Java Interview 04 - Static Binding Vs Dynamic Binding 

YOUR AWESOME! I've literally looked all day, just so I could understand this and you condensed it perfectly! Subbed!

The best tutorial on this topic! Short and concise! Thank you very much!

Thank you for simplifying this. You have no idea how helpful you have been.

I am sure it is the only video on youtube that gives to the point explanation about early and late binding

excellent explanation in such simple words, thank u nd want more of your videos on complex topics of java

Thanks Mahika. I’m glad I came across your tutorial. Finally understood this tricky concept.

just one word 'CONCISE'!!!

In Java dynamic method dispatch is the mechanism that allows the selection of what method will be invoked and hence dynamic binding occurs.

Awesome u are a born teacher

Thank you for the good content.

My dood, I have no idea who you are but God bless you. Like you explained this simple topic in such a simple way compared to other people. OMG like I can't even explain to you how good this was thank lmfao God bless.

Brilliant🙌

Well done, thanks!

Nicely explained! Thanks...

Thank you for so easy explanation 😊

thank you Mahika Tutorials! Very cool

simple and easy to understand i hope this is the same for C#

Brilliant, Got it!

Great great explanation❤

Very good explanation 👍

Absolute gem!

very helpful Nd good job

While it is true that overwritten methods always use dynamic binding, i found that an intuitive way to understand it is when you pass an object as a method parameter by typing it using an interface that it implements (and then call a method on the passed object in the body). In this way the method can only know the reference type at runtime, so clearly it has to use the dynamic binding.

good explanation thank you so much

bravo...ty mam

helpful video, keep it up!

Amazing tutorial..you're a lifesaver

Thanks 👍

thank you so much

Clear explanation of concept. 👍

Woahhhh...! thsnk you

I didn't want to ruin my awesome list of liked videos, but you deserve it ;)

Simple and direct! Thanks.

THANK YOU SO MUCH TTTTTTTTTTT

great explanation maam

precise and well explained. Thank you

static binding and dynamic binding can be done without the method overloading or overriding. class A { public void display(){ ......} public void invoke(A x){ x.display();} } In this code, I didn’t do method overloading or overriding but still the compiler will bind the display method as dynamic. You can test this code too by compile this and before execute it see the assembly instruction of A class.

thanks for the video, have a test tomorrow

Excellent explanation!

God bless you. Hoping some more videos will be published soon!!!

The binding of static, private and final methods is compile-time. Why? The reason is that the these method cannot be overridden and the type of the class is determined at the compile time.

Very nicely explained, ma'am!

very helpful! thank you.

Useful video

Thank you, it was very helpful!

Amazing explanation. Thank u!

Very clear explanation thnkew

very nice and easy explanation

Thx for your clear explanation on the topic.

Thank u mam very nice explanation....😇😇😇😇😇😄

Reference type based only the job of compiler and object type taken care by JVM

miss a square in calculating circle area ???

it was very easy to understand!!!! thank you

very very nice explanation mam. thank you mam

Good video But I have one doubt: Consider the code: Class A{ void f1(){ System.out.println("A"); } } Class B extends A{ void f1(){ System.out.println("B"); super.f1(); } } Class example { public static void main (string [ ]args){ B obj=new B(); obj.f1(); } } The output was: B A Although according to what you taught there should be recursion as function f1 of class B will be binded to the call super.f1(); in dynamic binding and the output should print "B" infinite times Kindly solve this problem

👌👌

Super madam. U have super knowledge on Java. Make vedios on all topics which are useful for interview.

Mam objectsbare created at runtime how can it allocate memory at compile time itself mam

this program shows error Shape s=new Circle(); error :shape can't convert to circle

can we call subclass methods with superclass object without downcasting?

Can we create same name reference of different class? Why are you take S multiple time.

I was expecting a sword fight scene

2x is still better 😁

RU-vid pe teacher wali feel derhe ho😅

explicity as hell

No clear view on vedio