func();  
}  
  
int main()  
{  
    Test(&Student::ShowInf);  
}  

Be careful ： Because static member functions do not need to depend on class objects , So we can treat static member functions as ordinary functions .

③ Use the public member data of the class as a non type template parameter

#include <iostream>  
using namespace std;  
  
class Student  
{  
public:  
    int age;  
public:  
    Student() :age(0) {};  
};  
  
template <int Student::*obj>  
void ShowInf()  
{  
    cout << obj << endl;  
}  
  
int main()  
{  
    ShowInf<&Student::age>();  //  Output 1 Not what we want to see 0
}  

Be careful ：

Why the output results are different , because “ Member pointer ” It is not a definite address, but relative to the class this The offset of the pointer , See ：

What is? “ Member pointer ”？

④ Take the static member data of the class as the non type parameter of the template

#include <iostream>  
using namespace std;  
  
class Student  
{  
public:  
    static int age;  
};  
  
int Student::age = 0;  
  
template <int& obj>  //  Remember to add & quote 
void ShowInf()  
{  
    cout << obj << endl;  
}  
  
int main()  
{  
    ShowInf<Student::age>();  
}  

Be careful ： Remember to add & quote ！

We can only quote / Pointer to access static variables , We use a pointer to access a static member variable ：

#include <iostream>  
using namespace std;  
  
class Student  
{  
public:  
    static int age;  
};  
  
int Student::age = 0;  
  
template <int* obj> //  The parameter is a pointer   
void ShowInf()  
{  
    cout << *obj << endl;  
}  
  
int main()  
{  
    ShowInf<&Student::age>();  
}  

About static Detailed resolution of static members ：

static Detailed explanation of static members