CODEWITHSUNDEEP
javaruntimepolymorphismcompile-time
Ravi
Ravi

Reputation: 31397

compile time vs run time

After going through 2-3 hour to know, what is the difference between compile-time and run-time. Lastly, i came up with this.

Memory allocated at runtime referred to run-time/dynamic binding and allocated at compile time referred to compile-time/static binding.

and then i tried this example

class myclass {

    void here() {
        System.out.println("Here from myclass !!");
    }

    void here(int i) {
        System.out.println("Here !!" + i);
    }
}

class thisclass extends myclass {

    void here() {
        System.out.println("Here from thisclass !!");
    }
}

public class poly {

    public static void main(String s[]) {
        myclass m= new myclass();
        myclass a= new thisclass();
        m.here();
        m.here(12);
        a.here();
        a.here(13);
    }
}

So, i also found that myclass a= new thisclass(); is considered to be run-time binding. Since, a is the object of myclass, but suddenly compiler found that, class mis-matched. So, it will be dynamically bind the space of thisclass object.

So, till here, i got the things. But, i found that, another common answer was overloading refer to compile time and overriding refer to run-time. I didn't get this point.

thisclass a= new thisclass();
a.here();

Is this also called to be run-time binding. ?? Please correct me, if wrote anything wrong here.

Upvotes: 0

Views: 6836

Answers (3)

Marko Topolnik
Marko Topolnik

Reputation: 200138

First of all, memory allocation is not in this picture. There is no compile-time memory allocation.

The question conflates compile-time with static binding and run-time with dynamic binding.

Static binding happens at compile time; dynamic binding happens at runtime.

Now, when you write

myclass m= new thisclass();
m.here(18);

what happens at compile-time is the resolution of the method signature: you are calling here(int) and that choice is final. This is termed "static binding". What happens at runtime is method dispatch: the runtime chooses a here(int) implementation appropriate to the runtime type of the object referenced by m. There are two methods to choose from: myclass.m(int) and thisclass.m(int), and the runtime chooses the latter in this particular example. This is termed "dynamic binding".

As to your question "is overriding compulsory for dynamic binding"... The Java Language Specification prescribes the rules on choosing the correct method to invoke at runtime. These rules imply a procedure known as "dynamic binding" for the general case. But if you are asking whether any specific process always happens at runtime, the story is different: an optimizing JIT compiler can see that there is only one method to choose from and output a "monomorphic call site" which hardcodes the single choice. Further, it can also inline the entire method into the caller, thereby removing even the call itself.

Upvotes: 4

alan.sambol
alan.sambol

Reputation: 265

But, i found that, another common answer was overloading refer to compile time and overriding refer to run-time. I didn't get this point.

Overloading means having multiple methods with different parameters in the same class. Which method is called is known at compile time, because the arguments are specified at this time.

Overriding means re-defining a method from parent-class in the subclass.

Upvotes: 0

wxyz
wxyz

Reputation: 707

This: 

thisclass a= new thisclass();
a.here();

is not run-time binding as the compiler knows what here() method to call (the one from thisclass). But if you would say:

myclass a= new thisclass();
a.here();

then would be a run-time binding.

P.S.: your classes names should start with a capital letter.

Upvotes: 0

Related Questions