Java String type

Question

I have this Java code:

String x = "abcgrfdhdrhfhtjfhtgdrfdjrtkkytjgykjgrthfd";
String y = "abcgrfdhdrhfhtjfhtgdrfdjrtkkytjgykjgrthfd";
System.out.println(x == y);

It prints out: true

How is this possible? Isn't String just a pointer to the first byte (character) of the string itself? Why does it act like a primitive type?

If you want to create a new instance, I think you need to do `new String(x)`. — Hugo Sousa, Jul 16 '14 at 12:27
Look at the documentation http://docs.oracle.com/javase/7/docs/api/java/lang/String.html — Adam Yost, Jul 16 '14 at 12:27
Give a look to this answer: http://stackoverflow.com/a/513839/3625883 — Filo, Jul 16 '14 at 12:29
It is a potential duplicate, but it's definitely not a duplicate of the one it was closed as a dupe of so have re-opened it. — Tim B, Jul 16 '14 at 12:34
possible duplicate of [How do I compare strings in Java?](http://stackoverflow.com/questions/513832/how-do-i-compare-strings-in-java) — William Morrison, Jul 16 '14 at 12:50

akash · Answer 1 · 2014-07-16T12:41:54.533

4

enter image description here

S4 and S3 are String literals while S1 and S2 are String Objects.

String s1="java";//literal
String s2=new String("java");//Object

As == is used for checking whether two Strings are the same Object or not.Means referencing the same memory location or not.

edited Jul 16 '14 at 12:41

answered Jul 16 '14 at 12:28

akash

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To sum up: s3 and s4 are created with `new String(...)` while s1 and s2 are created like `String a = "aaa"`. Is that it? – Hugo Sousa Jul 16 '14 at 12:31
2

@HugoSousa it's the other way around actually – Tim B Jul 16 '14 at 12:36
@TimB Yep, messed up there. – Hugo Sousa Jul 16 '14 at 12:39

score 0 · Answer 2 · answered Jul 16 '14 at 12:27

0

== operator check if the two references point to the same object or not.

Here both x and y pointing same value in Heap

answered Jul 16 '14 at 12:27

Ruchira Gayan Ranaweera

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score 0 · Answer 3 · answered Jul 16 '14 at 12:30

It prints out: true How is this possible?

String x = "abcgrfdhdrhfhtjfhtgdrfdjrtkkytjgykjgrthfd"; 
// one String object created on the String pool.
String y = "abcgrfdhdrhfhtjfhtgdrfdjrtkkytjgykjgrthfd"; 
// y is a reference which points to the same object created abovr on the String pool.
So,
System.out.println(x == y); // true

Isn't String just a pointer to the first byte (character) of the string itself? Why does it act like a primitive type?

Well, yes. String Objects are indeed backed by a character array. But that has nothing to do with equality

score 0 · Answer 4 · answered Jul 16 '14 at 12:39

As you know Java == compares the references, it asks whether two objects are actually the same object, not just having the same value.

String x = "abcgrfdhdrhfhtjfhtgdrfdjrtkkytjgykjgrthfd";
String y = "abcgrfdhdrhfhtjfhtgdrfdjrtkkytjgykjgrthfd";
System.out.println(x == y);

In this case you have assigned x and y to point to the exact same String. Internally the Java compiler has noticed that both x and y are being given the same value, so to save space it has created only one String object - and then both x and y point to that object.

This is why x == y..

If you changed one or the other (or both) to:

String x = new String("abcgrfdhdrhfhtjfhtgdrfdjrtkkytjgykjgrthfd");

Then you would get the result false.

Java String type

4 Answers4