Solved

C++ Object Instantiation

Posted on 2008-06-24
6
1,156 Views
Last Modified: 2013-11-12
What's the difference between the following statements in C++ and the proper situations in which one would be used over the others?

Base *obj = new Derived();
Derived *obj = new Base();
Derived *obj = new Derived();

Does it affect the way you would call a method? (i.e. obj->someMethod() )?  Let's say the method only existed in the derived class.  How does that change things?

Note:  I will be away from a computer for a while so I won't be able to respond to posts until tomorrow :-)
0
Comment
Question by:cuziyq
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
6 Comments
 
LVL 10

Expert Comment

by:margajet24
ID: 21856456
if your are using the "Base" as the data type and you want to call a "SubClass" function, you need to type cast it to the "SubClass" to be able to have the access right to the function.

you use the Base Class if you are not yet sure which subclass to type cast to.. since Subclass inherits from the Base Class, casting from BaseClass to SubClass is possible..

one example maybe for this is when you use a Factory Design Pattern
0
 
LVL 1

Accepted Solution

by:
Andrew_Lvov earned 65 total points
ID: 21857276
a) You won't be able to execute Derived *obj = new Base(), because object of Class base is not  Derived, vice versa is true.
b) The difference between 1. and 3. is in non-virtual functions call. Let's say void f() is non-virtual function defined in both Base and Derived classes. Then the next calls will execute f() from Base and from Derived respectively:

Base* b = new Derived();
Derived* d = new Derived();
b->f();     // would call Base::f()
d->f();     // would call Derived::f()

Open in new window

0
 
LVL 1

Expert Comment

by:Andrew_Lvov
ID: 21857314
Oh, and of course if der() is not defined in Base, you'll have a compilation error trying to call:
class Base
{
  void f();
};
 
class Derived : public Base
{
  void f();  // override, but non-virtual
  void der();
};
 
main()
{
  Base* b = new Derived();
  b -> der();     // would lead to compilation error
 
  return 0;
}

Open in new window

0
Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
LVL 17

Assisted Solution

by:rstaveley
rstaveley earned 60 total points
ID: 21857854
Best to experiment. The following shows you some static invocation too...
#include <iostream>
#include <string>
using std::string;
using std::cout;
 
class base;
std::ostream& operator<<(std::ostream& os,const base& b);
class base {
	string name;
	friend std::ostream& operator<<(std::ostream& os,const base& b);
public:
	void f() {
		cout << *this << ".base::f()\n";
	}
	virtual void v() {
		cout << *this << ".base::v()\n";
	}
	base(const string& name) : name(name) {}
	virtual ~base() {}
};
 
std::ostream& operator<<(std::ostream& os,const base& b)
{
	return os << b.name;
}
 
class derived : public base {
	string name;
	friend std::ostream& operator<<(std::ostream& os,const base& b);
public:
	void f() {
		cout << *this << ".derived::f()\n";
	}
	virtual void v() {
		cout << *this << ".derived::v()\n";
	}
	derived(const string& name) : base(name) {}
};
 
int main()
{
	{
		base *obj = new base("alpha-base");
		obj->f();
		obj->v();
		obj->base::f();
		obj->base::v();
		delete obj;
	}
	{
		derived *obj = new derived("beta-derived");
		obj->f();
		obj->v();
		obj->base::f();
		obj->base::v();
		obj->derived::f();
		obj->derived::v();
		delete obj;
	}
	{
		base *obj = new derived("gamma-derived");
		obj->f();
		obj->v();
		obj->base::f();
		obj->base::v();
		//obj->derived::f();	// Compiler error
		//obj->derived::v();	// Compiler error
		delete obj;
	}
	
}

Open in new window

0
 
LVL 17

Expert Comment

by:rstaveley
ID: 21857925
Seriously though my point is  that you'll understand and learn this stuff best by experimentation. See what happens when you invoke virtual and non-virtual functions in a polymorphic object. See what happens when the static type is base and when the static type is derived. It is much better learning that way than hearing about it from know-it-alls like me.
0
 
LVL 40

Expert Comment

by:evilrix
ID: 21857975
The behavior you see when calling a method on a base class will depend on whether that method is virtual or not. If it is you'll get polymorphic behavior.
http://www.cplusplus.com/doc/tutorial/polymorphism.html
0

Featured Post

On Demand Webinar: Networking for the Cloud Era

Did you know SD-WANs can improve network connectivity? Check out this webinar to learn how an SD-WAN simplified, one-click tool can help you migrate and manage data in the cloud.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

This article shows you how to optimize memory allocations in C++ using placement new. Applicable especially to usecases dealing with creation of large number of objects. A brief on problem: Lets take example problem for simplicity: - I have a G…
One of the most frequent problems a "newbie" developer may encounter is having to deal with different data formats. One for all: THE DATE We, as humans, need to "see" a date and then interpret it (much of the times this is an automatic operation)…
The goal of the tutorial is to teach the user how to use functions in C++. The video will cover how to define functions, how to call functions and how to create functions prototypes. Microsoft Visual C++ 2010 Express will be used as a text editor an…
The goal of the video will be to teach the user the difference and consequence of passing data by value vs passing data by reference in C++. An example of passing data by value as well as an example of passing data by reference will be be given. Bot…

690 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question