Question

In C++:

class Rectangle {
    int x, y;
  public:
    void set_values (int,int);
    int area () {return (x*y);}
};

int main () {
  Rectangle rect;
  rect.set_values (3,4);
}

In Java:

class Rectangle {
    int x, y;
    void set_values (int x,int y);
    int area () {return (x*y);}
}

public static void main(String[] args) {

Rectangle rect=new Rectangle(3,4);

}

In C++ compiler will create rect object & reserve memory 4 bytes. I want to know How Java is creating object?

Answer 1

When you do: –

Rectangle rect;

in C++, it invokes, the 0-arg default constructor, provided by the compiler.

If you wanted to use a 2-arg constructor, you would have to do: –

Rectangle rect(3, 4);

wherein, you would have to provide a 2-arg constructor explicitly, else it won’t compile.

---

Now, similar is the case in Java. If you do: –

new Rectangle();

then its ok, because, in that case, compiler will provide you with a default 0-arg constructor, as in C++, but when your create object like: –

new Rectangle(3, 4);

Then you would have to explicitly provide the class with a 2-arg constructor, else it won’t compile, as in your 2nd example, that won’t compile.

Also, once you provide a parameterized constructor, then in that case, Compiler won’t provide the default constructor. You would have to provide a 0-arg constructor explicitly, if you want to use one. This suffice in both, Java and C++

Now, as per the memory allocation is concerned, since Java is platform independent, so the size of int type is 32 bits, in all platforms. This is in contrast to C++ where size of data types are platform dependent.

See JVM Specification - The Structure of JVM for detaled information about the allocation of various types.

P.S.: –

I suggest you to go through the below link, for basics of Java classes and objects: –

• http://docs.oracle.com/javase/tutorial/java/javaOO/index.html