Java: String literal and + operator

Question

import java.util.*;
import java.lang.*;
import java.io.*;    

class Test {
   public static void main (String[] args)  {

      String a="hello"+"world";  //line 1
      String b="hello";
      String c="world";
      String d=b+c;
      String e="helloworld";
      System.out.println(e==a);
      System.out.println(a==d);
     }
}

output:

true

false

From this discussion How does the String class override the + operator? I understood that 'a' and 'e' will refer to same String literal object.

Could anybody please tell

1. Why a and d are not referring to same string literal object?

2. How many objects were creating in line 1

Answer 1

From JLS 4.3.3. The Class String

The string concatenation operator + (§15.18.1) implicitly creates a new String object when the result is not a compile-time constant expression (§15.28).

Because literal value of a is "helloworld" and literal value of d is a reference object of b + c that is not a literal as from above JLS 4.3.3.

From JLS 3.10.5 String Literals

A string literal consists of zero or more characters enclosed in double quotes.

From 15.28. Constant Expressions

A compile-time constant expression is an expression denoting a value of primitive type or a String that does not complete abruptly and is composed using only the following:

• Literals of primitive type and literals of type String (§3.10.1, §3.10.2, §3.10.3, §3.10.4, §3.10.5)

If you want to make System.out.println(a==d); true use final keyword see below code:

  String a="hello"+"world";  //line 1
  final String b="hello"; // now b is a literal
  final String c="world";// now c is a literal
  String d=b+c;
  String e="helloworld";
  System.out.println(e==a);
  System.out.println(a==d);// now it will be true because both are literals.

---

Answer for comment:
How many objects were creating in line 1: Answer is 3
String a="hello"+"world";

1st literal object hello
2nd literal object world
3rd literal object "hello"+"world" == "helloworld"

Answer 2

The == sign checks if references to variables are the same. To check if any object is equal to other use Object.equals(Object o) method like this:

e.equals(a);