Declaring a method when creating an object
Why first way is correct, but second isn't?
First way:
new Object() {
public void a() {
/*code*/
}
}.a();
Second way:
Object object = new Object() {
public void a() {
/*code*/
}
};
object.a();
And where can I find more information about it?
java object methods anonymous-class anonymous-inner-class
edited Dec 28 '18 at 18:14
Andrew Tobilko
26.9k104285
asked Dec 28 '18 at 15:21
Sekonoishi KamikiSekonoishi Kamiki
14212
add a comment |
Why first way is correct, but second isn't?
First way:
new Object() {
public void a() {
/*code*/
}
}.a();
Second way:
Object object = new Object() {
public void a() {
/*code*/
}
};
object.a();
And where can I find more information about it?
java object methods anonymous-class anonymous-inner-class
edited Dec 28 '18 at 18:14
Andrew Tobilko
26.9k104285
asked Dec 28 '18 at 15:21
Sekonoishi KamikiSekonoishi Kamiki
14212
add a comment |
Why first way is correct, but second isn't?
First way:
new Object() {
public void a() {
/*code*/
}
}.a();
Second way:
Object object = new Object() {
public void a() {
/*code*/
}
};
object.a();
And where can I find more information about it?
java object methods anonymous-class anonymous-inner-class
edited Dec 28 '18 at 18:14
Andrew Tobilko
26.9k104285
asked Dec 28 '18 at 15:21
Sekonoishi KamikiSekonoishi Kamiki
14212
Why first way is correct, but second isn't?
First way:
new Object() {
public void a() {
/*code*/
}
}.a();
Second way:
Object object = new Object() {
public void a() {
/*code*/
}
};
object.a();
And where can I find more information about it?
java object methods anonymous-class anonymous-inner-class
java object methods anonymous-class anonymous-inner-class
edited Dec 28 '18 at 18:14
Andrew Tobilko
26.9k104285
asked Dec 28 '18 at 15:21
Sekonoishi KamikiSekonoishi Kamiki
14212
edited Dec 28 '18 at 18:14
Andrew Tobilko
26.9k104285
asked Dec 28 '18 at 15:21
Sekonoishi KamikiSekonoishi Kamiki
14212
edited Dec 28 '18 at 18:14
Andrew Tobilko
26.9k104285
edited Dec 28 '18 at 18:14
Andrew Tobilko
26.9k104285
edited Dec 28 '18 at 18:14
Andrew Tobilko
26.9k104285
26.9k104285
asked Dec 28 '18 at 15:21
Sekonoishi KamikiSekonoishi Kamiki
14212
asked Dec 28 '18 at 15:21
Sekonoishi KamikiSekonoishi Kamiki
14212
asked Dec 28 '18 at 15:21
Sekonoishi KamikiSekonoishi Kamiki
14212
14212
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
java.lang.Object has no a methods declared (2), while the anonymous class returned by the class instance creation expression new Object() { public void a() {} } does (1).
Use Java 10's local variable type inference (var) to make the second option as valid as the first one.
var object = new Object() {
public void a() {}
};
object.a();
answered Dec 28 '18 at 15:36
Andrew TobilkoAndrew Tobilko
26.9k104285
9
Interesting to know that thevarkeyword changes things here...
– ernest_k
Dec 28 '18 at 15:44
1
Note thatvarwas introduced in Java 10 and will not be available in lower versions.
– nasch
Dec 28 '18 at 20:03
5
@ernest_k It's really not that hard or interesting. By sayingObject object =you are downcasting to anObject,vargivesobjectits actual correct type that hasa()defined.
– Sombrero Chicken
Dec 28 '18 at 20:41
2
@SombreroChicken I understand the effect ofvarhere. But beside the first syntax in the question, there was no other way (that I know) to have a static reference to the anonymous type. Now I know thatvarcan be used for that. Even more, it allows one to make multiple method calls from the anonymous class. I find that strangely new and interesting!
– ernest_k
Dec 28 '18 at 20:46
2
You can also declare an interface that has thea()method and create an anonymous object from that instead ofObject, which would work in lower versions that don't supportvar. E.g.,private interface HasA { public void a(); }thenHasA object = new HasA() { public void a() { ... } };.
– jpmc26
Dec 28 '18 at 23:33
|
show 1 more comment
In the second option, you assign your new object to a reference of type Object. Because of this, only methods defined in java.lang.Object could be called on that reference.
And in the first option, you basically create new object of anonymous class that extends java.lang.Object. That anonymous class has the additional method a(), which is why you can call it.
edited Dec 28 '18 at 20:20
MTCoster
3,06621940
answered Dec 28 '18 at 15:24
IvanIvan
5,0831821
add a comment |
Java is statically typed. When you say object.a() it is looking for the method a in the Object class which is not present. Hence it does not compile.
What you can do is get the method of object using reflection as shown below :
Object object = new Object() {
public void a() {
System.out.println("In a");
}
}
Method method = object.getClass().getDeclaredMethod("a");
method.invoke(object, null);
This would print
In a
answered Dec 28 '18 at 19:31
fastcodejavafastcodejava
24k19109162
2
That's not true. Java uses static typing (you can see it in your example -Object object- you just declared its type!). It uses polymorphism to determine which method should be called, but that's a different thing than dynamic typing. Some debate about dynamic vs static can be found here
– MatthewRock
Dec 28 '18 at 19:39
2
you should have pointed out that calling the method through reflection is the last thing OP wants :)
– Andrew Tobilko
Dec 28 '18 at 20:17
@MatthewRock The JVM does support dynamic dispatch, however.
– Solomon Ucko
Dec 29 '18 at 19:27
@SolomonUcko JVM is not Java. Dynamic dispatch is also not something that makes a language a static one. C++ has dynamic dispatch for virtual methods, but is still a static language.
– MatthewRock
Dec 29 '18 at 20:59
1
@MatthewRock What exactly was wrong?
– fastcodejava
Dec 30 '18 at 4:35
add a comment |
Don't worry, you will have to do a bit of correction
Both are ways to access the private member of a class. By using first way you don' have to pre-declare the method.ex: -
public class demo {
public static void main(String args) {
new Object() {
public void a() {
/*code*/
System.out.println("Hello");
}
}.a();
}
}
But by using second way you will have to explicitly declare the method a(); either in abstract class or in interface then you can override it. like: -
interface Object
{
public void a();
}
class demo {
public static void main(String args) {
Object object = new Object() {
public void a() {
System.out.println("Hello");
}
}; object.a();
}
}
I hope it will help a bit.
answered Dec 30 '18 at 12:28
MAYANKMAYANK
13
1
"to access the private member of a class" what's the private member?
– Andrew Tobilko
Dec 30 '18 at 12:46
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
java.lang.Object has no a methods declared (2), while the anonymous class returned by the class instance creation expression new Object() { public void a() {} } does (1).
Use Java 10's local variable type inference (var) to make the second option as valid as the first one.
var object = new Object() {
public void a() {}
};
object.a();
answered Dec 28 '18 at 15:36
Andrew TobilkoAndrew Tobilko
26.9k104285
9
Interesting to know that thevarkeyword changes things here...
– ernest_k
Dec 28 '18 at 15:44
1
Note thatvarwas introduced in Java 10 and will not be available in lower versions.
– nasch
Dec 28 '18 at 20:03
5
@ernest_k It's really not that hard or interesting. By sayingObject object =you are downcasting to anObject,vargivesobjectits actual correct type that hasa()defined.
– Sombrero Chicken
Dec 28 '18 at 20:41
2
@SombreroChicken I understand the effect ofvarhere. But beside the first syntax in the question, there was no other way (that I know) to have a static reference to the anonymous type. Now I know thatvarcan be used for that. Even more, it allows one to make multiple method calls from the anonymous class. I find that strangely new and interesting!
– ernest_k
Dec 28 '18 at 20:46
2
You can also declare an interface that has thea()method and create an anonymous object from that instead ofObject, which would work in lower versions that don't supportvar. E.g.,private interface HasA { public void a(); }thenHasA object = new HasA() { public void a() { ... } };.
– jpmc26
Dec 28 '18 at 23:33
|
show 1 more comment
java.lang.Object has no a methods declared (2), while the anonymous class returned by the class instance creation expression new Object() { public void a() {} } does (1).
Use Java 10's local variable type inference (var) to make the second option as valid as the first one.
var object = new Object() {
public void a() {}
};
object.a();
answered Dec 28 '18 at 15:36
Andrew TobilkoAndrew Tobilko
26.9k104285
9
Interesting to know that thevarkeyword changes things here...
– ernest_k
Dec 28 '18 at 15:44
1
Note thatvarwas introduced in Java 10 and will not be available in lower versions.
– nasch
Dec 28 '18 at 20:03
5
@ernest_k It's really not that hard or interesting. By sayingObject object =you are downcasting to anObject,vargivesobjectits actual correct type that hasa()defined.
– Sombrero Chicken
Dec 28 '18 at 20:41
2
@SombreroChicken I understand the effect ofvarhere. But beside the first syntax in the question, there was no other way (that I know) to have a static reference to the anonymous type. Now I know thatvarcan be used for that. Even more, it allows one to make multiple method calls from the anonymous class. I find that strangely new and interesting!
– ernest_k
Dec 28 '18 at 20:46
2
You can also declare an interface that has thea()method and create an anonymous object from that instead ofObject, which would work in lower versions that don't supportvar. E.g.,private interface HasA { public void a(); }thenHasA object = new HasA() { public void a() { ... } };.
– jpmc26
Dec 28 '18 at 23:33
|
show 1 more comment
java.lang.Object has no a methods declared (2), while the anonymous class returned by the class instance creation expression new Object() { public void a() {} } does (1).
Use Java 10's local variable type inference (var) to make the second option as valid as the first one.
var object = new Object() {
public void a() {}
};
object.a();
answered Dec 28 '18 at 15:36
Andrew TobilkoAndrew Tobilko
26.9k104285
java.lang.Object has no a methods declared (2), while the anonymous class returned by the class instance creation expression new Object() { public void a() {} } does (1).
Use Java 10's local variable type inference (var) to make the second option as valid as the first one.
var object = new Object() {
public void a() {}
};
object.a();
answered Dec 28 '18 at 15:36
Andrew TobilkoAndrew Tobilko
26.9k104285
edited Jan 2 at 19:27
answered Dec 28 '18 at 15:36
Andrew TobilkoAndrew Tobilko
26.9k104285
answered Dec 28 '18 at 15:36
Andrew TobilkoAndrew Tobilko
26.9k104285
answered Dec 28 '18 at 15:36
Andrew TobilkoAndrew Tobilko
26.9k104285
26.9k104285
9
Interesting to know that thevarkeyword changes things here...
– ernest_k
Dec 28 '18 at 15:44
1
Note thatvarwas introduced in Java 10 and will not be available in lower versions.
– nasch
Dec 28 '18 at 20:03
5
@ernest_k It's really not that hard or interesting. By sayingObject object =you are downcasting to anObject,vargivesobjectits actual correct type that hasa()defined.
– Sombrero Chicken
Dec 28 '18 at 20:41
2
@SombreroChicken I understand the effect ofvarhere. But beside the first syntax in the question, there was no other way (that I know) to have a static reference to the anonymous type. Now I know thatvarcan be used for that. Even more, it allows one to make multiple method calls from the anonymous class. I find that strangely new and interesting!
– ernest_k
Dec 28 '18 at 20:46
2
You can also declare an interface that has thea()method and create an anonymous object from that instead ofObject, which would work in lower versions that don't supportvar. E.g.,private interface HasA { public void a(); }thenHasA object = new HasA() { public void a() { ... } };.
– jpmc26
Dec 28 '18 at 23:33
|
show 1 more comment
9
Interesting to know that thevarkeyword changes things here...
– ernest_k
Dec 28 '18 at 15:44
1
Note thatvarwas introduced in Java 10 and will not be available in lower versions.
– nasch
Dec 28 '18 at 20:03
5
@ernest_k It's really not that hard or interesting. By sayingObject object =you are downcasting to anObject,vargivesobjectits actual correct type that hasa()defined.
– Sombrero Chicken
Dec 28 '18 at 20:41
2
@SombreroChicken I understand the effect ofvarhere. But beside the first syntax in the question, there was no other way (that I know) to have a static reference to the anonymous type. Now I know thatvarcan be used for that. Even more, it allows one to make multiple method calls from the anonymous class. I find that strangely new and interesting!
– ernest_k
Dec 28 '18 at 20:46
2
You can also declare an interface that has thea()method and create an anonymous object from that instead ofObject, which would work in lower versions that don't supportvar. E.g.,private interface HasA { public void a(); }thenHasA object = new HasA() { public void a() { ... } };.
– jpmc26
Dec 28 '18 at 23:33
9
9
Interesting to know that the
var keyword changes things here...– ernest_k
Dec 28 '18 at 15:44
Interesting to know that the
var keyword changes things here...– ernest_k
Dec 28 '18 at 15:44
1
1
Note that
var was introduced in Java 10 and will not be available in lower versions.– nasch
Dec 28 '18 at 20:03
Note that
var was introduced in Java 10 and will not be available in lower versions.– nasch
Dec 28 '18 at 20:03
5
5
@ernest_k It's really not that hard or interesting. By saying
Object object = you are downcasting to an Object, var gives object its actual correct type that has a() defined.– Sombrero Chicken
Dec 28 '18 at 20:41
@ernest_k It's really not that hard or interesting. By saying
Object object = you are downcasting to an Object, var gives object its actual correct type that has a() defined.– Sombrero Chicken
Dec 28 '18 at 20:41
2
2
@SombreroChicken I understand the effect of
var here. But beside the first syntax in the question, there was no other way (that I know) to have a static reference to the anonymous type. Now I know that var can be used for that. Even more, it allows one to make multiple method calls from the anonymous class. I find that strangely new and interesting!– ernest_k
Dec 28 '18 at 20:46
@SombreroChicken I understand the effect of
var here. But beside the first syntax in the question, there was no other way (that I know) to have a static reference to the anonymous type. Now I know that var can be used for that. Even more, it allows one to make multiple method calls from the anonymous class. I find that strangely new and interesting!– ernest_k
Dec 28 '18 at 20:46
2
2
You can also declare an interface that has the
a() method and create an anonymous object from that instead of Object, which would work in lower versions that don't support var. E.g., private interface HasA { public void a(); } then HasA object = new HasA() { public void a() { ... } };.– jpmc26
Dec 28 '18 at 23:33
You can also declare an interface that has the
a() method and create an anonymous object from that instead of Object, which would work in lower versions that don't support var. E.g., private interface HasA { public void a(); } then HasA object = new HasA() { public void a() { ... } };.– jpmc26
Dec 28 '18 at 23:33
|
show 1 more comment
In the second option, you assign your new object to a reference of type Object. Because of this, only methods defined in java.lang.Object could be called on that reference.
And in the first option, you basically create new object of anonymous class that extends java.lang.Object. That anonymous class has the additional method a(), which is why you can call it.
edited Dec 28 '18 at 20:20
MTCoster
3,06621940
answered Dec 28 '18 at 15:24
IvanIvan
5,0831821
add a comment |
In the second option, you assign your new object to a reference of type Object. Because of this, only methods defined in java.lang.Object could be called on that reference.
And in the first option, you basically create new object of anonymous class that extends java.lang.Object. That anonymous class has the additional method a(), which is why you can call it.
edited Dec 28 '18 at 20:20
MTCoster
3,06621940
answered Dec 28 '18 at 15:24
IvanIvan
5,0831821
add a comment |
In the second option, you assign your new object to a reference of type Object. Because of this, only methods defined in java.lang.Object could be called on that reference.
And in the first option, you basically create new object of anonymous class that extends java.lang.Object. That anonymous class has the additional method a(), which is why you can call it.
edited Dec 28 '18 at 20:20
MTCoster
3,06621940
answered Dec 28 '18 at 15:24
IvanIvan
5,0831821
In the second option, you assign your new object to a reference of type Object. Because of this, only methods defined in java.lang.Object could be called on that reference.
And in the first option, you basically create new object of anonymous class that extends java.lang.Object. That anonymous class has the additional method a(), which is why you can call it.
edited Dec 28 '18 at 20:20
MTCoster
3,06621940
answered Dec 28 '18 at 15:24
IvanIvan
5,0831821
edited Dec 28 '18 at 20:20
MTCoster
3,06621940
edited Dec 28 '18 at 20:20
MTCoster
3,06621940
edited Dec 28 '18 at 20:20
MTCoster
3,06621940
3,06621940
answered Dec 28 '18 at 15:24
IvanIvan
5,0831821
answered Dec 28 '18 at 15:24
IvanIvan
5,0831821
answered Dec 28 '18 at 15:24
IvanIvan
5,0831821
5,0831821
add a comment |
add a comment |
Java is statically typed. When you say object.a() it is looking for the method a in the Object class which is not present. Hence it does not compile.
What you can do is get the method of object using reflection as shown below :
Object object = new Object() {
public void a() {
System.out.println("In a");
}
}
Method method = object.getClass().getDeclaredMethod("a");
method.invoke(object, null);
This would print
In a
answered Dec 28 '18 at 19:31
fastcodejavafastcodejava
24k19109162
2
That's not true. Java uses static typing (you can see it in your example -Object object- you just declared its type!). It uses polymorphism to determine which method should be called, but that's a different thing than dynamic typing. Some debate about dynamic vs static can be found here
– MatthewRock
Dec 28 '18 at 19:39
2
you should have pointed out that calling the method through reflection is the last thing OP wants :)
– Andrew Tobilko
Dec 28 '18 at 20:17
@MatthewRock The JVM does support dynamic dispatch, however.
– Solomon Ucko
Dec 29 '18 at 19:27
@SolomonUcko JVM is not Java. Dynamic dispatch is also not something that makes a language a static one. C++ has dynamic dispatch for virtual methods, but is still a static language.
– MatthewRock
Dec 29 '18 at 20:59
1
@MatthewRock What exactly was wrong?
– fastcodejava
Dec 30 '18 at 4:35
add a comment |
Java is statically typed. When you say object.a() it is looking for the method a in the Object class which is not present. Hence it does not compile.
What you can do is get the method of object using reflection as shown below :
Object object = new Object() {
public void a() {
System.out.println("In a");
}
}
Method method = object.getClass().getDeclaredMethod("a");
method.invoke(object, null);
This would print
In a
answered Dec 28 '18 at 19:31
fastcodejavafastcodejava
24k19109162
2
That's not true. Java uses static typing (you can see it in your example -Object object- you just declared its type!). It uses polymorphism to determine which method should be called, but that's a different thing than dynamic typing. Some debate about dynamic vs static can be found here
– MatthewRock
Dec 28 '18 at 19:39
2
you should have pointed out that calling the method through reflection is the last thing OP wants :)
– Andrew Tobilko
Dec 28 '18 at 20:17
@MatthewRock The JVM does support dynamic dispatch, however.
– Solomon Ucko
Dec 29 '18 at 19:27
@SolomonUcko JVM is not Java. Dynamic dispatch is also not something that makes a language a static one. C++ has dynamic dispatch for virtual methods, but is still a static language.
– MatthewRock
Dec 29 '18 at 20:59
1
@MatthewRock What exactly was wrong?
– fastcodejava
Dec 30 '18 at 4:35
add a comment |
Java is statically typed. When you say object.a() it is looking for the method a in the Object class which is not present. Hence it does not compile.
What you can do is get the method of object using reflection as shown below :
Object object = new Object() {
public void a() {
System.out.println("In a");
}
}
Method method = object.getClass().getDeclaredMethod("a");
method.invoke(object, null);
This would print
In a
answered Dec 28 '18 at 19:31
fastcodejavafastcodejava
24k19109162
Java is statically typed. When you say object.a() it is looking for the method a in the Object class which is not present. Hence it does not compile.
What you can do is get the method of object using reflection as shown below :
Object object = new Object() {
public void a() {
System.out.println("In a");
}
}
Method method = object.getClass().getDeclaredMethod("a");
method.invoke(object, null);
This would print
In a
answered Dec 28 '18 at 19:31
fastcodejavafastcodejava
24k19109162
edited Dec 28 '18 at 19:51
answered Dec 28 '18 at 19:31
fastcodejavafastcodejava
24k19109162
answered Dec 28 '18 at 19:31
fastcodejavafastcodejava
24k19109162
answered Dec 28 '18 at 19:31
fastcodejavafastcodejava
24k19109162
24k19109162
2
That's not true. Java uses static typing (you can see it in your example -Object object- you just declared its type!). It uses polymorphism to determine which method should be called, but that's a different thing than dynamic typing. Some debate about dynamic vs static can be found here
– MatthewRock
Dec 28 '18 at 19:39
2
you should have pointed out that calling the method through reflection is the last thing OP wants :)
– Andrew Tobilko
Dec 28 '18 at 20:17
@MatthewRock The JVM does support dynamic dispatch, however.
– Solomon Ucko
Dec 29 '18 at 19:27
@SolomonUcko JVM is not Java. Dynamic dispatch is also not something that makes a language a static one. C++ has dynamic dispatch for virtual methods, but is still a static language.
– MatthewRock
Dec 29 '18 at 20:59
1
@MatthewRock What exactly was wrong?
– fastcodejava
Dec 30 '18 at 4:35
add a comment |
2
That's not true. Java uses static typing (you can see it in your example -Object object- you just declared its type!). It uses polymorphism to determine which method should be called, but that's a different thing than dynamic typing. Some debate about dynamic vs static can be found here
– MatthewRock
Dec 28 '18 at 19:39
2
you should have pointed out that calling the method through reflection is the last thing OP wants :)
– Andrew Tobilko
Dec 28 '18 at 20:17
@MatthewRock The JVM does support dynamic dispatch, however.
– Solomon Ucko
Dec 29 '18 at 19:27
@SolomonUcko JVM is not Java. Dynamic dispatch is also not something that makes a language a static one. C++ has dynamic dispatch for virtual methods, but is still a static language.
– MatthewRock
Dec 29 '18 at 20:59
1
@MatthewRock What exactly was wrong?
– fastcodejava
Dec 30 '18 at 4:35
2
2
That's not true. Java uses static typing (you can see it in your example -
Object object - you just declared its type!). It uses polymorphism to determine which method should be called, but that's a different thing than dynamic typing. Some debate about dynamic vs static can be found here– MatthewRock
Dec 28 '18 at 19:39
That's not true. Java uses static typing (you can see it in your example -
Object object - you just declared its type!). It uses polymorphism to determine which method should be called, but that's a different thing than dynamic typing. Some debate about dynamic vs static can be found here– MatthewRock
Dec 28 '18 at 19:39
2
2
you should have pointed out that calling the method through reflection is the last thing OP wants :)
– Andrew Tobilko
Dec 28 '18 at 20:17
you should have pointed out that calling the method through reflection is the last thing OP wants :)
– Andrew Tobilko
Dec 28 '18 at 20:17
@MatthewRock The JVM does support dynamic dispatch, however.
– Solomon Ucko
Dec 29 '18 at 19:27
@MatthewRock The JVM does support dynamic dispatch, however.
– Solomon Ucko
Dec 29 '18 at 19:27
@SolomonUcko JVM is not Java. Dynamic dispatch is also not something that makes a language a static one. C++ has dynamic dispatch for virtual methods, but is still a static language.
– MatthewRock
Dec 29 '18 at 20:59
@SolomonUcko JVM is not Java. Dynamic dispatch is also not something that makes a language a static one. C++ has dynamic dispatch for virtual methods, but is still a static language.
– MatthewRock
Dec 29 '18 at 20:59
1
1
@MatthewRock What exactly was wrong?
– fastcodejava
Dec 30 '18 at 4:35
@MatthewRock What exactly was wrong?
– fastcodejava
Dec 30 '18 at 4:35
add a comment |
Don't worry, you will have to do a bit of correction
Both are ways to access the private member of a class. By using first way you don' have to pre-declare the method.ex: -
public class demo {
public static void main(String args) {
new Object() {
public void a() {
/*code*/
System.out.println("Hello");
}
}.a();
}
}
But by using second way you will have to explicitly declare the method a(); either in abstract class or in interface then you can override it. like: -
interface Object
{
public void a();
}
class demo {
public static void main(String args) {
Object object = new Object() {
public void a() {
System.out.println("Hello");
}
}; object.a();
}
}
I hope it will help a bit.
answered Dec 30 '18 at 12:28
MAYANKMAYANK
13
1
"to access the private member of a class" what's the private member?
– Andrew Tobilko
Dec 30 '18 at 12:46
add a comment |
Don't worry, you will have to do a bit of correction
Both are ways to access the private member of a class. By using first way you don' have to pre-declare the method.ex: -
public class demo {
public static void main(String args) {
new Object() {
public void a() {
/*code*/
System.out.println("Hello");
}
}.a();
}
}
But by using second way you will have to explicitly declare the method a(); either in abstract class or in interface then you can override it. like: -
interface Object
{
public void a();
}
class demo {
public static void main(String args) {
Object object = new Object() {
public void a() {
System.out.println("Hello");
}
}; object.a();
}
}
I hope it will help a bit.
answered Dec 30 '18 at 12:28
MAYANKMAYANK
13
1
"to access the private member of a class" what's the private member?
– Andrew Tobilko
Dec 30 '18 at 12:46
add a comment |
Don't worry, you will have to do a bit of correction
Both are ways to access the private member of a class. By using first way you don' have to pre-declare the method.ex: -
public class demo {
public static void main(String args) {
new Object() {
public void a() {
/*code*/
System.out.println("Hello");
}
}.a();
}
}
But by using second way you will have to explicitly declare the method a(); either in abstract class or in interface then you can override it. like: -
interface Object
{
public void a();
}
class demo {
public static void main(String args) {
Object object = new Object() {
public void a() {
System.out.println("Hello");
}
}; object.a();
}
}
I hope it will help a bit.
answered Dec 30 '18 at 12:28
MAYANKMAYANK
13
Don't worry, you will have to do a bit of correction
Both are ways to access the private member of a class. By using first way you don' have to pre-declare the method.ex: -
public class demo {
public static void main(String args) {
new Object() {
public void a() {
/*code*/
System.out.println("Hello");
}
}.a();
}
}
But by using second way you will have to explicitly declare the method a(); either in abstract class or in interface then you can override it. like: -
interface Object
{
public void a();
}
class demo {
public static void main(String args) {
Object object = new Object() {
public void a() {
System.out.println("Hello");
}
}; object.a();
}
}
I hope it will help a bit.
answered Dec 30 '18 at 12:28
MAYANKMAYANK
13
answered Dec 30 '18 at 12:28
MAYANKMAYANK
13
answered Dec 30 '18 at 12:28
MAYANKMAYANK
13
answered Dec 30 '18 at 12:28
MAYANKMAYANK
13
13
1
"to access the private member of a class" what's the private member?
– Andrew Tobilko
Dec 30 '18 at 12:46
add a comment |
1
"to access the private member of a class" what's the private member?
– Andrew Tobilko
Dec 30 '18 at 12:46
1
1
"to access the private member of a class" what's the private member?
– Andrew Tobilko
Dec 30 '18 at 12:46
"to access the private member of a class" what's the private member?
– Andrew Tobilko
Dec 30 '18 at 12:46
add a comment |
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