Overridden methods and dynamic binding

Question

Is there a possibility for an overridden method to be resolved statically?

And what is the relation between Invoke virtual and dynamic binding? Are all invoke virtual methods dynamically bound?

class Dynamic
{
    public void display()
    {
        System.out.println("in dynamic");
    }
    public static void invoke(Dynamic x)
    {
        x.display();
    }
}

here x.display shows invoke virtual? what does it mean?does it mean it is dynamically bound?

What do you want to achieve and which *concrete* coding problems did you stumble upon? — Smutje, Oct 21 '14 at 08:38
Was just doing some research,and stumbled on this topic,where someone wrote this code without extending another class,and on parsing it showed invoke virtual,and said even this is dynamically bind — curositykiller, Oct 21 '14 at 08:40
@Rustam its a bytecode, you have invoke static,virtual ,invokespecial as well — curositykiller, Oct 21 '14 at 08:41

score 1 · Accepted Answer · answered Oct 21 '14 at 08:43

1

No. Overriden methods cannot be resolved at compile time. They are resolved during Runtime based on type of object.

InvokeVirtual is the Byte-code way of telling you that a method has been called.

answered Oct 21 '14 at 08:43

TheLostMind

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SO why there is 4 different ways? invoke static,invoke special,invoke virtual and invoke interface? – curositykiller Oct 21 '14 at 08:45
@curositykiller - invoke static is for *static methods* `String.valueO()`, *invoke Specia* is for *constructors* , *invoke virtual* is for any non-static method (SOP). – TheLostMind Oct 21 '14 at 08:54
so invoke virtual doesn't have to do anything with dynamic binding? – curositykiller Oct 21 '14 at 09:03
@curositykiller - Not necessarily. Even if you call a normal method, you will *Invoke Virtual* will be present in the byte code. – TheLostMind Oct 21 '14 at 09:20

Overridden methods and dynamic binding

1 Answers1