CODEWITHSUNDEEP
c++operator-overloadingoverloading
nmd_07
nmd_07

Reputation: 706

Accessing data members of a data member(itself is object) as if they were class members

Given the scenerio,

template <bool T>
struct A{
    struct B{
       int x; 
       double y;
    }
    struct C{
       char c;
       std::string k;
    }
 using type = std::conditional<T, B, C>::type;
 type b;
}

I would like to access to data members such as:

int main(){
    A<true> object_1;
    A<false> object_2;
    cout << object_1.x;
    cout << object_2.k;
}

This would be possible if .(dot) operator could be overloaded. However, this is not possible(at least now). Is there a workaround to make the example work?

Upvotes: 0

Views: 35

Answers (2)

nmd_07
nmd_07

Reputation: 706

Seems that it is easier than I thought : 

struct B{
    int x; 
    double y;
}
struct C{
   char c;
   std::string k;
}
template <bool T>
struct A : public std::conditional <T, B, C>::type{

}

Upvotes: 0

sebrockm
sebrockm

Reputation: 6002

You cannot overload ., that is correct. You can, however, overload ->, if that is also an option for you?

operator-> must return a type that itself supports operator->, e.g. a raw pointer suits this:

template <bool T> struct A
{
    struct B
    {
        int x; 
        double y; 
    }; 
    struct C
    { 
        char c; 
        std::string k;
    };
    using type = std::conditional<T, B, C>::type;
    type b;

    type* operator->() { return &b; }
};

You can use it like so:

int main()
{ 
    A<true> object_1; 
    A<false> object_2; 
    cout << object_1->x; 
    cout << object_2->k;
}

Upvotes: 1

Related Questions