If Strings in java are immutable, why I can do this?

Question

I realize that the hashCode of the variable name, is different after the "update", but objectively what makes a String object in fact immutable ?

public static void main(String[] args) {

        String str = "AB";
         System.out.println(str ); // AB

         str = str .replace(str .charAt(0) ,'W');
         System.out.println(str );//WB

 }

EDIT 1 : The hashCode is based on the value of the variable and have no relation with memory adress.

EDIT 2 : I now understand that Strings are references and not Objects in it self. I read back all the answers for this same question and found out good answers in topics like [this] (Immutability of Strings in Java). Thank you whos tried to help me and my excuses for any silly misunderstood.

I also recommend this articles here to who wants better understand how Strings works in Java :

https://www.pushkarrajpujari.com/article/strings-in-java/

and how references works :

https://javaranch.com/campfire/StoryPassBy.jsp

EDIT 3: I cannot DELETE this topic anymore, according with Stackoverflow "You cannot delete this question as others have invested time and effort into answering it." which I agree.

The string returned from replace is a different string. The variable is not the object. You just overwrite the reference to one string with a reference to another string. — Nathan Hughes, Sep 19 '19 at 21:02
You're not mutating the String. You're assigning a **new** String to the same variable. — Jordan, Sep 19 '19 at 21:02
Variables are always mutable, in the sense that they can be made to reference new objects, even when the objects themselves are immutable. — Dawood ibn Kareem, Sep 19 '19 at 21:10
BTW that is not the address of the variable, but the address of the object the variable is pointing to (or the content of the variable (since it is a reference/pointer)) ((haven't checked the documentation, but doubt that is not explained)) — user85421, Sep 23 '19 at 19:32
`addressOf(str)` does not give you the address of the variable `str`, but instead the address of the object it is referencing. Keep in mind that `31864662320` == `0x76b482930`. — Progman, Sep 23 '19 at 19:39
@Progman, thanks for your answer man. I already erase my misguided edit. — basquiatraphaeu, Sep 24 '19 at 00:26
@Progman, thanks for your answer man. I already erase my misguided edit. — basquiatraphaeu, Sep 24 '19 at 01:01
@JohannesKuhn Well, there are also *constant variables*, like ones marked `static final`. — MC Emperor, Apr 01 '20 at 22:05
@Progman Well, it's not a memory address per se. It is implementation specific, and *appears* to be a memory address or a part of it. See [here for more details](https://stackoverflow.com/questions/1961146/memory-address-of-variables-in-java). — MC Emperor, Apr 01 '20 at 22:14

score 1 · Accepted Answer · answered Sep 19 '19 at 21:03

1

If you look at the documentation of replace(), it mentions:

Returns a new string resulting from replacing all occurrences of oldChar in this string with newChar.

Therefore, the replaced String is an entirely new String.

answered Sep 19 '19 at 21:03

Anoop R Desai

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You are right man, my bad don´t look at documentation first. – basquiatraphaeu Sep 20 '19 at 18:07
But, I know that hashCode does not represent the memory adress of the variable, so how can I check or CONFIRM the immutability of a String ? – basquiatraphaeu Sep 20 '19 at 18:16
@RaphaelSilva [This answer](https://stackoverflow.com/a/580987/1898437) (and perhaps other answers for the same question) may be of help. – Anoop R Desai Sep 20 '19 at 20:08
Hey @Anoop, thanks for your answer. Before close this topic, I did my research I´d like to answer my own question here for the posterity. – basquiatraphaeu Sep 23 '19 at 19:02
@RaphaelSilva Sure, go ahead! – Anoop R Desai Sep 24 '19 at 08:56

If Strings in java are immutable, why I can do this?

1 Answers1