Why don't two variables referencing the same object change when one of them does?

Question

Question

The below code show that when affecting an array to an another one, the two arrays becomes depending to each others.

int tab [] = {1,2,3};

int tab2 [] = tab;

tab2[0] = 5;

System.out.print(tab[0]); // 5

I want to know why this isn't the same with the type String, since if we had the following :

String ch1 = "hello";

String ch2 = ch1;

ch2 = "hi";

System.out.print(ch1); // hello

The two variables ch1 and ch2 are referencing the same string, so changing one will affect the other.

The two variables are clearly not referencing the same string, as your snippet demonstrates. — jsheeran, Aug 26 '16 at 14:21
In the first example you're modifying the value of something referenced by two variables. In the second example you're changing the reference of `ch2`. — Dave Newton, Aug 26 '16 at 14:21
changing String in Java is not possible, String provides immutable objects. — Jean-Baptiste Yunès, Aug 26 '16 at 14:42

f1sh · Answer 1 · 2016-08-26T15:24:09.343

3

When you use

int tab2 [] = tab;

then tab2 is a reference to the same array. Changing the value of tab2[0] meaning you change a value inside the data structure that both tab and tab2 are pointing at.

Using

ch2 = "hi";

create a new instance of String, it does not change the object that ch2 was referring to before.

edited Aug 26 '16 at 15:24

answered Aug 26 '16 at 14:22

f1sh

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Please don't bring the immutability of `String` into this, it's irrelevant. The same concept applies to any type. This issue is a re-assignment of the variable. – Sotirios Delimanolis Aug 26 '16 at 14:44
Yeah, immutability is not the cause. It works the same way even for mutable objects. – Govardhan Aug 26 '16 at 15:12
@SotiriosDelimanolis that is correct, i removed it. – f1sh Aug 26 '16 at 15:23

score 3 · Accepted Answer · answered Aug 26 '16 at 14:26

Think like this.

Both array and String operate based on Addresses.

When you say arr1 = arr2, you are telling that arr1 is going to be pointing to same address location as arr2.

With strings, "XYZ" is also a string, which contains an address location. when you say str1 = str2, str1 will be pointing to same address as str2. But when you say str1 = "XYZ", "XYZ" is another object which is stored at a different address. So, str1 will be reassigned to the new address.

Even with arrays, if you say arr1 = arr3, arr2 won't be changing. arr1 will be repointed to a new address pointed by arr3. But when you say arr1[0], you are actually trying to change the value stored in location, so it will also influence arr2 or arr3 depending on which assignment is latest.

After analysing the situation i concluded the same thing that you explained. Thanks for the answer. — user3078643, Aug 26 '16 at 14:54

puhlen · Answer 3 · 2016-08-26T14:58:07.610

2

You aren't changing the String itself, you are just setting another value for ch2 reference, so it refers to a new String object, ch1 still refers to the old String object.

To make your array code similar to the string code, you need to change it a bit

 int tab [] = {1,2,3};

int tab2 [] = tab;

tab2 = new int[]{5,6,7};

System.out.print(tab[0]); // 1

edited Aug 26 '16 at 14:58

answered Aug 26 '16 at 14:21

puhlen

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The `System.out.println` will actually print 1, I would get rid of that comment at the end. – Orin Aug 26 '16 at 14:40
I think he means tab2[0] – user3078643 Aug 26 '16 at 14:48
@user3078643 no it was `tab[0]` I just made a mistake with the comment, it would be 1, not 5 because tab is still the original array. – puhlen Aug 26 '16 at 14:59
i see, thanks for the answer. – user3078643 Aug 26 '16 at 15:01

Why don't two variables referencing the same object change when one of them does?

3 Answers3