How string literals are created?

Question

I am trying to understand the string constant pool, how string literal objects are managed in constant pool, i am not able to understand why I am getting false from below code where s2 == s4

 public static void main(String[] args) {
    String s1 = "abc";
    String s2 = "abcd";
    String s3 = "abc" +"d";
    String s4 = s1 + "d";
    System.out.println(s2 == s3); //  OP:  true
    System.out.println(s2 == s4); // OP:  false
 }

Answer 1

The expression "abc" + "d" is a constant expression, so the concatenation is performed at compile-time, leading to code equivalent to:

String s1 = "abc";
String s2 = "abcd";
String s3 = "abcd";
String s4 = s1 + "d";

The expression s1 + "d" is not a constant expression, and is therefore performed at execution time, creating a new string object. Therefore although s2 and s3 refer to the same string object (due to constant string interning), s2 and s4 refer to different (but equal) string objects.

See section 15.28 of the JLS for more details about constant expressions.