When I compile this snippet.
public class InternTest {
    public static void main(String...strings ){
        final String str1="str";
        final String str2="ing";
        String str= str1+str2;
    }
}
Which produces the following byte code
public static void main(java.lang.String...);
   flags: ACC_PUBLIC, ACC_STATIC, ACC_VARARGS
   Code:
     stack=1, locals=4, args_size=1
        0: ldc           #16                 // String str
        2: astore_1
        3: ldc           #18                 // String ing
        5: astore_2
        6: ldc           #20                 // String string
        8: astore_3
        9: return
so string literal "string" is already there in the constant pool which gets pushed 6: ldc      #20  // String string on stack at this line.
Quoting JSL
From JLS §4.12.4 - final Variables:
A variable of primitive type or type String, that is final and initialized with a compile-time constant expression (§15.28), is called a constant variable.
Also from JLS §15.28 - ConstantExpression:
Compile-time constant expressions of type String are always "interned" so as to share unique instances, using the method String#intern().
So i know str1 and str2 will be interned the moment it has been created."str" and "ing" will share the same memory at line String str= str1+str2;
But how str1+str2 directly produces "string" in the constant string pool. Without invoking any String Builder class like it does when i don't write final. ? To see if it has got anything to do with intern things 
I wrote this snippet
public class IntermTest {
    public static void main(String...strings ){
         String str1=("str").intern();
        String str2=("ing").intern();
        String str= str1+str2;
    }
}
But when i generated the byte code i got this
public static void main(java.lang.String...);
    flags: ACC_PUBLIC, ACC_STATIC, ACC_VARARGS
    Code:
      stack=3, locals=4, args_size=1
         0: ldc           #16                 // String str
         2: invokevirtual #18                 // Method java/lang/String.intern:
()Ljava/lang/String;
         5: astore_1
         6: ldc           #24                 // String ing
         8: invokevirtual #18                 // Method java/lang/String.intern:
()Ljava/lang/String;
        11: astore_2
        12: new           #26                 // class java/lang/StringBuilder
        15: dup
        16: aload_1
        17: invokestatic  #28                 // Method java/lang/String.valueOf
:(Ljava/lang/Object;)Ljava/lang/String;
        20: invokespecial #32                 // Method java/lang/StringBuilder.
"<init>":(Ljava/lang/String;)V
        23: aload_2
        24: invokevirtual #35                 // Method java/lang/StringBuilder.
append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
        27: invokevirtual #39                 // Method java/lang/StringBuilder.
toString:()Ljava/lang/String;
        30: astore_3
        31: return
Indeed it also uses stringBuilder for concatenation. So it has do something with final.  Is there something special about final Strings that I'm definitely not aware of ? 
http://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.28 says that
Simple names (§6.5.6.1) that refer to constant variables (§4.12.4) are constant expressions.
http://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.28 also says:
A 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)
 - [...]
 - The additive operators + and - (§15.18)
 - [...]
 - Simple names (§6.5.6.1) that refer to constant variables (§4.12.4).
 Example 15.28-1. Constant Expressions
[...]
"The integer " + Long.MAX_VALUE + " is mighty big."
Since those two variables are constant expressions, the compiler does the concatenation:
String str = str1 + str2;
is compiled the same way as
String str = "str" + "ing";
which is compiled the same way as
String str = "string";
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