I read this question and thought that would easily be solved (not that it isn't solvable without) if one could write:
#Override
public String toString() {
return super.super.toString();
}
I'm not sure if it is useful in many cases, but I wonder why it isn't and if something like this exists in other languages.
What do you guys think?
EDIT:
To clarify: yes I know, that's impossible in Java and I don't really miss it. This is nothing I expected to work and was surprised getting a compiler error. I just had the idea and like to discuss it.
It violates encapsulation. You shouldn't be able to bypass the parent class's behaviour. It makes sense to sometimes be able to bypass your own class's behaviour (particularly from within the same method) but not your parent's. For example, suppose we have a base "collection of items", a subclass representing "a collection of red items" and a subclass of that representing "a collection of big red items". It makes sense to have:
public class Items
{
public void add(Item item) { ... }
}
public class RedItems extends Items
{
#Override
public void add(Item item)
{
if (!item.isRed())
{
throw new NotRedItemException();
}
super.add(item);
}
}
public class BigRedItems extends RedItems
{
#Override
public void add(Item item)
{
if (!item.isBig())
{
throw new NotBigItemException();
}
super.add(item);
}
}
That's fine - RedItems can always be confident that the items it contains are all red. Now suppose we were able to call super.super.add():
public class NaughtyItems extends RedItems
{
#Override
public void add(Item item)
{
// I don't care if it's red or not. Take that, RedItems!
super.super.add(item);
}
}
Now we could add whatever we like, and the invariant in RedItems is broken.
Does that make sense?
I think Jon Skeet has the correct answer. I'd just like to add that you can access shadowed variables from superclasses of superclasses by casting this:
interface I { int x = 0; }
class T1 implements I { int x = 1; }
class T2 extends T1 { int x = 2; }
class T3 extends T2 {
int x = 3;
void test() {
System.out.println("x=\t\t" + x);
System.out.println("super.x=\t\t" + super.x);
System.out.println("((T2)this).x=\t" + ((T2)this).x);
System.out.println("((T1)this).x=\t" + ((T1)this).x);
System.out.println("((I)this).x=\t" + ((I)this).x);
}
}
class Test {
public static void main(String[] args) {
new T3().test();
}
}
which produces the output:
x= 3
super.x= 2
((T2)this).x= 2
((T1)this).x= 1
((I)this).x= 0
(example from the JLS)
However, this doesn't work for method calls because method calls are determined based on the runtime type of the object.
I think the following code allow to use super.super...super.method() in most case.
(even if it's uggly to do that)
In short
create temporary instance of ancestor type
copy values of fields from original object to temporary one
invoke target method on temporary object
copy modified values back to original object
Usage :
public class A {
public void doThat() { ... }
}
public class B extends A {
public void doThat() { /* don't call super.doThat() */ }
}
public class C extends B {
public void doThat() {
Magic.exec(A.class, this, "doThat");
}
}
public class Magic {
public static <Type, ChieldType extends Type> void exec(Class<Type> oneSuperType, ChieldType instance,
String methodOfParentToExec) {
try {
Type type = oneSuperType.newInstance();
shareVars(oneSuperType, instance, type);
oneSuperType.getMethod(methodOfParentToExec).invoke(type);
shareVars(oneSuperType, type, instance);
} catch (Exception e) {
throw new RuntimeException(e);
}
}
private static <Type, SourceType extends Type, TargetType extends Type> void shareVars(Class<Type> clazz,
SourceType source, TargetType target) throws IllegalArgumentException, IllegalAccessException {
Class<?> loop = clazz;
do {
for (Field f : loop.getDeclaredFields()) {
if (!f.isAccessible()) {
f.setAccessible(true);
}
f.set(target, f.get(source));
}
loop = loop.getSuperclass();
} while (loop != Object.class);
}
}
I don't have enough reputation to comment so I will add this to the other answers.
Jon Skeet answers excellently, with a beautiful example. Matt B has a point: not all superclasses have supers. Your code would break if you called a super of a super that had no super.
Object oriented programming (which Java is) is all about objects, not functions. If you want task oriented programming, choose C++ or something else. If your object doesn't fit in it's super class, then you need to add it to the "grandparent class", create a new class, or find another super it does fit into.
Personally, I have found this limitation to be one of Java's greatest strengths. Code is somewhat rigid compared to other languages I've used, but I always know what to expect. This helps with the "simple and familiar" goal of Java. In my mind, calling super.super is not simple or familiar. Perhaps the developers felt the same?
There's some good reasons to do this. You might have a subclass which has a method which is implemented incorrectly, but the parent method is implemented correctly. Because it belongs to a third party library, you might be unable/unwilling to change the source. In this case, you want to create a subclass but override one method to call the super.super method.
As shown by some other posters, it is possible to do this through reflection, but it should be possible to do something like
(SuperSuperClass this).theMethod();
I'm dealing with this problem right now - the quick fix is to copy and paste the superclass method into the subsubclass method :)
In addition to the very good points that others have made, I think there's another reason: what if the superclass does not have a superclass?
Since every class naturally extends (at least) Object, super.whatever() will always refer to a method in the superclass. But what if your class only extends Object - what would super.super refer to then? How should that behavior be handled - a compiler error, a NullPointer, etc?
I think the primary reason why this is not allowed is that it violates encapsulation, but this might be a small reason too.
I think if you overwrite a method and want to all the super-class version of it (like, say for equals), then you virtually always want to call the direct superclass version first, which one will call its superclass version in turn if it wants.
I think it only makes rarely sense (if at all. i can't think of a case where it does) to call some arbitrary superclass' version of a method. I don't know if that is possible at all in Java. It can be done in C++:
this->ReallyTheBase::foo();
At a guess, because it's not used that often. The only reason I could see using it is if your direct parent has overridden some functionality and you're trying to restore it back to the original.
Which seems to me to be against OO principles, since the class's direct parent should be more closely related to your class than the grandparent is.
Calling of super.super.method() make sense when you can't change code of base class. This often happens when you are extending an existing library.
Ask yourself first, why are you extending that class? If answer is "because I can't change it" then you can create exact package and class in your application, and rewrite naughty method or create delegate:
package com.company.application;
public class OneYouWantExtend extends OneThatContainsDesiredMethod {
// one way is to rewrite method() to call super.method() only or
// to doStuff() and then call super.method()
public void method() {
if (isDoStuff()) {
// do stuff
}
super.method();
}
protected abstract boolean isDoStuff();
// second way is to define methodDelegate() that will call hidden super.method()
public void methodDelegate() {
super.method();
}
...
}
public class OneThatContainsDesiredMethod {
public void method() {...}
...
}
For instance, you can create org.springframework.test.context.junit4.SpringJUnit4ClassRunner class in your application so this class should be loaded before the real one from jar. Then rewrite methods or constructors.
Attention: This is absolute hack, and it is highly NOT recommended to use but it's WORKING! Using of this approach is dangerous because of possible issues with class loaders. Also this may cause issues each time you will update library that contains overwritten class.
#Jon Skeet Nice explanation.
IMO if some one wants to call super.super method then one must be want to ignore the behavior of immediate parent, but want to access the grand parent behavior.
This can be achieved through instance Of. As below code
public class A {
protected void printClass() {
System.out.println("In A Class");
}
}
public class B extends A {
#Override
protected void printClass() {
if (!(this instanceof C)) {
System.out.println("In B Class");
}
super.printClass();
}
}
public class C extends B {
#Override
protected void printClass() {
System.out.println("In C Class");
super.printClass();
}
}
Here is driver class,
public class Driver {
public static void main(String[] args) {
C c = new C();
c.printClass();
}
}
Output of this will be
In C Class
In A Class
Class B printClass behavior will be ignored in this case.
I am not sure about is this a ideal or good practice to achieve super.super, but still it is working.
Look at this Github project, especially the objectHandle variable. This project shows how to actually and accurately call the grandparent method on a grandchild.
Just in case the link gets broken, here is the code:
import lombok.val;
import org.junit.Assert;
import org.junit.Test;
import java.lang.invoke.*;
/*
Your scientists were so preoccupied with whether or not they could, they didn’t stop to think if they should.
Please don't actually do this... :P
*/
public class ImplLookupTest {
private MethodHandles.Lookup getImplLookup() throws NoSuchFieldException, IllegalAccessException {
val field = MethodHandles.Lookup.class.getDeclaredField("IMPL_LOOKUP");
field.setAccessible(true);
return (MethodHandles.Lookup) field.get(null);
}
#Test
public void test() throws Throwable {
val lookup = getImplLookup();
val baseHandle = lookup.findSpecial(Base.class, "toString",
MethodType.methodType(String.class),
Sub.class);
val objectHandle = lookup.findSpecial(Object.class, "toString",
MethodType.methodType(String.class),
// Must use Base.class here for this reference to call Object's toString
Base.class);
val sub = new Sub();
Assert.assertEquals("Sub", sub.toString());
Assert.assertEquals("Base", baseHandle.invoke(sub));
Assert.assertEquals(toString(sub), objectHandle.invoke(sub));
}
private static String toString(Object o) {
return o.getClass().getName() + "#" + Integer.toHexString(o.hashCode());
}
public class Sub extends Base {
#Override
public String toString() {
return "Sub";
}
}
public class Base {
#Override
public String toString() {
return "Base";
}
}
}
Happy Coding!!!!
I would put the super.super method body in another method, if possible
class SuperSuperClass {
public String toString() {
return DescribeMe();
}
protected String DescribeMe() {
return "I am super super";
}
}
class SuperClass extends SuperSuperClass {
public String toString() {
return "I am super";
}
}
class ChildClass extends SuperClass {
public String toString() {
return DescribeMe();
}
}
Or if you cannot change the super-super class, you can try this:
class SuperSuperClass {
public String toString() {
return "I am super super";
}
}
class SuperClass extends SuperSuperClass {
public String toString() {
return DescribeMe(super.toString());
}
protected String DescribeMe(string fromSuper) {
return "I am super";
}
}
class ChildClass extends SuperClass {
protected String DescribeMe(string fromSuper) {
return fromSuper;
}
}
In both cases, the
new ChildClass().toString();
results to "I am super super"
It would seem to be possible to at least get the class of the superclass's superclass, though not necessarily the instance of it, using reflection; if this might be useful, please consider the Javadoc at http://java.sun.com/j2se/1.5.0/docs/api/java/lang/Class.html#getSuperclass()
public class A {
#Override
public String toString() {
return "A";
}
}
public class B extends A {
#Override
public String toString() {
return "B";
}
}
public class C extends B {
#Override
public String toString() {
return "C";
}
}
public class D extends C {
#Override
public String toString() {
String result = "";
try {
result = this.getClass().getSuperclass().getSuperclass().getSuperclass().newInstance().toString();
} catch (InstantiationException ex) {
Logger.getLogger(D.class.getName()).log(Level.SEVERE, null, ex);
} catch (IllegalAccessException ex) {
Logger.getLogger(D.class.getName()).log(Level.SEVERE, null, ex);
}
return result;
}
}
public class Main {
public static void main(String... args) {
D d = new D();
System.out.println(d);
}
}
run:
A
BUILD SUCCESSFUL (total time: 0 seconds)
I have had situations like these when the architecture is to build common functionality in a common CustomBaseClass which implements on behalf of several derived classes.
However, we need to circumvent common logic for specific method for a specific derived class. In such cases, we must use a super.super.methodX implementation.
We achieve this by introducing a boolean member in the CustomBaseClass, which can be used to selectively defer custom implementation and yield to default framework implementation where desirable.
...
FrameworkBaseClass (....) extends...
{
methodA(...){...}
methodB(...){...}
...
methodX(...)
...
methodN(...){...}
}
/* CustomBaseClass overrides default framework functionality for benefit of several derived classes.*/
CustomBaseClass(...) extends FrameworkBaseClass
{
private boolean skipMethodX=false;
/* implement accessors isSkipMethodX() and setSkipMethodX(boolean)*/
methodA(...){...}
methodB(...){...}
...
methodN(...){...}
methodX(...){
if (isSkipMethodX()) {
setSKipMethodX(false);
super.methodX(...);
return;
}
... //common method logic
}
}
DerivedClass1(...) extends CustomBaseClass
DerivedClass2(...) extends CustomBaseClass
...
DerivedClassN(...) extends CustomBaseClass...
DerivedClassX(...) extends CustomBaseClass...
{
methodX(...){
super.setSKipMethodX(true);
super.methodX(...);
}
}
However, with good architecture principles followed in framework as well as app, we could avoid such situations easily, by using hasA approach, instead of isA approach. But at all times it is not very practical to expect well designed architecture in place, and hence the need to get away from solid design principles and introduce hacks like this.
Just my 2 cents...
IMO, it's a clean way to achieve super.super.sayYourName() behavior in Java.
public class GrandMa {
public void sayYourName(){
System.out.println("Grandma Fedora");
}
}
public class Mama extends GrandMa {
public void sayYourName(boolean lie){
if(lie){
super.sayYourName();
}else {
System.out.println("Mama Stephanida");
}
}
}
public class Daughter extends Mama {
public void sayYourName(boolean lie){
if(lie){
super.sayYourName(lie);
}else {
System.out.println("Little girl Masha");
}
}
}
public class TestDaughter {
public static void main(String[] args){
Daughter d = new Daughter();
System.out.print("Request to lie: d.sayYourName(true) returns ");
d.sayYourName(true);
System.out.print("Request not to lie: d.sayYourName(false) returns ");
d.sayYourName(false);
}
}
Output:
Request to lie: d.sayYourName(true) returns Grandma Fedora
Request not to lie: d.sayYourName(false) returns Little girl Masha
I think this is a problem that breaks the inheritance agreement.
By extending a class you obey / agree its behavior, features
Whilst when calling super.super.method(), you want to break your own obedience agreement.
You just cannot cherry pick from the super class.
However, there may happen situations when you feel the need to call super.super.method() - usually a bad design sign, in your code or in the code you inherit !
If the super and super super classes cannot be refactored (some legacy code), then opt for composition over inheritance.
Encapsulation breaking is when you #Override some methods by breaking the encapsulated code.
The methods designed not to be overridden are marked
final.
In C# you can call a method of any ancestor like this:
public class A
internal virtual void foo()
...
public class B : A
public new void foo()
...
public class C : B
public new void foo() {
(this as A).foo();
}
Also you can do this in Delphi:
type
A=class
procedure foo;
...
B=class(A)
procedure foo; override;
...
C=class(B)
procedure foo; override;
...
A(objC).foo();
But in Java you can do such focus only by some gear. One possible way is:
class A {
int y=10;
void foo(Class X) throws Exception {
if(X!=A.class)
throw new Exception("Incorrect parameter of "+this.getClass().getName()+".foo("+X.getName()+")");
y++;
System.out.printf("A.foo(%s): y=%d\n",X.getName(),y);
}
void foo() throws Exception {
System.out.printf("A.foo()\n");
this.foo(this.getClass());
}
}
class B extends A {
int y=20;
#Override
void foo(Class X) throws Exception {
if(X==B.class) {
y++;
System.out.printf("B.foo(%s): y=%d\n",X.getName(),y);
} else {
System.out.printf("B.foo(%s) calls B.super.foo(%s)\n",X.getName(),X.getName());
super.foo(X);
}
}
}
class C extends B {
int y=30;
#Override
void foo(Class X) throws Exception {
if(X==C.class) {
y++;
System.out.printf("C.foo(%s): y=%d\n",X.getName(),y);
} else {
System.out.printf("C.foo(%s) calls C.super.foo(%s)\n",X.getName(),X.getName());
super.foo(X);
}
}
void DoIt() {
try {
System.out.printf("DoIt: foo():\n");
foo();
Show();
System.out.printf("DoIt: foo(B):\n");
foo(B.class);
Show();
System.out.printf("DoIt: foo(A):\n");
foo(A.class);
Show();
} catch(Exception e) {
//...
}
}
void Show() {
System.out.printf("Show: A.y=%d, B.y=%d, C.y=%d\n\n", ((A)this).y, ((B)this).y, ((C)this).y);
}
}
objC.DoIt() result output:
DoIt: foo():
A.foo()
C.foo(C): y=31
Show: A.y=10, B.y=20, C.y=31
DoIt: foo(B):
C.foo(B) calls C.super.foo(B)
B.foo(B): y=21
Show: A.y=10, B.y=21, C.y=31
DoIt: foo(A):
C.foo(A) calls C.super.foo(A)
B.foo(A) calls B.super.foo(A)
A.foo(A): y=11
Show: A.y=11, B.y=21, C.y=31
It is simply easy to do. For instance:
C subclass of B and B subclass of A. Both of three have method methodName() for example.
public abstract class A {
public void methodName() {
System.out.println("Class A");
}
}
public class B extends A {
public void methodName() {
super.methodName();
System.out.println("Class B");
}
// Will call the super methodName
public void hackSuper() {
super.methodName();
}
}
public class C extends B {
public static void main(String[] args) {
A a = new C();
a.methodName();
}
#Override
public void methodName() {
/*super.methodName();*/
hackSuper();
System.out.println("Class C");
}
}
Run class C Output will be:
Class A
Class C
Instead of output:
Class A
Class B
Class C
If you think you are going to be needing the superclass, you could reference it in a variable for that class. For example:
public class Foo
{
public int getNumber()
{
return 0;
}
}
public class SuperFoo extends Foo
{
public static Foo superClass = new Foo();
public int getNumber()
{
return 1;
}
}
public class UltraFoo extends Foo
{
public static void main(String[] args)
{
System.out.println(new UltraFoo.getNumber());
System.out.println(new SuperFoo().getNumber());
System.out.println(new SuperFoo().superClass.getNumber());
}
public int getNumber()
{
return 2;
}
}
Should print out:
2
1
0
public class SubSubClass extends SubClass {
#Override
public void print() {
super.superPrint();
}
public static void main(String[] args) {
new SubSubClass().print();
}
}
class SuperClass {
public void print() {
System.out.println("Printed in the GrandDad");
}
}
class SubClass extends SuperClass {
public void superPrint() {
super.print();
}
}
Output: Printed in the GrandDad
The keyword super is just a way to invoke the method in the superclass.
In the Java tutorial:https://docs.oracle.com/javase/tutorial/java/IandI/super.html
If your method overrides one of its superclass's methods, you can invoke the overridden method through the use of the keyword super.
Don't believe that it's a reference of the super object!!! No, it's just a keyword to invoke methods in the superclass.
Here is an example:
class Animal {
public void doSth() {
System.out.println(this); // It's a Cat! Not an animal!
System.out.println("Animal do sth.");
}
}
class Cat extends Animal {
public void doSth() {
System.out.println(this);
System.out.println("Cat do sth.");
super.doSth();
}
}
When you call cat.doSth(), the method doSth() in class Animal will print this and it is a cat.
I have followed this link and successfully made singleton class in Android.
http://www.devahead.com/blog/2011/06/extending-the-android-application-class-and-dealing-with-singleton/
Problem is that i want a single object. like i have Activity A and Activity B. In Activity A I access the object from Singleton class. I use the object and made some changes to it.
When I move to Activity B and access the object from Singleton Class it gave me the initialized object and does not keep the changes which i have made in Activity A.
Is there any other way to save the changing?
Please help me Experts.
This is MainActivity
public class MainActivity extends Activity {
protected MyApplication app;
private OnClickListener btn2=new OnClickListener() {
#Override
public void onClick(View arg0) {
Intent intent=new Intent(MainActivity.this,NextActivity.class);
startActivity(intent);
}
};
#Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
//Get the application instance
app = (MyApplication)getApplication();
// Call a custom application method
app.customAppMethod();
// Call a custom method in MySingleton
Singleton.getInstance().customSingletonMethod();
Singleton.getInstance();
// Read the value of a variable in MySingleton
String singletonVar = Singleton.customVar;
Log.d("Test",singletonVar);
singletonVar="World";
Log.d("Test",singletonVar);
Button btn=(Button)findViewById(R.id.button1);
btn.setOnClickListener(btn2);
}
}
This is NextActivity
public class NextActivity extends Activity {
#Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_next);
String singletonVar = Singleton.customVar;
Log.d("Test",singletonVar);
}
}
Singleton Class
public class Singleton
{
private static Singleton instance;
public static String customVar="Hello";
public static void initInstance()
{
if (instance == null)
{
// Create the instance
instance = new Singleton();
}
}
public static Singleton getInstance()
{
// Return the instance
return instance;
}
private Singleton()
{
// Constructor hidden because this is a singleton
}
public void customSingletonMethod()
{
// Custom method
}
}
and MyApplication
public class MyApplication extends Application
{
#Override
public void onCreate()
{
super.onCreate();
// Initialize the singletons so their instances
// are bound to the application process.
initSingletons();
}
protected void initSingletons()
{
// Initialize the instance of MySingleton
Singleton.initInstance();
}
public void customAppMethod()
{
// Custom application method
}
}
When i run this code, i get Hello which i have initialized in Singleton then World which i gave it in MainActivity and again shows Hello in NextActivity in logcat.
I want it to show world again in NextActivity.
Please help me to correct this.
Tip: To create singleton class In Android Studio, right click in your project and open menu:
New -> Java Class -> Choose Singleton from dropdown menu
EDIT :
The implementation of a Singleton in Android is not "safe" (see here) and you should use a library dedicated to this kind of pattern like Dagger or other DI library to manage the lifecycle and the injection.
Could you post an example from your code ?
Take a look at this gist : https://gist.github.com/Akayh/5566992
it works but it was done very quickly :
MyActivity : set the singleton for the first time + initialize mString attribute ("Hello") in private constructor and show the value ("Hello")
Set new value to mString : "Singleton"
Launch activityB and show the mString value. "Singleton" appears...
It is simple, as a java, Android also supporting singleton. -
Singleton is a part of Gang of Four design pattern and it is categorized under creational design patterns.
-> Static member : This contains the instance of the singleton class.
-> Private constructor : This will prevent anybody else to instantiate the Singleton class.
-> Static public method : This provides the global point of access to the Singleton object and returns the instance to the client calling class.
create private instance
create private constructor
use getInstance() of Singleton class
public class Logger{
private static Logger objLogger;
private Logger(){
//ToDo here
}
public static Logger getInstance()
{
if (objLogger == null)
{
objLogger = new Logger();
}
return objLogger;
}
}
while use singleton -
Logger.getInstance();
answer suggested by rakesh is great but still with some discription
Singleton in Android is the same as Singleton in Java:
The Singleton design pattern addresses all of these concerns. With the Singleton design pattern you can:
1) Ensure that only one instance of a class is created
2) Provide a global point of access to the object
3) Allow multiple instances in the future without affecting a
singleton class's clients
A basic Singleton class example:
public class MySingleton
{
private static MySingleton _instance;
private MySingleton()
{
}
public static MySingleton getInstance()
{
if (_instance == null)
{
_instance = new MySingleton();
}
return _instance;
}
}
As #Lazy stated in this answer, you can create a singleton from a template in Android Studio. It is worth noting that there is no need to check if the instance is null because the static ourInstance variable is initialized first. As a result, the singleton class implementation created by Android Studio is as simple as following code:
public class MySingleton {
private static MySingleton ourInstance = new MySingleton();
public static MySingleton getInstance() {
return ourInstance;
}
private MySingleton() {
}
}
You are copying singleton's customVar into a singletonVar variable and changing that variable does not affect the original value in singleton.
// This does not update singleton variable
// It just assigns value of your local variable
Log.d("Test",singletonVar);
singletonVar="World";
Log.d("Test",singletonVar);
// This actually assigns value of variable in singleton
Singleton.customVar = singletonVar;
I put my version of Singleton below:
public class SingletonDemo {
private static SingletonDemo instance = null;
private static Context context;
/**
* To initialize the class. It must be called before call the method getInstance()
* #param ctx The Context used
*/
public static void initialize(Context ctx) {
context = ctx;
}
/**
* Check if the class has been initialized
* #return true if the class has been initialized
* false Otherwise
*/
public static boolean hasBeenInitialized() {
return context != null;
}
/**
* The private constructor. Here you can use the context to initialize your variables.
*/
private SingletonDemo() {
// Use context to initialize the variables.
}
/**
* The main method used to get the instance
*/
public static synchronized SingletonDemo getInstance() {
if (context == null) {
throw new IllegalArgumentException("Impossible to get the instance. This class must be initialized before");
}
if (instance == null) {
instance = new SingletonDemo();
}
return instance;
}
#Override
protected Object clone() throws CloneNotSupportedException {
throw new CloneNotSupportedException("Clone is not allowed.");
}
}
Note that the method initialize could be called in the main class(Splash) and the method getInstance could be called from other classes. This will fix the problem when the caller class requires the singleton but it does not have the context.
Finally the method hasBeenInitialized is uses to check if the class has been initialized. This will avoid that different instances have different contexts.
The most clean and modern way to use singletons in Android is just to use the Dependency Injection framework called Dagger 2. Here you have an explanation of possible scopes you can use. Singleton is one of these scopes. Dependency Injection is not that easy but you shall invest a bit of your time to understand it. It also makes testing easier.
Is it possible that the base class member function can access child class member function directly?
I found the code from Androind, the BufferQueue inherits BnSurfaceTexture, and has one member function "requestBuffer".
In the base class BnSurfaceTexture, I found it just call requestBuffer directly.
How does the base class BnSurfaceTexture know the function "requestBuffer"?
Thanks
The base class member function:
status_t BnSurfaceTexture::onTransact(
uint32_t code, const Parcel& data, Parcel* reply, uint32_t flags)
{
switch(code) {
case REQUEST_BUFFER: {
CHECK_INTERFACE(ISurfaceTexture, data, reply);
int bufferIdx = data.readInt32();
sp<GraphicBuffer> buffer;
/* it call requestBuffer directly */ <--------
int result = requestBuffer(bufferIdx, &buffer);
reply->writeInt32(buffer != 0);
The child class declaration & implementation:
class BufferQueue : public BnSurfaceTexture {
private:
virtual status_t requestBuffer(int slot, sp<GraphicBuffer>* buf);
status_t BufferQueue::requestBuffer(int slot, sp<GraphicBuffer>* buf) {
ATRACE_CALL();
ST_LOGV("requestBuffer: slot=%d", slot);
Mutex::Autolock lock(mMutex);
...
return NO_ERROR;
}
How does the base class BnSurfaceTexture know the function "requestBuffer"?
This function must have been at least declared (it could have been defined with some default implementation) in the base class and it must be virtual.
So, at compile time, the compiler sees that such function exists.
The function call is resolved run-time. This is polymorphism.
Tiny example:
class Base
{
public:
virtual void f() { /* Base */ } // could be pure virtual
virtual void g() { f(); };
};
class Derived: public Base
{
public:
virtual void f() { /* derived */ }
};
When you have
Base* pB = new Derived;
pB->g();
g() will call Derived::f;
Just make sure the base class function is declared, and the derived class is define before definition of base class function definition. Here is an example
#include <iostream>
class Base
{
public :
void call_derived ();
};
class Derived : public Base
{
public :
void derived_fun ()
{
std::cout << "Derived class member function called!" << std::endl;
}
};
void Base::call_derived ()
{
static_cast<Derived *>(this)->derived_fun();
}
int main ()
{
Derived dobj;
dobj.call_derived();
return 0;
}
However, this use of static_cast is unsafe, compiler won't complain if you try call_derived on an incomplete object. However, you can add an assertion assert(dynamic_cast<Derived *>(this), at least in debug mode, for debugging purpose. Alternatively, you can declare Base class' constructor, destructor protected, to prevent creation of incomplete object, or attempting to destroy a derived object while the base is not polymorphic
However, the above code are not really common in real world. Other more advanced techniques exists, like CRTP, which also use static_cast and provide static polymorphism without virtual functions.
This answer may not be exactly what you want. It only shows that it is possible to call derived class members in base class even it is not defined in base at all. But to call it directly without any cast, you still need virtual functions.
I have implemented a simple ostream and streambuf class. For some reason, it crashes when I try to instantiate my AndroidLogOStream object.
Note: I have stlport_static in my Application.mk
class AndroidLogStreamBuf : public std::streambuf
{
public:
inline AndroidLogStreamBuf() : std::streambuf()
{
//std::cout << "asdfg";
}
inline ~AndroidLogStreamBuf()
{
}
};
class AndroidLogOStream : public std::ostream
{
public:
inline AndroidLogOStream() : std::ostream(&mBuf)
{
}
inline ~AndroidLogOStream()
{
}
private:
AndroidLogStreamBuf mBuf;
};
It's barebones, and it runs fine on windows. It compiles fine on android, but it crashes for some reason. The last line it tries to execute is in _streambuf.c:46:
template <class _CharT, class _Traits>
locale
basic_streambuf<_CharT, _Traits>::pubimbue(const locale& __loc) {
this->imbue(__loc); <---- crash
locale __tmp = _M_locale;
_M_locale = __loc;
return __tmp;
}
Granted I am still quite confused on iostreams, but it must be something wrong with the constructor, I suppose it is not valid?
In a constructor, the base class is initialized first, followed by all of the members. When you call the base class constructor std::ostream, you're passing it the address of mBuf, which has not yet been constructed. Accessing an object that hasn't yet been constructed has undefined behaviour.
To get around this, you could redesign your classes as follows:
class AndroidLogStreamBuf : public std::streambuf
{
public:
AndroidLogStreamBuf() : std::streambuf()
{ }
~AndroidLogStreamBuf()
{ }
};
class AndroidLogOStream : public std::ostream
{
public:
AndroidLogOStream(AndroidLogStreamBuf *buf) :
std::ostream(buf),
mBuf(buf)
{ }
~AndroidLogOStream()
{ }
private:
AndroidLogStreamBuf *mBuf;
};
class AndroidLogOStreamWithBuf
{
private:
AndroidLogStreamBuf mBuf;
AndroidLogOStream mStream;
public:
AndroidLogOStreamWithBuf() :
mBuf(&mStream),
mStream()
{ }
virtual ~AndroidLogOStreamWithBuf()
{ }
AndroidLogOStream& getOStream()
{
return mStream;
}
};
Notice the order I've declared mBuf and mStream in AndroidLogOStreamWithBuf: the two fields will be initialized in that order, regardless of the order they appear in the constructor initializer list. As an aside, marking the member functions as inline in your original code was superfluous: when you define a member function within the class definition, it's automatically marked as inlinable.
Whether this is a sensible design for your system depends on how you're intending to use these classes, but the answer is probably "no".
As was pointed out, the base class is constructed first and from the looks of it, the base class constructor seems to do something. I don't think it is meant to but the base class destructor would also create a problem and that will call pubsync() on the stream buffer.
Of course, this explains the problem but doesn't provide a solution: the solution to this initialization problem is to make the the stream buffer (or a custom class containing the stream buffer as member) a virtual base class:
class oandroidligstream:
virtual AndroidLogStream,
public std::ostringstream {
...
}
};
the reason the base has to be virtual is that the stream buffer is an argument to the virtual base std::ios. To make sure your stream buffer is initialized first it has to be the left-most virtual base.