<?php
class A {
public static function who() {
echo __CLASS__;
}
public static function test() {
static::who(); // Here comes Late Static Bindings
}
}
class B extends A {
public static function who() {
echo __CLASS__;
}
}
B::test(); // Outputs "B"
?>
I want to get an equivalent in Java...so something like
class A {
public static void who(){
System.out.println("A");
};
public static void test(){
who(); //<<< How to implement a static:: thing here???
}
}
class B extends A {
public static void who(){
System.out.println("B");
};
public static void main(String[] args){
B.test(); // Outputs "A" but I want "B"
}
}
I want the who() call inside A::test to resolve as in PHP 5.3 by calling B::who.
EDIT: I know there is no "standard way" of doing this in most popular languages. I'm looking for hacks and such. Also, is this possible in C/C++, or any other popular OOP language?
This is not for any real design on anything. I'm just being curious.
Not possible in Java. (At least not without ugly reflection hacks.)
I encourage you to rethink your design and rely on proper objects.
Related question:
Can I override and overload static methods in Java?
Edit: B.test() will (or can at least according to spec) be compiled into a call to A.test(), so there's no way to discover how the call was made from within A.test(). In other words, there's no way to let the behaviour of A.test depend on if it was called through A.test() or B.test().
Since you're asking out of curiosity, here's AFAIK the closest "solution".
Overload test with a test(Class<?> c) which takes as argument the class which defines the intended who method.
Hide (note that you can't override) test() in class B.
And change the implementation of A.test slightly.
In code:
class A {
public static void who() {
System.out.println("A");
}
public static void test() {
test(A.class);
}
public static void test(Class<?> c) {
//who(); //<<< How to implement a static:: thing here???
try {
c.getMethod("who").invoke(null); // Call static who on given class.
} catch (Exception e) {
}
}
}
public class B extends A {
public static void who(){
System.out.println("B");
}
public static void test() {
test(B.class);
}
public static void main(String[] args){
A.test(); // Outputs "A"
B.test(); // Outputs "B"
}
}
It seems that the compiler generates a call to B.test in the bytecode even though B doesn't declare a method named test.
Bytecode of main method:
invokestatic #5 = Method B.test(()V)
return
Given the names of a class and method ("B" and "who") you can easily use reflection to call the method. So the question becomes
Can you extract B by combining the call stack and bytecode inside A.test?
You'll need to use the return address stored on the stack to locate the call to B.test in the bytecode and extract the declared call. There are plenty of bytecode manipulation libraries, but I don't know if any of them allow you to tie that to the execution stack in the JVM.
You can't override static methods in java.
http://geekexplains.blogspot.co.uk/2008/06/can-you-override-static-methods-in-java.html
Here's an example from Java. It uses Java 8 default methods and getClass(). I bet it works with classes too:
interface A {
default String name() {
return getClass().getName();
}
}
class B implements A {}
public class LateBinding {
public static void main(String[] args) {
// Create an anonymous class in `LateBinding` (called `$1`)
System.out.println(new A(){}.name());
// Instantiate a new `B`
B b = new B();
System.out.println(b.name());
}
}
Results:
$ javac LateBinding.java && java LateBinding
LateBinding$1
B
As you can see the method knows in both cases where it's running, although it's defined in A. This example is not really static, because you can't call getClass() statically, but LSB in PHP is not really limited to static contexts.
There is no elegant way to do it with static method declaration (Only Delphi from what I'm aware of supports override for static methods). However if static is not necessary for you you can write something like this:
class A {
public void who(){
System.out.println("A");
};
public void test(){
who(); //<<< How to implement a static:: thing here???
}
}
class B extends A {
#Override
public void who(){
System.out.println("B");
};
public void main(String[] args){
A instance = new A();
instance.test(); // prints 'A'
instance = new B();
instance.test(); // prints 'B'
}
}
EDIT after clarification:
Pretty hacky way of doing this: Thread.currentThread().getStackTrace() then from top-most record get method and class this method belongs to. Having Class c - you could write c.getMethod("who").invoke(null); to call the correspond who() method.
Related
In java, I'd like to do something like this
public class Tata{
public static void f(){
//something
}
public static void g(){
//something
}
}
public class Titi{
public static void f(){
//something
}
public static void g(){
//something
}
}
public class Toto{
private Class c = Tata.class; //or Titi.class
public static void main(String[] args) {
c.f();
c.g();
}
}
To be precise, I'd like to be able to freely switch between classes Tata and Titi, to use their respective methods f or g.
This doesn't work as intended, as I get the cannot resolve method 'f()' error. Simply replacing c.f(); and c.g(); with Tata.f(); and Tata.g(); works fine, but defeats the purpose of using a parameter. How to solve this?
Will turn the comment into answer after all.. The correct (Java) way to deal with what you want is the use of interface. So in your demo code the implementation would be the following:
public interface TheFGFunctions {
void f();
void g();
}
public class Tata implements TheFGFunctions {
#Override
public void f() {
//something
}
#Override
public void g() {
//something
}
}
public class Titi implements TheFGFunctions {
#Override
public void f() {
//something
}
#Override
public void g() {
//something
}
}
public class Toto {
private TheFGFunctions c;
public Toto(TheFGFunctions c) {
this.c = c;
}
public void notStaticFunction() {
c.f();
c.g();
}
}
This way is totally typesafe with zero exceptions to deal with!
You cannot access a static method polymorphically. The Java language doesn't support it.
The reason your current approach fails is that c is an instance of the class Class, and the class Class doesn't define methods f() or g().
(The methods that it does define are listed in the javadoc for Class. Note that Class is final so you can't create a custom subclass with extra methods.)
The simple alternative is to use reflection; e.g.
Class c =
Method f = c.getMethod("f");
f.invoke(null); // because it is static
But note:
This is not statically type-safe. The compiler cannot tell when you make the mistake of trying to use a static f() on a class that doesn't have such a method.
There are a few exceptions that you need to deal with; e.g. missing methods, incorrect signatures, methods that are not static, methods that don't have the correct access.
Other answers have proposed creating an interface and wrapper classes to make certain static methods dispatchable. It will work and it will be compile-time type-safe (!) but there is a lot of boiler plate code to write.
#Michael Michailidis commented:
Thus interfaces!
Yea ... kind of. You can only dispatch polymorphically on instance methods declared on an interface. That implies that you must have an instance of Tata or Titi, and call the methods on it. My reading of the Question is that the author wants to avoid that.
(IMO, the avoidance is the real problem. You are better of not trying to avoid instance methods.)
FWIW, you can declare static methods in an interface (since Java 8), but they would behave the same as if you declared them in a class. You wouldn't be able to dispatch ...
You could use reflections:
private Class c = Tata.class;
public Toto() throws Exception {
c.getMethod("f").invoke(null);
c.getMethod("g").invoke(null);
}
Here my Tata class
public class Tata {
public static void f() {
System.out.println("ffff");
}
public static void g() {
System.out.println("gggg");
}
}
Output on new Toto() call:
ffff
gggg
Update (call with parameters):
public Toto() throws Exception {
c.getMethod("f", String.class).invoke(null, "paramValue1");
c.getMethod("g", String.class).invoke(null, "paramValue2");
}
public class Tata {
public static void f(String param1) {
System.out.println("ffff " + param1);
}
public static void g(String param2) {
System.out.println("gggg " + param2);
}
}
Output:
ffff paramValue1
gggg paramValue2
Write a wrapper interface
interface CWrapper {
void f();
void g();
}
and wrapper class factory method for each Class containing the methods
class CWrappers {
CWrapper forTiti(Class<Titi> titiClass) {
return new CWrapper() {
void f() { Titi.f(); }
void g() { Titi.g(); }
}
}
// another factory method for Tata
}
Then you can use that:
public class Toto {
private CWrapper c = CWrappers.forTata(Tata.class); //or forTiti(Titi.class)
public static void main(String[] args) {
c.f();
c.g();
}
}
So, I want to be able to get an instance of a subclass that is being run when it calls a method from the super class. For example, if I had this class:
public class A {
public void aMethod() {
//Here is where I want to see if class B is calling the code
}
}
public class B extends A {
}
public class C {
B b = new B();
b.aMethod();
}
And, like the comment says, I want to check, in aMethod, if class B, the subclass of class A, is calling the code.
As has been pointed out to you, there is almost never a good reason to do this and I agree that you should be using polymorphism instead. However, if you "need" to do this or just want to know how to go about doing something like this, you can use instanceof on this inside of the method:
class A {
public void aMethod() {
if (this instanceof B) {
System.out.println("I'm a B!");
}
}
}
public class B extends A {
public static void main(String[] args) {
B b = new B();
b.aMethod();
}
}
public class A {
public void aMethod() {
if(this.getClass() == B.class){
System.out.println("huhuhuuuuuuuuuuuuuuuuu");
}
}
}
public class B extends A {
}
public class C {
public static void main(String[] args) {
B b = new B();
b.aMethod();
}
}
Check here: How to get the caller class in Java
The 2nd part of the answer from #dystroy is probably a start.
Note that this finds a call at any depth:
for(final StackTraceElement element : Thread.currentThread().getStackTrace()) {
if (element.getClassName().equals(B.class.getName())) {
System.out.println("BINGO");
}
}
If you want to check only a limited depth, don't iterate through all of the array.
This can be useful e.g. if some framework forces you to have a special method or a no-arg constructor to be present, but you don't want any developer to call this method directly. (Yes, it is a hack, but sometimes odd frameworks force you to do odd things). Then you can have an assertion in the unwanted method that just throws an exception if it is called by the wrong corner of your code.
Anyway you should try do avoid things like this if possible.
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?
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.
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. ;)
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. :)
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;
}
}
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.
I have FinanceRequests and CommisionTransactions in my domain.
If I have a list of FinanceRequests each FinanceRequest could contain multiple CommisionTransactions that need to be clawed back. Dont worry how exactly that is done.
The class below (very bottom) makes me feel all fuzzy and warm since its succint and reuses existing code nicely. One problem Type erasure.
public void clawBack(Collection<FinanceRequest> financeRequestList)
public void clawBack(Collection<CommissionTrns> commissionTrnsList)
They both have the same signature after erasure, ie:
Collection<FinanceRequest> --> Collection<Object>
Collection<CommissionTrns> --> Collection<Object>
So eclipse complainst that:
Method clawBack(Collection) has the same erasure clawBack(Collection) as another method in type CommissionFacade
Any suggestions to restructure this so that it still an elegant solution that makes good code reuse?
public class CommissionFacade
{
/********FINANCE REQUESTS****************/
public void clawBack(FinanceRequest financeRequest)
{
Collection<CommissionTrns> commTrnsList = financeRequest.getCommissionTrnsList();
this.clawBack(commTrnsList);
}
public void clawBack(Collection<FinanceRequest> financeRequestList)
{
for(FinanceRequest finReq : financeRequestList)
{
this.clawBack(finReq);
}
}
/********COMMISSION TRANSACTIOS****************/
public void clawBack(CommissionTrns commissionTrns)
{
//Do clawback for single CommissionTrns
}
public void clawBack(Collection<CommissionTrns> commissionTrnsList)
{
for(CommissionTrns commTrn : commissionTrnsList)
{
this.clawBack(commTrn);
}
}
}
Either rename the methods, or use polymorphism: use an interface, and then either put the clawback code in the objects themselves, or use double-dispatch (depending on your design paradigm and taste).
With code in objects that would be:
public interface Clawbackable{
void clawBack()
}
public class CommissionFacade
{
public <T extends Clawbackable> void clawBack(Collection<T> objects)
{
for(T object: objects)
{
object.clawBack();
}
}
}
public class CommissionTrns implements Clawbackable {
public void clawback(){
// do clawback for commissions
}
}
public class FinanceRequest implements Clawbackable {
public void clawBack(){
// do clwaback for FinanceRequest
}
}
I prefer this approach, since I'm of the belief your domain should contain your logic; but I'm not fully aware of your exact wishes, so I'll leave it up to you.
With a double dispatch, you would pass the "ClawbackHandler" to the clawback method, and on the handler call the appropriate method depending on the type.
I think your best option is to simply name the method differently.
public void clawBackFinReqs(Collection<FinanceRequest> financeRequestList) {
}
public void clawBackComTrans(Collection<CommissionTrns> commissionTrnsList) {
}
In fact, it's not too bad, since you don't get anything extra out of having the same name on them.
Keep in mind, that the JVM will not decide which method to call at runtime. As opposed to virtual methods / method overriding resolution of overloaded methods are done at compile time. The Java Tutorials on method overloading even points out that "Overloaded methods should be used sparingly...".
Here is a trick with overloading by the second varargs parameter for the CommissionFacade class from the question:
public class CommissionFacade {
public void clawBack(Collection<FinanceRequest> financeRequestList, FinanceRequestType ...ignore) {
// code
}
public void clawBack(Collection<CommissionTrns> commissionTrnsList, CommissionTrnsType ...ignore) {
// code
}
/*******TYPES TO TRICK TYPE ERASURE*******/
private static class FinanceRequestType {}
private static class CommissionTrnsType {}
}
The code snippet to fast-check this trick works:
import java.util.ArrayList;
class HelloType {
public static void main(String[] args) {
method(new ArrayList<Integer>());
method(new ArrayList<Double>());
}
static void method(ArrayList<Integer> ints, IntegerType ...ignore) {
System.out.println("Hello, Integer!");
}
static void method(ArrayList<Double> dbs, DoubleType ...ignore) {
System.out.println("Hello, Double!");
}
static class IntegerType {}
static class DoubleType {}
}
In Python, class methods can be inherited. e.g.
>>> class A:
... #classmethod
... def main(cls):
... return cls()
...
>>> class B(A): pass
...
>>> b=B.main()
>>> b
<__main__.B instance at 0x00A6FA58>
How would you do the equivalent in Java? I currently have:
public class A{
public void show(){
System.out.println("A");
}
public void run(){
show();
}
public static void main( String[] arg ) {
new A().run();
}
}
public class B extends A{
#Override
public void show(){
System.out.println("B");
}
}
I'd like to call B.main() and have it print "B", but clearly it will print "A" instead, since "new A()" is hardcoded.
How would you change "new A()" so that it's parameterized to use the class it's in when called, and not the hard-coded class A?
Static methods in java are not classmethods they are staticmethods. In general it is not possible to know which class reference the static method was called from.
Your class B does not have a main method and static methods are not inherited.
The only way I can see this happening is to find whatever is calling A.main( String[] arg ) and change it to call B.main instead.
B.main:
public static void main( String[] arg ) {
new B().run();
}
How is your program started? Is there a batch file, shortcut, etc? Something you can change? Where does A.main get called?
I think this isn't possible. Here's why:
In Java, the implementation of a method is determined by the instance's run-time type. So, to execute B.show(), you need to have an instance of B. The only way I could see to do this, if the method that constructs the instance is supposed to be inherited, is to use Class.newInstance() to construct an instance of a type that's not known at runtime.
The problem with that is that within a static method, you have no reference to the containing class, so you don't know whose newInstance method to call.
Why do you want to do this, though? There may be some better way to achieve whatever it is you want to achieve.
In your example I wouldn't put your main method inside of A. This is setup as the entry point into the system (you can't be in B if you are specifically entering into A).
In the example below I created class A, B, and C. Class C instantiates A and B and runs them. Notice that in C I created an A, a B, and another A that I instantiate as a B. My output is:
A
B
B
Hopefully this makes sense.
public class A {
public void show(){
System.out.println("A");
}
public void run(){
show();
}
}
public class B extends A {
#Override
public void show(){
System.out.println("B");
}
}
public class C {
public static void main(String[] args) {
A a = new A();
B b = new B();
A anothera = new B();
a.show();
b.show();
anothera.show();
}
}