Java Tutorial: Method Overloading in Java

[@hsetips2574]

Now 32 out of 113.... 😊 Notes + video + explaination in easiest way = No doubt all clear 👍👍👍❤️🙏

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It's been 38 days since I have started learning, some days were too difficult for me to concentrate but still I'm on the go untill I finish this whole playlist and anyone learning or has just started or whoever gonna start in future, we all gonna make it. Take rest, take your own time but don't quit.

[@SandipanSom]

bro how much have u covered ?

[@mrunknown855]

u watched just 0.77 video per day?

[@jatindubey7892]

@@mrunknown855 its about consistency ; consistency is not everyone's cup of tea !!

[@AshishKhetwal]

@@jatindubey7892 true in starting i was watching 5-6 videos daily and now I can't sit for one video but still trying to be consistent atleast

[@SAUMYASAHU-n9j]

did u start java as your 1st programming language? In which year rn?

[@hwaiting3573]

11:45 Change function will update the value of arr at index 0 because when we pass array to function/method it will actually pass the address of the value , therefore it is call by reference and it will definitely change the value of array at index 0 and print 98

[@karthikramkoru]

Tq bhai now I understood

[@ranvirkumar1472]

thanks bhai.😇

[@TabraizKhan-iv1cl]

If thats the case then for the integer , we passed x which is reference for 45 ,shouldn't a get x and x is the address of 45 so it should change based on your logic right..? I listened to it many times but couldn't understand it properly. I would appreciate you if you could elaborate it for me

[@bothighlights4179]

@@TabraizKhan-iv1cl suppose you have an array which is array1 = {1,2,3,4} and have a variable, int x = 6 so if you print the variable x it will give you 6 : system.out.println(x) = 6 and when you print the array it will give you the address of the array like this : [I@4517d9a3 so when you're passing x (int variable) in a method the thing you're passing is 6, because x = 6 you are not passing the address or reference of int x but when you're passing an array you are not passing the integers in array but the address of the array : [I@4517d9a3 and because you're passing the address of array the method is changing the actual array not a copy. in int x = 6 case you are passing 6 not the address of x that's why it is changing the copy not the actual x I hope you understood i can further help you if you didn't understand

[@harilalchauhan9069]

Harry I usually refrain from putting up comment... But Your explanation is spectacularly awesome......that i couldn't stop myself to praise your way of teaching.... I LOVE YOU BROTHER

[@dorababuarig765]

Call by value vs call by reference is clearly explained. Thankyou Harry bayya ❤️

[@parinkoralwala3653]

Time stamp?

[@DK-js8cz]

@@parinkoralwala3653 case1 explains the call by value concept and case 2 explains call by reference concept bro.15:00

[@Yogiiii__]

Wow!!!!... Amazing Harry bhai.. All concepts cleared. I came here after listening so called paid course lectures.. You are doing amazing work.. i have taken printouts of all your notes & they are helping me a lot while doing this java course. God bless you Harry bhai..!

[@codemc7376]

I respect your notes and also appreciate you hard work. I try my 100% to learn java and till now I understand all things. 23:33

[@sahilgupta7001]

11:52 the value of marks[0] will be changed because the address of the array will be passed in the function and all the changes will be reflected in the original array

[@madhursikarwar8531]

I never thought That I will Be knowing this much number of Programming langugae.... You are my true programming teacher. I never used to think That I will become a that sort of programmer . Thanks a lot Harry Sir ....May you get 100 Billion Subscribers and Views .. I have a wish to meet you one day 😀😀 and Sir pls bring Constructors fast

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@CodeWithHarry Thanks for hearting My comment sir

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[@roxyisthebestwaifu712]

12:07 It will print 98. The value of array is changed because the reference of array is passed to the function. It didn't change during the integer because while passing an integer, the value was passed to the method.

[@TabraizKhan-iv1cl]

How is that same case then. He said it in the video that for integer it is no different compared to arra

[@kanavatri267]

Hey Harry,you’re doing an awesome job .Keep up the good work and best of luck for future !Love from Melbourne ,Australia .

[@a.kzestos2085]

@Ledger Kalel WHAT...!! How

[@sushantkumarkapri3542]

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[@sushantkumarkapri3542]

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[@dnyaneshwarigund9616]

11:46 Output: The value of x after running change is: 98 Reason: Because Here in case of array the reference of array is pass thus it can change the value of arr[0]

[@gold8780]

In the example of (12:00), *value* is changed because "array" is not a primitive data type. (Harry said it previously that, this happens with primitive data types only). Same thing happens in integer as well as array, but in case of array only the reference is copied. Hope you are getting the point! It's alright if you got it wrong. Even I got it wrong at first 🙃

[@zainabkazi4102]

i rewind the video thrice but still not able to understand the difference in call by value n call by reference

[@atifasadkhan]

@@zainabkazi4102 inshort the value changed for variables because they are premetive data type but array is not premetive data type so you were able to change !!! premetive data types are (Boolean, char, byte, int, short, long, float, and double.)

[@MAD2ON_Talkroaster]

​@@atifasadkhanbut we can change the value of primitive data type through call by reference.

[@Vedicrystal]

Thanks for this course

[@parthsharma4996]

I have seen all your c and c++ playlist videos good going dude And thanks a lot for helping collage students like me from these videos

[@chaitanyadave92650]

video Number 32 11:57 Not Change in Actual Array Thank You Harry Bhai For Lunch java Best Course 😊🙏🙏🙏🙏

[@anonymoususer4148]

Thank you for teaching us so nicely. You're the best programming teacher on youtube

[@harshvardhanshinde9648]

11:51 Yes the value will change... As we are passing the name of array which is actually the base address of the array

[@aruntak3054]

Hello sir, your teaching style is awesome. I am beginner but after watching your tutorials I have created my hobby as a android dovelopper. So thanks a lot sir for providing these good stuff for beginners.

[@karthikt3957]

11:50 the value at index 0 gets changed as whenever array is passed , array referrences are passed changing the value formally as well as in main method

[@rickk3300]

The concepts of call by value and call by reference are perfectly clear now..❤️❤️❤️

[@TabraizKhan-iv1cl]

so what happens actually if you call by value, copy of value is taken and then..? He said that it is no different from what happened to array .

[@rickk3300]

@@TabraizKhan-iv1cl yes, in call by value, copy of value is taken. Arrays in Java are always created in the heap memory. When an array is passed to a method, it is always passed as a reference, the array will not be copied, two variables will refer to the same array.

[@IshmeetSinghBaweja]

His notes are really outstanding and very much helping + perfect !

[@harshrana9664]

You are my true programming teacher.

[@harshitverma2809]

12:05 Value of marks[0] is 98. So '98' will be the output

[@IshmeetSinghBaweja]

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[@suyashmukherjee2332]

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[@hitarthshah5549]

timestamp 11:58 According to me, the given line will print 52.

[@HarshKumar66743]

At 12:05 it will print " 52 ", because even though the method has changed the copied value but has not returned it to the main function.

[@achievewithgyanu632]

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[@abheygupta2332]

11:48 Sir I think the value will change cuz I learned in your C in one video that in case of passing an array, the address is passed which is capable of changing the value of an array Baaki pata ni😅😅😅😅

[@chiragkhatri7183]

same lets see what happen

[@AnkushSaral]

11:45 It will change the value of array at index 0 because you said it will pass the reference of array rather than passing copy of the original array.

[@ishansingh306]

@harry 21:44 You said method overloading cannot be performed by chnaging the return type but in your source code site you have mentioned it can be perfomed by changing the return type. Please clear the confusion.

[@sourikkarmakar1200]

11:56 The function will update the value and print 98

[@amitghanwat8625]

Great course! Thanks for the efforts Harry!

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[@marcus9013]

Time 11:56 Output The value of X after running change is : 98

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[@saumyapandey8940]

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[@Dxiag123]

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[@prajwalsarangi2601]

yes

[@soumyaranjan580]

11:58 the value wil change as the reference is passed to the methood.

[@ayushdeveloper]

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[@saturn-6-66]

Yes sir

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[@trueknowledge2065]

Yeah

[@harshvachheta7811]

I have learnt this but I was not this much clear with it thanks to you .Now I know the logic behind method and method overloading

[@sujatajena267]

11:45 Haan bhai change hoga kyunki agar nahi hota to aap ye nahi karwate 🤣🤣🤣 it will print 98

[@fellincode2607]

sahi bole bhai or aapne is comment ko edit kiya hai

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@ sujata Jena aap jabhi bhi meri pehlli comment read karoge to mujhe reply bhejo

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Y O U A R E L E G E N D

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[@nilaychatterjee54]

Your Notes are the best.. Short and crisp.

[@anandrajbind7898]

11:48 marks 0 value is not change because marks is pass copy value (It is also known as call by value )

[@adityagiri1285]

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[@eshikarya]

for personal understanding: When a primitive data is passed to a method in this case a copy of the primitive data is made and any changes that happen inside the method affect the copy and not the original primitive data due to this we will not see any change in the original primitive data however in terms of object the reference of the object is passed in the method and not the original object itself due to this a copy of the reference is made in the method and not a copy of the object. Now, since the original reference of the object actually pointed to the address of the object, the copy that is made in the method i.e. the copied reference; this reference will also hold the same address in it. Since both references point to the same address, due to this any changes that are made in the copied reference end up affecting the actual object.

[@tiktiktikjohn3526]

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[@RehanAliArman-mt8pr]

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[@ayushsinha9689]

12:09 98 print hoga kyunki array k case me reference pass hota hai isiliye

[@anmolsaggu7895]

12:07 the value of arr[0] will be change to 98 from 52.

[@xpert__ngel8857]

Harry Bhai your teaching techniques are awesome!!!

[@BhullaRangeela]

12:09 the value of marks[0] should change as array will pass address rather than making a copy of it's value.

[@soumyachaudharyy]

you are a such hardworking teacher, thanks a lot!

[@snehareddy5188]

It's like finally I found you. Thank you 🙏.

[@dhairyathakran5695]

MethodOverloading 11:58 YEs function will change the value of array of marks [0]

[@aniketsingh5570]

You are really a good tutor!! THANK YOU HARRY

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you are god for coding students.....

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[@Ralhirastogi]

7:43 thanks for solving our confusion.🤗🤗🤗🤗🤗

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I like his videos even before watching it ❤️

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Hey Harry!!! you're doing great, keep it up. Love from London🤗🤗♥♥

[@Sourabh_koli]

12:05 The value marks[0] is updated and printed 98. Because we passing reference of array .

[@munirasaeed7847]

AMAZINGGGG....lecture as always.....👍👍👍

[@AmarjeetKumar-mj9iq]

In simple terms when you create a array with a variable ,the variable get the address of the first index of that array. and when you pass the array the same address is send to method so changes done are permanent and effect the original copy.

[@vickys40]

11:42 In my opinion Change nhi honi chahiye kyunki marks[0] ki value change2() method k ander change hua hai aur hume print kr rahe hai usse uske method k bahar which is main() method.

[@mandarbuchake2430]

11:48 yes it change value because of call by referring because I already know c language

[@arvindtiwari4605]

Thank you Harry Bhai 🙂 dil se. I got all the concepts easily

[@pluto2089]

12:00 it will change to 52 because you have passed marks that tell from where you have to change value. So the answer will be - 52

[@DK-js8cz]

Method overloading can also be performed by changing the return type of methods, but you have to change the type of parameters also. public class Main { static int plusMethod(int x, int y) { return x + y; } static double plusMethod(double x, double y) { return x + y; } public static void main(String[] args) { int myNum1 = plusMethod(8, 5); double myNum2 = plusMethod(4.3, 6.26); System.out.println("int: " + myNum1); System.out.println("double: " + myNum2); } }

[@nishantkarn5311]

Thank U sir for providing this wonderful video with notes❤

[@blinx_edits]

Thank you so much bhai !!!!! Great video ,mera kal exam hai yeh dekh ke bhot sikha mene

[@mitrajsinh7081]

12:05 yes it will as array is kinda like pointer.

[@sona_772]

Learning java our efforts 20 % harry sir putting efforts 80%

[@MdSamir-hv2kr]

Thank you very much for your great lectures

[@NileshPanchal-nx4pw]

TIme Stamp 11:45. Yes. It will change value of marks[0].

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Thank you for being so kind to the community 😊😊

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Thank you so much sir for this High Quality Content!!!!!!!!! 💖

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than you sir you are such a great teacher better than uni teachers

[@gameon9922]

For primitive data type - Methode Called by value. For Object/ Non primitive data type- Methode Called by reference/address.

[@dhruvjain2874]

11:49 marks[0] = 98; kyuki Reference is Passed in Case of Array

[@NileshPanchal-nx4pw]

TIme Stamp 11:45. Yes. It will change value of marks[0] to 98.

[@Creative_Things_25]

12:06 yes the change2 method will change the first value of the array as the actual location of the array is passed to the method....

[@SantoshShriwas-z6u]

Nice explanation 💥 in 2024

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[@DANABOYINAVARSHAVANI]

such an excellent explanation👌👌👌👌🙌🙌