Java String comparision depending on object initialisation

Question

Following code:

    String a = new String("aaa");
    String a2 = new String("aaa");
    System.out.println(a == a2);

    String b = "bbb";
    String b2 = "bbb";
    System.out.println(b == b2);

Produces following output:

false
true

Why there is difference in output for comparision a==a2 and b==b2 depending from type of String creation ?

[What is the difference between "text" and new String("text")?](http://stackoverflow.com/q/3052442) — Tom, Dec 15 '15 at 00:04

score -1 · Accepted Answer · edited May 23 '17 at 12:04

-1

When you declare a and a2 you explicitly create new (different) Strings. The use of the constructor causes a copy to be made. Hence == fails as a and a2 point to different values.

When you declare b and b2, b2 can re-use the same string from the pool. Hence they actually point to the same value and == returns true.

Take a look here or here for a detailed answer.

edited May 23 '17 at 12:04

Community

1
1

answered Dec 15 '15 at 00:05

dave

11,641
5
47
65

Thanks ! Two years of experience and I didnt know that functionallitz about string :D Shame – user109447 Dec 15 '15 at 00:07
No worries. That's was SO is for. Feel free to accept the answer :) – dave Dec 15 '15 at 00:09

Java String comparision depending on object initialisation

1 Answers1