assignment operator String object

Question

I am new to JAVA programming. I have read it in my book

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

This should return false as a & b refer to different instances of String objects.

Bcoz the assignments operator compares the instances of objects but Still I am getting a true.
I am using Eclipse IDE.

Example in book goes as this:

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

That bit of code prints “false” for s == sToo. That's because s and sToo are references to different instances of the String object. So, even though they have the same value, they are not equal in the eyes of the equality operators. Also, s == “s” prints false, because the string literal produces yet another instance of the String class.

Name of book: JAVA 7 for Absolute Beginners

Answer 1

This is an optimisation called string pooling in which compile-time constant Strings (aka known to be identical at compile time) can be set such that they really are the same object in memory (saving space for one of the most used types of object). Or in the words of the docs;

"All literal strings and string-valued constant expressions are interned."

Note that this only applies to Strings that are defined at compile time, so the following truly would print false.

String a="Hello";
String b=new String("Hello");
System.out.println(a==b); //prints false because a new string was forced

or

String a="Hello";
String b1="He";
String b2="llo";
String b=b1+b2;

System.out.println(a==b); //prints false because b wasn't know to be "Hello" at compile time so could not use string pooling

N.B. It is possible to cause the second snippet to print true by making b1 and b2 final, allowing b1+b2 to be known at compile time. All in all you need to be very careful and treat string==string with considerable respect, in the vast majority of cases you want string.equals(string) in which this behaviour does not exist.