CODEWITHSUNDEEP
c++c++11
painterflap
painterflap

Reputation: 15

Base class has no new method error

This is follow up question to this. Now I have this code:

class A
{
protected:
    string name;
public:
    A(string a) {name = a;}
    virtual string getName() {return "A name: " + name;}
};


class B: public A
{
public:
    using A::A;
    string getName() {return "B name: " + name;}
    string newMethod() {return "B name new: " + name;}
};


void print_name(A & obj)
{
    // HOW to check if obj is B, and call newMethod then.
    cout << obj.newMethod() << endl; // THIS LINE HAS ERROR below
}


int main()
{


    A a("a");
    B b("b");

    print_name(a);
    print_name(b);

    return 0;
}

I get error 'class A' has no member named 'newMethod'. I know its becasue newMethod is only in B.

But how can I check if obj in print_name is A or B, and call newMethod only if obj is B?

Upvotes: 0

Views: 89

Answers (5)

M.M
M.M

Reputation: 141618

Well, this is not a great design. But you can write:

cout << dynamic_cast<B &>(obj).newMethod() << endl;

This will throw an exception if obj is not actually a B. If you don't want an exception then you can cast to a pointer:

B *ptr = dynamic_cast<B *>(&obj);

if ( ptr )
    cout << ptr->newMethod() << endl;

Upvotes: 2

Beta
Beta

Reputation: 99124

This is the way to use a virtual method:

class B: public A
{
public:
  ...
  // new implementation of virtual method, old signature
  string getName() {return "B name new: " + name;}
};    

void print_name(A & obj)
{
  cout << obj.getName() << endl;
}

Upvotes: 2

Marcin
Marcin

Reputation: 238299

You could use dynamic casting:

void print_name(A & obj)
{
    B * b = dynamic_cast<B*>(&obj);

    if (b != nullptr) {            
        cout << b->newMethod() << endl; 
    } else { 
       cout << obj.getName() << endl;
    }
}

Upvotes: 0

Ankur
Ankur

Reputation: 1473

Here is trick, use dynamic_cast

TYPE& dynamic_cast<TYPE&> (object);
TYPE* dynamic_cast<TYPE*> (object);

dynamic_cast can only be used with pointers and references to classes (or with void*). Its purpose is to ensure that the result of the type conversion points to a valid complete object of the destination pointer type.

dynamic_cast Operator

Upvotes: 2

nikhil mehta
nikhil mehta

Reputation: 1032

it is failing since the newMethod is not declared in class A,obviously the object reference of the class A will not find the method named newMethod. To achieve what you want you should use concepts of RTTI use typeof operator.

Upvotes: 0

Related Questions