46

When I tried to write something like this:

public interface MyInterface {
    static {
        System.out.println("Hello!");
    }
}

the compiler could not compile it.

But when I wrote something like this:

interface MyInterface {
    Integer iconst = Integer.valueOf(1);
}

and decompiled it, I saw static initialization:

public interface MyInterface{
    public static final java.lang.Integer i;

    static {};
      Code:
      0:   iconst_1
      1:   invokestatic    #1; //Method java/lang/Integer.valueOf:(I)Ljava/lang/Integer;
      4:   putstatic       #2; //Field i:Ljava/lang/Integer;
      7:   return
}

Could you please explain this behavior to me?

Sergey Morozov
  • 4,398
  • 3
  • 24
  • 37

5 Answers5

24

Interfaces should not have side-effects and that even applies to static intializers. They would have highly JVM-implementation dependent behavior. Look at the following code

public class InterfaceSideEffects {
  public static void main(String[] args) {
    System.out.println("InterfaceSideEffects.main()");
    Impl i=new Impl();
    System.out.println("Impl initialized");
    i.bla();
    System.out.println("Impl instance method invoked");
    Foo f=new Impl();
    System.out.println("Impl initialized and assigned to Foo");
    f.bla();
    System.out.println("Foo interface method invoked");
  }
}
interface Foo {
  int dummy=Bar.haveSideEffect();
  void bla();
}
class Bar {
  static int haveSideEffect() {
    System.out.println("interface Foo initialized");
    return 0;
  }
}
class Impl implements Foo {
  public void bla() {
  }
}

What do you think, when will interface Foo initialized be printed? Try to guess and run code afterwards. The answer might surprise you.

Holger
  • 243,335
  • 30
  • 362
  • 661
  • 2
    I'm working as a java developer for the last 5 years and the answer truly surprised me. What a shame. Thanks! – hsestupin May 15 '16 at 15:48
  • 1
    That being said, though, the side effects can just as well be brought about via functions called by initializer expressions, just as you demonstrated. Forbidding `static {}` blocks does nothing to prevent that. – Dolda2000 Oct 27 '16 at 17:30
  • 3
    @Dolda2000: you can’t make a programming language bullet-proof, well, at least not without dramatically reducing its usefulness. So they only thing you can try when designing, is to make correct things easy and incorrect things harder to achieve. – Holger Oct 27 '16 at 17:38
  • @Holger: Perhaps, but in this case, it also creates hassles when one wants to initialize static fields in a more complex way than simple expressions allow. Not too seldom I find myself having to create useless classes containing `private static` methods just for initializing some field that requires a `for` loop or something. Makes one hold a grudge. – Dolda2000 Oct 27 '16 at 17:40
  • @Dolda2000: maybe you’re overusing static fields in interfaces… – Holger Oct 27 '16 at 17:42
  • @Holger: Or maybe the language is just being too opinionated. Anyway, I take my simple revenge by naming the useless classes things such as `JavaIsStupid`, so it's alright; it's a situation of mutual abuse. :) – Dolda2000 Oct 27 '16 at 17:46
  • @Holger where should the surprise come from btw? that the static field initialization in interface is deferred only when actually used? – Eugene Apr 10 '18 at 07:40
  • @Eugene there is no individual “static field initialization”, but only one initialization for the entire class/interface. The fact that the field has not been initialized indicates that the entire interface initialization did not happen (in the context of this Q&A, this implies that hypothetical `static { … }` blocks would not have been executed as well) So the behavior that may surprise developers is that neither, using a class implementing the interface nor even invoking an interface method on it, will trigger the interface initialization (even less intuitive, `default` methods change that). – Holger Apr 10 '18 at 07:56
20

You can have static initialisation, but you cannot have a static block. The fact the static initialisation needs a static code block to implement does change the Java syntax.

The point is you are not meant to have code in an interface (before Java 8) but you are allowed to initialise fields.

BTW you can have a nested class or enum which has as much code as you like and you can call this while initialising a field. ;)

Peter Lawrey
  • 498,481
  • 72
  • 700
  • 1,075
  • Thanks for answer. Maybe you know why that did(disable static block in interface)? – Sergey Morozov Nov 01 '13 at 08:34
  • 4
    @frostjogla interfaces should not have side-effects. – Holger Nov 01 '13 at 08:45
  • 2
    @frostjogla interfaces are only there to define a contract, not provide any implementation. Constants are allowed on the basis they might be used in a method. – Peter Lawrey Nov 01 '13 at 08:47
  • 2
    Sometimes interfaces contain public static final constants... if you have a Set or Map or something, the initialization (to be readable) might require a static block, without introducing any side effects – Kip Jul 27 '16 at 18:39
7

You can get around the problem - if you see it as a problem - by putting a second non-public class in the same file.

public interface ITest {
  public static final String hello = Hello.hello();
}

// You can have non-public classes in the same file.
class Hello {
  static {
    System.out.println("Static Hello");
  }
  public static String hello() {
    System.out.println("Hello again");
    return "Hello";
  }
}

Testing this with:

public class Test {
  public void test() {
    System.out.println("Test Hello");
    System.out.println(ITest.hello);
  }

  public static void main(String args[]) {
    try {
      new Test().test();
    } catch (Throwable t) {
      t.printStackTrace(System.err);
    }
  }

}

prints:

Test Hello
Static Hello
Hello again
Hello

Java is such a clever language - it makes it difficult to do stupid things but not impossible. :)

OldCurmudgeon
  • 60,862
  • 15
  • 108
  • 197
0

Intefaces does not have any initialization blocks. Following code snippet may be helpful..

public interface MyInterface {
public static final int a;// Compilation error as there is no way for 
                          // explicit initialization

}

public class MyClass {
public static final int a;// Still no error as there is another way to 
                          //initialize variable even though they are final.
 static{
    a=10;
   }

}
Siddappa Walake
  • 245
  • 4
  • 12
-1

There is never a point to declaring a static method in an interface. They cannot be executed by the normal call MyInterface.staticMethod(). (EDIT:Since that last sentence confused some people, calling MyClass.staticMethod() executes precisely the implementation of staticMethod on MyClass, which if MyClass is an interface cannot exist!) If you call them by specifying the implementing class MyImplementor.staticMethod() then you must know the actual class, so it is irrelevant whether the interface contains it or not.

More importantly, static methods are never overridden, and if you try to do:

MyInterface var = new MyImplementingClass();
var.staticMethod();

the rules for static say that the method defined in the declared type of var must be executed. Since this is an interface, this is impossible.

You can of course always remove the static keyword from the method. Everything will work fine. You may have to suppress some warnings if it is called from an instance method.

To answer some of the comments below, the reason you can't execute "result=MyInterface.staticMethod()" is that it would have to execute the version of the method defined in MyInterface. But there can't be a version defined in MyInterface, because it's an interface. It doesn't have code by definition.

Manish Doshi
  • 1,157
  • 1
  • 8
  • 16
  • 5
    The entire answer is missing the point of the question. It’s not about static methods, it’s about static initializers. – Holger Nov 01 '13 at 08:42