Java compiler handles inline Strings efficiently?

Question

1.

static final String memFriendly = "Efficiently stored String";
System.out.println(memFriendly);

2.

System.out.println("Efficiently stored String");

Will Java compiler treat both (1 and 2) of these in the same manner?

FYI: By efficiently I am referring to runtime memory utilization as well as code execution time. e.g. can the 1st case take more time on stack loading the variable memFriendly?

Answer 1

Yes.

OK, I need at least 15 characters in my answer.

Answer 2

In this case, the compiler will handle both the same.

Anytime a String is defined at compile time, Java will optimize the storage of Strings.

If a string is defined at runtime, Java has no way of doing the same optomization.

Answer 3

The code you have is equivilant because string literals are automatically interned by the compiler.

If you are really concerned about String performance and will be reusing the same strings over and over you should take a look at the intern method on the string class.

http://java.sun.com/javase/6/docs/api/java/lang/String.html#intern()

Answer 4

This is covered in the Java Language Spec:

Each string literal is a reference (§4.3) to an instance (§4.3.1, §12.5) of class String (§4.3.3). String objects have a constant value. String literals-or, more generally, strings that are the values of constant expressions (§15.28)-are "interned" so as to share unique instances, using the method String.intern.

You can also see for yourself using the javap tool.

For this code:

System.out.println("Efficiently stored String");
final String memFriendly = "Efficiently stored String";
System.out.println(memFriendly);

javap gives the following:

0:   getstatic     #2; //Field java/lang/System.out:Ljava/io/PrintStream;
3:   ldc       #3; //String Efficiently stored String
5:   invokevirtual #4; //Method java/io/PrintStream.println:(Ljava/lang/String;)V
8:   getstatic     #2; //Field java/lang/System.out:Ljava/io/PrintStream;
11:  ldc       #3; //String Efficiently stored String
13:  invokevirtual #4; //Method java/io/PrintStream.println:(Ljava/lang/String;)V

Answer 5

public static void main(String[] args) {
 System.out.println("Hello world!");

 String hola = "Hola, mundo!";
 System.out.println(hola);
}

Here is what javap shows as the disassembly for this code:

0:   getstatic       #16; //Field java/lang/System.out:Ljava/io/PrintStream;
3:   ldc     #22; //String Hello world!
5:   invokevirtual   #24; //Method java/io/PrintStream.println:(Ljava/lang/String;)V
8:   ldc     #30; //String Hola, mundo!
10:  astore_1
11:  getstatic       #16; //Field java/lang/System.out:Ljava/io/PrintStream;
14:  aload_1
15:  invokevirtual   #24; //Method java/io/PrintStream.println:(Ljava/lang/String;)V
18:  return

Looks like the second string is being stored, whereas the first one is simply passed to the method directly.

This was built with Eclipse's compiler, which may explain differences in my answer and McDowell's.

Update: Here are the results if hola is declared as final (results in no aload_1, if I'm reading this right then it means this String is both stored and inlined, as you might expect):

0:   getstatic       #16; //Field java/lang/System.out:Ljava/io/PrintStream;
3:   ldc     #22; //String Hello world!
5:   invokevirtual   #24; //Method java/io/PrintStream.println:(Ljava/lang/String;)V
8:   ldc     #30; //String Hola, mundo!
10:  astore_1
11:  getstatic       #16; //Field java/lang/System.out:Ljava/io/PrintStream;
14:  ldc     #30; //String Hola, mundo!
16:  invokevirtual   #24; //Method java/io/PrintStream.println:(Ljava/lang/String;)V
19:  return