Java returning false on comparing two equal strings

Question

When I execute:

String a = "hello";
String b = "hello";
System.out.println(a==b);

I get output as "true".

But when I run:

String a = new String("hello");
String b = "hello";
System.out.println(a==b);

I get output as "false".

I understand that in the first case, Java makes 'b' point to the same object where 'a' pointed, but why can't it do that in the second case?

Because you are explicitly telling it to create a new object. — racraman, Jan 31 '20 at 05:58
@racraman Yes, but only for 'a'. While creating 'b', Java should search for an object already containing "hello" and that to 'b' - as it did it in the first case. — Kunal Jain, Jan 31 '20 at 06:00
In short, `==` compares references, `.equals()` compares your String content. — user3437460, Jan 31 '20 at 06:06
`Java should search ... as it did it in the first case. ` No, it shouldn't, since that would have a (big) impact on runtime speed. In the first case, the "search" does NOT happen at runtime but at compile time against compile-time constants, and so there is no impact on runtime speed. — racraman, Jan 31 '20 at 06:09

score 2 · Answer 1 · answered Jan 31 '20 at 05:59

String in Java are immutable. Meaning:

String a = "hello";
String b = "hello";

a and b are litterally pointing to the same object in memory

Here you explicitly create a new object

String a = new String("hello");
String b = "hello";

Hence the 2 are not equal.

== compares the memory address. To check if an Object is equal to another in Java you should use the equals method that is on all objects.

If you rewrite System.out.println(a==b); to System.out.println(a.equals(b)); it will be true for both cases

score 0 · Accepted Answer · answered Jan 31 '20 at 06:00

0

Please read the following well explained answer on stackoverflow:
What is the difference between "text" and new String("text")?

answered Jan 31 '20 at 06:00

Tarun

2 Answers2