Comparing strings with == which are declared final in Java

When you declare a String (which is immutable) variable as final, and initialize it with a compile-time constant expression, it also becomes a compile-time constant expression, and its value is inlined by the compiler where it is used. So, in your second code example, after inlining the values, the string concatenation is translated by the compiler to:

String concat = "str" + "ing";  // which then becomes `String concat = "string";`

which when compared to "string" will give you true, because string literals are interned.

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 - Constant Expression:

Compile-time constant expressions of type String are always "interned" so as to share unique instances, using the method String#intern().

This is not the case in your first code example, where the String variables are not final. So, they are not a compile-time constant expressions. The concatenation operation there will be delayed till runtime, thus leading to the creation of a new String object. You can verify this by comparing byte code of both the codes.

The first code example (non-final version) is compiled to the following byte code:

  Code:   0:   ldc     #2; //String str   2:   astore_1   3:   ldc     #3; //String ing   5:   astore_2   6:   new     #4; //class java/lang/StringBuilder   9:   dup   10:  invokespecial   #5; //Method java/lang/StringBuilder."<init>":()V   13:  aload_1   14:  invokevirtual   #6; //Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;   17:  aload_2   18:  invokevirtual   #6; //Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;   21:  invokevirtual   #7; //Method java/lang/StringBuilder.toString:()Ljava/lang/String;   24:  astore_3   25:  getstatic       #8; //Field java/lang/System.out:Ljava/io/PrintStream;   28:  aload_3   29:  ldc     #9; //String string   31:  if_acmpne       38   34:  iconst_1   35:  goto    39   38:  iconst_0   39:  invokevirtual   #10; //Method java/io/PrintStream.println:(Z)V   42:  return

Clearly it is storing str and ing in two separate variables, and using StringBuilder to perform the concatenation operation.

Whereas, your second code example (final version) looks like this:

  Code:   0:   ldc     #2; //String string   2:   astore_3   3:   getstatic       #3; //Field java/lang/System.out:Ljava/io/PrintStream;   6:   aload_3   7:   ldc     #2; //String string   9:   if_acmpne       16   12:  iconst_1   13:  goto    17   16:  iconst_0   17:  invokevirtual   #4; //Method java/io/PrintStream.println:(Z)V   20:  return

So it directly inlines the final variable to create String string at compile time, which is loaded by ldc operation in step 0. Then the second string literal is loaded by ldc operation in step 7. It doesn't involve creation of any new String object at runtime. The String is already known at compile time, and they are interned.

As per my research, all the final String are interned in Java. From one of the blog post:

So, if you really need to compare two String using == or != make sure you call String.intern() method before making comparison. Otherwise, always prefer String.equals(String) for String comparison.

So it means if you call String.intern() you can compare two strings using == operator. But here String.intern() is not necessary because in Java final String are internally interned.

You can find more information String comparision using == operator and Javadoc for String.intern() method.

Also refer this Stackoverflow post for more information.

If you take a look at this methods

public void noFinal() {    String str1 = "str";    String str2 = "ing";    String concat = str1 + str2;    System.out.println(concat == "string");}public void withFinal() {    final String str1 = "str";    final String str2 = "ing";    String concat = str1 + str2;    System.out.println(concat == "string");}

and its decompiled with javap -c ClassWithTheseMethods versions you will see

  public void noFinal();    Code:       0: ldc           #15                 // String str       2: astore_1             3: ldc           #17                 // String ing       5: astore_2             6: new           #19                 // class java/lang/StringBuilder       9: dup                 10: aload_1             11: invokestatic  #21                 // Method java/lang/String.valueOf:(Ljava/lang/Object;)Ljava/lang/String;      14: invokespecial #27                 // Method java/lang/StringBuilder."<init>":(Ljava/lang/String;)V      17: aload_2             18: invokevirtual #30                 // Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;      21: invokevirtual #34                 // Method java/lang/StringBuilder.toString:()Ljava/lang/String;      ...

and

  public void withFinal();    Code:       0: ldc           #15                 // String str       2: astore_1             3: ldc           #17                 // String ing       5: astore_2             6: ldc           #44                 // String string       8: astore_3             ...

So if Strings are not final compiler will have to use StringBuilder to concatenate str1 and str2 so

String concat=str1+str2;

will be compiled to

String concat = new StringBuilder(str1).append(str2).toString();

which means that concat will be created at runtime so will not come from String pool.

Also if Strings are final then compiler can assume that they will never change so instead of using StringBuilder it can safely concatenate its values so

String concat = str1 + str2;

can be changed to

String concat = "str" + "ing";

and concatenated into

String concat = "string";

which means that concate will become sting literal which will be interned in string pool and then compared with same string literal from that pool in if statement.