CODEWITHSUNDEEP
c++templatespointerspointer-to-member
tower120
tower120

Reputation: 5255

C++ template point to member

I have theoretical question about templates in C++.
Let's say I have the following code:

Code

struct C{
   int val = 15;
};

struct B{
    C c;
    int k = 9;
};

struct A{
    template<typename T, typename FType, FType T::* Field = nullptr>
    void test(T& d){
        if (Field != nullptr){
            int a = d.*Field;
        }
    }
};


int main(int argc, char *argv[])
{

    A a;
    B be;

    a.test<B, int, &B::c::val>(be);
    //a.test<B, int, &B::k>(be);
}

The question is: why I can't access to B::c::val field, when I can access B::k?
B::c::val is also member of B. Or not?

Upvotes: 2

Views: 87

Answers (1)

Arne Mertz
Arne Mertz

Reputation: 24606

B::c is a member of B, and C::val is a member of C, but there is nothing like B::c::val

What you are wanting to do here could be achieved like this: 

struct A {
  template<typename T, typename OType, typename IType, OType T::* OField = nullptr, IType OType::* IField = nullptr>
  void test(T& d){
    if (OField != nullptr && IField != nullptr) {
      auto a = d.*OField.*IField; 
    }
  }
};

int main(int argc, char *argv[])
{

    A a;
    B be;

    a.test<B, C, int, &B::c, &C::val>(be);
}

http://coliru.stacked-crooked.com/a/f89b96274218e223

Of course, you could provide the member pointers as function parameters and let the compiler deduce the type parameters for you.

Upvotes: 1

Related Questions