What is the difference between override and overload?
Overloading: picking a method signature at compile time based on the number and type of the arguments specified
Overriding: picking a method implementation at execution time based on the actual type of the target object (as opposed to the compile-time type of the expression)
For example:
class Base
{
void foo(int x)
{
System.out.println("Base.foo(int)");
}
void foo(double d)
{
System.out.println("Base.foo(double)");
}
}
class Child extends Base
{
#Override void foo (int x)
{
System.out.println("Child.foo(int)");
}
}
...
Base b = new Child();
b.foo(10); // Prints Child.foo(int)
b.foo(5.0); // Prints Base.foo(double)
Both calls are examples of overloading. There are two methods called foo, and the compiler determines which signature to call.
The first call is an example of overriding. The compiler picks the signature "foo(int)" but then at execution time, the type of the target object determines that the implementation to use should be the one in Child.
Overloading of methods is a compiler trick to allow you to use the same name to perform different actions depending on parameters.
Overriding a method means that its entire functionality is being replaced. Overriding is something done in a child class to a method defined in a parent class.
src: http://www.jchq.net/tutorial/06_02Tut.htm
Overloading :
public Bar foo(int some);
public Bar foo(int some, boolean x); // Same method name, different signature.
Overriding :
public Bar foo(int some); // Defined in some class A
public Bar foo(int some); // Same method name and signature. Defined in subclass of A.
If the second method was not defined it would have inherited the first method. Now it will be replaced by the second method in the subclass of A.
Overload - similar signature - same name, different parameters
void foo() {
/** overload */
}
void foo( int a ) {
/** overload */
}
int foo() {
/** this is NOT overloading, signature is for compiler SAME like void foo() */
}
Override - you can redefine method body when you inherit it.
class A {
void foo() {
/** definition A */
}
}
class B extends A {
void foo() {
/** definition B, this definition will be used when you have instance of B */
}
}
On interesting thing to mention:
public static doSomething(Collection<?> c) {
// do something
}
public static doSomething(ArrayList<?> l) {
// do something
}
public static void main(String[] args) {
Collection<String> c = new ArrayList<String> ();
doSomething(c); // which method get's called?
}
One would suppose the method with the ArrayList argument would be called but it doesn't. The first method is called since the proper method is selected at compile time.
Override
Is when a method which is inherited by a subclass from a superclass is replaced (overridden) in the subclass.
class A {
void foo() {
/** definition A of foo */
}
}
class B extends A {
void foo() {
/** definition B of foo */
}
}
Now if you call foo using:
A a = new B();
a.foo();
The B definition of foo will be run. This is not so intuitive since you will get a compile error if the class A didn't have a method called foo. So the type of the object a which is A has to have the method foo, then you can call it, and the method foo of the instance will be executed, which is that of class B, hence 'execution time'.
Overload
When you create a method with the same name as an existing method. To avoid a compile time error, you have to define the new method with different parameters than the existing one. This way the methods will be distinguishable. Have a method with the same name and parameters, but a different return type is still vague and will therefore cause a compile error. Example of overloading:
class A {
void bar(int i) {}
// The following method is overloading the method bar
void bar(Object a) {}
// The following will cause a compile error.
// Parameters should differ for valid overload
boolean bar(int i) {
return true;
}
}
Related
public class DemoParent {
public void m1(int i) {
System.out.println("parent");
}
}
public class DemoChild extends DemoParent{
public void m1(int... i) {
System.out.println("child");
}
public static void main(String[] args) {
DemoParent p = new DemoChild();
p.m1(10);
DemoChild c = new DemoChild();
c.m1(10);
}
}
Here I have two class DemoParent and DemoChild which extends DemoParent, in DemoParent i have general method m1(int i) but in DemoChild i have varargs method m1(int... i). When i am creating object of child class and calling m1 method,it's giving me parent class method output. See below the output.
o/p- parent
parent
Can anyone explain me why parent class method is always calling from the Child class reference even if child has the same method?
but if we reverse the code i.e. parent class have varargs method and child has normal method then it will acts as an overloading instead of Overriding . So the output is
o/p - parent
child
but in the above 1st scenario i didn't understand why always parent class method is calling from child reference.
You're not overriding. You're overloading.
DemoChild has two methods called m1, and it has to choose which of the methods to invoke, based on the fact that you pass a single int as a parameter.
The non-varargs method will always be matched before the varargs overload, because the language spec rules around method invocation say that it will be (to preserve backwards compatibility with pre-varargs code).
Is there a way to get the methods of an Class object, without getting the methods defined by the Object class?
Right now I am using getDeclaredMethods() to look for a specific method with a list of parameters.
My problem is that this also returns functions like
"equals", "hashCode", etc...
and thus it could be ambiguous between these functions and the one I'm looking for.
Looking at the documentation, it says that it only returns the public methods defined by this class (or in my case an interface), and my objects never override these methods
Is there any workaround for this?
Example:
class Test implements ITest {
void myMethod() {...}
}
and in my code I have something like
Object o = new Test();
for (Method m : o.getClass().getDeclaredMethods()) {
System.out.println(m.getName()...);
}
and this prints me methods defined in the Object class
getMethods()returns all methods declared by a class, plus methods declared in the object class. You can use getDeclaredMethods() to access methods that are class specific or implemented from super class. Example:
interface IChild {
int child1(int a, int b);
}
static class ChildImpl implements IChild {
#Override
public int child1(int a, int b) {
return a*b;
}
public int child2(int x, int y){
return x+x*y;
}
}
public static void main(String[] args) {
IChild i=new ChildImpl();
System.out.println("===Non Object Methods===");
for(Method m:i.getClass().getDeclaredMethods()){
System.out.println("Declared Method"+m.getName());
}
System.out.println("===All Methods===");
for(Method m:i.getClass().getMethods()){
System.out.println("Method: "+m.getName());
}
}
}
Sample output:
===Non Object Methods===
Declared Method: child2
Declared Method: child1
===All Methods===
Method: child2
Method: child1
Method: wait
Method: wait
Method: wait
Method: equals
Method: toString
Method: hashCode
Method: getClass
Method: notify
Method: notifyAll
public class A
{
public void display(int i)
{
System.out.println("Inside A");
}
}
public class B extends A
{
public void display(Integer i)
{
System.out.println("Inside B");
}
}
public class starter
{
public static void main (String args[])
{
A a = new B();
a.display(5);
System.out.println("So now you know or not");
}
}
Output : Inside A
Can somebody explain this output? Normally child method should be called. How does Java behave here when we have a wrapper class and a primitive class using inheritance?
B#display does not override A#display, as the signature is different.
The fact ints can be boxed into Integers is not relevant here.
You could easily verify this by using the #Override annotation.
Since the reference type of a is A, the method is resolved with the exact match for a literal integer (your given 5 argument), which is int, therefore A#display is invoked.
You can still force the invocation of B#display by using this idiom (not for production code):
((B)a).display(new Integer(5));
This casts your a variable as a B type, hence allowing visibility of B's display method within context.
It also passes an Integer rather than an int, thus employing the signature of B's display method and allowing resolution to that method.
We know that method signatures include only method name and parameter lists but not method return types. So why am I getting compiler error for the following code since java does not differentiate between methods with same signature.
public class InterfaceTest implements I2{
public void hello(){ }
public void world(){ }
}
interface I1{
public int hello();
}
interface I2 extends I1{
public void world();
}
You have not overloading here, you are overriding and also hidding methods but not in a correct way.... there are 2 possibilities to solve your problem:
public class InterfaceTest implements I2{
public void hello(int a){ } // overloaded method
#Override
public int hello(){ return 1; } // overriden method
public void world(){ } // this hides I1 method
}
Point is if you try this in a single class:
public void hello() {}
public int hello() {return 1;}
You will get Duplicate method hello() in type YourClass error, because to overloading a method you must change the FormalParameterListopt of the signature:
If two methods of a class [...] have the same name but signatures that are not override-equivalent, then the method name is said to be overloaded.
Last but not Least point:
method signatures include only method name and parameter lists but not method return types
According JSL §8.4, when you declare a method:
MethodDeclaration:
MethodHeader MethodBody
MethodHeader:
MethodModifiersopt TypeParametersopt Result MethodDeclarator Throwsopt
MethodDeclarator:
Identifier ( FormalParameterListopt )
So when you do this:
public int hellow(int number) throws Exception;
//| | | | └ throwing an exception (Throwsopt)
//| | | └──────────── receiving one int argument (MethodDeclarator FormalParameterListopt )
//| | └─────────────────── name hellow (MethodDeclarator Identifier)
//| └─────────────────────── returning an int (Result)
//└────────────────────────────── is a public method (MethodModifiersopt)
You are referring method overriding here not overloading. The return type must be the same as, or a subtype, of the return type declared in the original overridden method in the superclass/interface.
According to this article the return-type is part of the declaration. And the declaration has to be the same in the implementing class. Also an extending interface inherits all the methods from the parent-interface so your hellow-method must match the hellow-method from i1
The problem is exactly because java cannot differentiate between methods with same signature (but different return types).
Let's see what happens in your case (assuming it was possible),
You can do the following,
InterfaceTest obj = new InterfaceTest();
obj.hellow(); // The compiler knows it returns void here.
i1 obj1 = new InterfaceTest();
obj1.hellow(); // The compiler thinks it return an int, when it doesn't return anything.
The return type of a method is part of its signature, your compile error is because your return type doesn't match the one declared in your intereface.
You are implementing an interface in a class, which have the same name of method as the interface "hello()", so there are two case
1) If you are implementing a interface you should write concrete implementation of its methods in this case
public int hello();
public void world();
but you provided implementation of only one method
"public void world()"
so it will ask you to implement the other method which which will become part of this class.
2)When you provide the concrete implementation of the public void world() method it will collide with your method public void hello(), now as the overridden method become a member of your class and your method have same parameter list so it will fails the condition of the overloading.
so this will fail both the overriding and overloading functionality.
I've heard that all Java functions are implicitly virtual, but I'm still not sure if this will run how I want.
Suppose I have a class A, with child B.
both A and B have functions called foo(), so B's definition is overriding A's.
Suppose also that A has a function called that takes an instance of A as a parameter:
If I pass in an instance of B to the function, which definition of foo() will it call, A's or B's?
As I mentioned in my comment private functions are not virtual and I want to demonstrate it using following example:
class A {
public void foo() {
System.out.println("A#foo()");
}
public void bar() {
System.out.println("A#bar()");
qux();
}
private void qux() {
System.out.println("A#qux()");
}
}
class B extends A {
public void foo() {
System.out.println("B#foo()");
}
private void qux() {
System.out.println("B#qux()");
}
}
Now lets run following code:
A foobar = new B();
foobar.foo(); // outputs B#foo() because foobar is instance of B
foobar.bar(); // outputs A#bar() and A#qux() because B does not have method bar
// and qux is not virtual
B's implementation will be called.
That's exactly what virtual means.