Can anyone tell me whether the return type of the method will also be considered when you are overriding a method? what i mean exactly is whether the return type can be different when you are overriding a method.and somewhere i read like overloading will not come under polymorphism. is this true?
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Have you read anything about generics yet?Originally posted by ramadeviirrigir eddy -
Just try something like this:Originally posted by ramadeviirrigir eddy
[CODE=java]
public class Overriding {
public void myFunction()
{
System.out.prin tln("I'm in the void.");
}
public int myFunction()
{
System.out.prin tln("I'm in the int.");
return 0;
}
}
[/CODE]It will tell you
(or similar). So the answer is: No, you can't have several functions like that. To override a function, it MUST have different parameters. Then the type of the result doesn't matter anyway. SoCode:Duplicate method myFunction() in type Overriding
[CODE=java]
public class Overriding {
public static void main(String[] args)
{
myFunction();
int i = (Integer) myFunction(3);
}
// This could be a void without the return line
public static Object myFunction()
{
System.out.prin tln("I'm in the void.");
return new Object();
}
// This could be "public static int myFunction(int i)" too
public static Object myFunction(int i)
{
System.out.prin tln("I'm in the int.");
return i;
}
}
[/CODE]works fine, while it won't without the argument in the second function. What you can do is have a function, which takes and returns an int, one that takes and returns a double, etc. and call them the same. As long as the arguments are different, it'll work.
Example:
[CODE=java]
public int myFunction(int i)
{
System.out.prin tln("I'm in the int.");
return i;
}
public double myFunction(doub le d)
{
System.out.prin tln("I'm in the double.");
return d;
}
[/CODE]
Greetings,
Nepomuk
*Edit*
Of course, when using Generics, you can do some wonderful things. Read about them. However, that's not exactly what you're looking for, if I understood correctly. The compiler or jvm doesn't find out the type itself.Comment
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You cannot talk about overriding when you have only one class.
You need a method in a class with the same signature as a method in a super class of that classComment
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Hi,
The code you mentioned above is an example of overloading right. it is not overriding.i mean to say this code.
[CODE=java]
public int myFunction(int i)
{
System.out.prin tln("I'm in the int.");
return i;
}
public double myFunction(doub le d)
{
System.out.prin tln("I'm in the double.");
return d;
}
[/CODE]
but overriding means you should have same method signature like the following code
[CODE=java]
class One
{
int a=10,b=20;
public int add()
{
System.out.prin tln("a+b is:" + (a+b));
return a+b;
}
}
class Two extends One
{
int a=20,b=20;
public int add()
{
System.out.prin tln("a+b is:" + (a+b));
return a+b;
}
}
public class Three {
/**
* @param args
*/
public static void main(String[] args) {
Two x = new Two();
x.add();
One y = new One();
y.add();
}
}
[/CODE]
If you mention the return type as different in the above method then it'll give compilation error saying that "add() in Two cannot override add() in One; attempting to use incompatible return type" if u mention the return type in Two as double.Comment
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Hi,Originally posted by nepomuk
I think you are talking about overloading here not overriding.Comment
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Yes i too think that you are talking about overloading.
if you see the above code the class Two is returning double where as One is returning integer.Code:class One { int a=10,b=20; public int add() { System.out.println("a+b is:" + (a+b)); return a+b; } } class Two extends One { int a=20,b=20; public double add() { System.out.println("a+b is:" + (a+b)); return a+b; } } public class Three { /** * @param args */ public static void main(String[] args) { Two x = new Two(); x.add(); One y = new One(); y.add(); } }
if you compile this program it'll give error saying that "add() in Two cannot override add() in One; attempting to use incompatible return type"
by this example can i say that "overriding method should have same return type also including the method signature."???Comment
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Overloaded methods need to have different parameter lists (types and or number)
while for overridden methods the parameter lists need to be assignable and the
return type of the overridden method must be a supertype or identical to the
return type of the overriding method; here's a small example:
[code=java]
import java.util.Colle ction;
import java.util.List;
class Base {
public Collection func(int i) { return null; }
public Object func(String s) { return null; }
}
public class Derived extends Base {
public List func(int i) { return null; }
public String func(String s) { return null; }
}[/code]
kind regards,
JosComment
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ok thanks a lot. now i understood the difference clearly.Originally posted by JosAHComment
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[CODE=java]class Foo {
public Number print() {
System.out.prin tln("Foo");
return 0;
}
}
class Bar extends Foo {
public Integer print() {
System.out.prin t("Bar");
return 0;
}
public static void main(String[] args) {
Foo foo = new Foo();
Bar bar = new Bar();
foo.print();
bar.print();
}
}[/CODE]
Y A T F E
Edit: and Jos beat me to the punch againComment
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And I didn't even realize at the time, that I was making the Overloading/Overriding mistake. Shame on me. ^^Originally posted by ramadeviirrigir eddy
So, thanks to ramadeviirrigir eddy, madhoriya22 and JosAH for pointing out my mistake. ^^
Greetings,
NepomukComment
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Haha!Originally posted by r035198x
kind regards,
Jos ;-)Comment
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Hi,Originally posted by r035198x
I run ur example and got these two compile time errors ..
Are you trying to prove some point here or something else .. :)Code:Foo.java:8: The method java.lang.Integer print() declared in class Bar cannot override the method of the same signature declared in class Foo. They must have the same return type. public Integer print() { ^ Foo.java:4: Incompatible type for return. Can't convert int to java.lang.Number. return 0; ^ 2 errors
Confused !Comment
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What's your compiler version?Originally posted by madhoriya22Comment
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Hi,Originally posted by r035198x
My compiler version is 1.5.0_04.
Note : I am not trying to prove anything wrong. Just trying to resolve my confusion :(Comment
Comment