Related to String interning

Question

public static void main(String[] args) {

    String a = new String("lo").intern();
    final String d = a.intern();
    String b = "lo";
    final String e = "lo";
    String c = "Hello";
    System.out.println(b==a);//true
    System.out.println(d==a);//true
    System.out.println(e==a);//true
    System.out.println(c=="Hel"+a); //why is this false? when e==a is true
    System.out.println(c=="Hel"+d); //why is this false?
    System.out.println(c=="Hel"+b); //why is this false?
    System.out.println(c=="Hel"+e); //this is true

}

This results in

true
true
true
false
false
false
true

The expression e==a is true implies same reference. So why the last expression is true but the 4th to last ie c== "Hel"+a is false?

Answer 1

The expression

"Hel" + a

Is not a compile time constant. Actually, it compiles to:

new StringBuilder().append("Hel").append(a).toString()

(or similar) which creates a new String object at runtime.

However, because e is final, the compiler can determine that the concatenation of "Hel" and e's value is a constant value, and so interns it.

Answer 2

all these strings are calculated in runtime, this is why they are different

System.out.println(c=="Hel"+a); //why is this false? when e==a is true
System.out.println(c=="Hel"+d); //why is this false?
System.out.println(c=="Hel"+b); //why is this false?

this one calculated during compile time, because e is final:

System.out.println(c=="Hel"+e); //this is true

if you change code to this:

    System.out.println(c==("Hel"+a).intern()); //why is this false? when e==a is true
    System.out.println(c==("Hel"+d).intern()); //why is this false?
    System.out.println(c==("Hel"+b).intern()); //why is this false?

all of them will produce true