C - upcasting and downcasting

0 votes

As an example:

Shouldn't the second d.print() call during upcasting print "base"?

Isn't it a "d" derived object that has been upcasted to a base class object?

And what advantages does downcasting have?

Could you describe upcast and downcast in more detail?

#include <iostream>
using namespace std;

class Base {
public:
    void print() { cout << "base" << endl; }
};

class Derived :public Base{
public:
    void print() { cout << "derived" << endl; }

};

void main()
{
    // Upcasting
    Base *pBase;
    Derived d;
    d.print();
    pBase = &d;
    d.print();

    // Downcasting
    Derived *pDerived;
    Base *b;
    pDerived = (Derived*)b;
}
Jul 26, 2022 in C++ by Nicholas
• 7,760 points
781 views

No answer to this question. Be the first to respond.

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.

Related Questions In C++

0 votes
1 answer

The Definitive C++ Book Guide and List

For Beginner (includes those without coding experience) Programming: ...READ MORE

answered Jun 6, 2022 in C++ by pranav
• 2,590 points
964 views
0 votes
1 answer

Cases of static and dynamic binding in C++

When an object's static type is used to associate it with a member function, this is known as static binding (understand the type of its class). When a pointer or reference is associated with a member function based on the dynamic type of the object, this is known as dynamic binding (understand the instance of the variable at runtime). Before continuing, keep in mind that dynamic binding only works with pointers, references, and virtual functions for the base class. Because everything needed to call the function is known at compile time, the first call is a static binding (also known as early binding). Derived1 d1(1, 10); d1.display_data(); You already know that the d1 instance is a Derived1 automatic variable, and that it will call the Derived1::display data method (). The first condition is incorrect: d1 is neither a pointer nor a reference. The second condition isn't acceptable:  There is no virtual Derived1::display data. The second call is for ...READ MORE

answered Jun 7, 2022 in C++ by Damon
• 4,960 points
698 views
0 votes
1 answer

functions in c++ call by value and call by reference

Calling a function by value copies the argument and stores it in a local variable for use by the function, so the argument is unaffected if the value of the local variable changes.  The argument is passed to the function as a reference rather than a copy, so if the function changes the value of the argument, the argument is also changed.   The void change(int); function prototype informs the compiler that there is a function named change that takes a single int argument and returns void (i.e. nothing).  Because there is no & with the argument, it is called by value.  You have the line change(orig); later in your code, which actually calls the function with. Take a look at the output of ...READ MORE

answered Jun 7, 2022 in C++ by Damon
• 4,960 points
634 views
0 votes
0 answers

Use of min and max functions in C++

Are std::min and std::max better than fmin ...READ MORE

Jun 2, 2022 in C++ by Nicholas
• 7,760 points
595 views
0 votes
0 answers

functions in c++ call by value and call by reference

The code below shows how to call a function in both methods.  Please explain the major differences or meanings of call by value and call by reference to me.  1.Make a value-based call.  2.Call based on a reference.  The call by value method is demonstrated in the following code. In a comment, I expressed my reservations. #include<iostream> int main(){ void change(int);//why function prototype is before ...READ MORE

Jun 6, 2022 in C++ by Nicholas
• 7,760 points
505 views
0 votes
1 answer

Use of min and max functions in C++

The functions fmin and fmax are designed ...READ MORE

answered Jun 21, 2022 in C++ by Damon
• 4,960 points
11,336 views
0 votes
1 answer

Python class inherits object

Python 3.x: class MyClass(object): = new-style class class MyClass: = new-style ...READ MORE

answered Aug 30, 2018 in Python by Priyaj
• 58,020 points
814 views
0 votes
1 answer

How is inheritance in C++ different than that in Java?

The purpose of inheritance is same for ...READ MORE

answered Feb 6, 2019 in Java by Priyaj
• 58,020 points
999 views
0 votes
1 answer

C++ "Object" class

No, there is no generic base class&nb ...READ MORE

answered Jun 2, 2022 in C++ by Damon
• 4,960 points
512 views
0 votes
1 answer

C++ - Overloading vs Overriding in Inheritance

In C++, a derived class's method only overrides the base class's method if their declarations match (I say "match," but I'm not sure what the formal term is).  That is, all arguments must be of the same type, with the same const qualification.  If there are any mismatches, the derived class's method hides all methods with the same name rather than overriding.  This is what the "ERROR" in your image is attempting to convey.  So, in that image, / overrides in a comment is incorrect and misleading. Yes, many C++ instructors are unaware of these somewhat esoteric details. Furthermore, if you want to override, your base class's method must be virtual; otherwise, polymorphism will not work. . We could also say that the derived-class method hides the base-class method if it wasn't virtual.  The part about hiding, on the other hand, has almost no meaning here; what this term really means is that you're not in charge. Furthermore, overloading is the presence of multiple methods with the same name but different signatures, as you may have noticed. To be useful, they must all be present in the derived class; otherwise, they will be hidden if the derived class only has one method, fa1, and the other fa1 are in the base. There is, however, a syntax sugar that "copies" all fa1 from base to derived, removing all the hidden semantics: class A { public: void fa1(); ...READ MORE

answered Jun 7, 2022 in C++ by Damon
• 4,960 points
833 views
webinar REGISTER FOR FREE WEBINAR X
REGISTER NOW
webinar_success Thank you for registering Join Edureka Meetup community for 100+ Free Webinars each month JOIN MEETUP GROUP