Why String is Immutable in Java

Lets break it into some parts

String s1 = “hello”;
This Statement creates string containing hello and occupy space in memory i.e. in Constant String Pool and and assigned it to reference object s1

String s2 = s1;
This statement assigns the same string hello to new reference s2

        |       |
s1 ---->| hello |<----- s2

Both references are pointing to the same string so output the same value as follows.

out.println(s1); // o/p: hello
out.println(s2); // o/p: hello
Though String is immutable, assignment can be possible so the s1 will now refer to new value stack.

s1 = “stack”;

        |         |
s1 ---->| stack   |

But what about s2 object which is pointing to hello it will be as it is.

        |          |
s2 ---->| hello    |

out.println(s1); // o/p: stack
out.println(s2); // o/p: hello
Since String is immutable Java Virtual Machine won’t allow us to modify string s1 by its method. It will create all new String object in pool as follows.

s1.concat(” overflow”);

                |                   |
s1.concat ----> | stack overflow    |

out.println(s1); // o/p: stack
out.println(s2); // o/p: hello
out.println(s1.concat); // o/p: stack overflow
Note if String would be mutable then the output would have been

out.println(s1); // o/p: stack overflow
Now you might be surprised why String has such methods like concat() to modify. Following snippet will clear your confusion.

s1 = s1.concat(” overflow”);
Here we are assigning modified value of string back to s1 reference.

        |                   |
s1 ---->| stack overflow    |

out.println(s1); // o/p: stack overflow
out.println(s2); // o/p: hello
That’s why Java decided String to be a final class Otherwise anyone can modify and change the value of string. Hope this will help little bit.

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