up vote
7
down vote

favorite

I have actually 2 questions.

QUESTION 1

Consider the below code.

public class Test
 private static final List<String> var = new ArrayList<String>() 
 add("A");
 add("B");
 System.out.println("INNER : "+var);
 ;

 public static void main(String... args)
 System.out.println("OUTER : "+var);

When i run this code it gives me below output

INNER : null
OUTER : [A, B]

Can any one elaborate why INNER is null and execution flow at a time when exactly "A" & "B" added to collection.

QUESTION 2

I made some changes in above code and modified to below one(just put add method within first bracket)

public class Test
 private static final List<String> var = new ArrayList<String>() 
 public boolean add(String e) 
 add("A");
 add("B");
 System.out.println("INNER : " + var); return true; 
 ;
 ;

 public static void main(String... args)
 System.out.println("OUTER : "+var);

After running above code i got below result

OUTER :

After viewing this I'm totally clueless here about what is going on ... where did INNER gone ??? why it's not printing ? is it not called ?

Great help if anyone can explain this behavior and fundamental behind it.

edited 51 mins ago

maspinu

7032922

asked 56 mins ago

RBS

434

add a commentÂ |Â

up vote
7
down vote

favorite

I have actually 2 questions.

QUESTION 1

Consider the below code.

public class Test
 private static final List<String> var = new ArrayList<String>() 
 add("A");
 add("B");
 System.out.println("INNER : "+var);
 ;

 public static void main(String... args)
 System.out.println("OUTER : "+var);

QUESTION 2

I made some changes in above code and modified to below one(just put add method within first bracket)

public class Test
 private static final List<String> var = new ArrayList<String>() 
 public boolean add(String e) 
 add("A");
 add("B");
 System.out.println("INNER : " + var); return true; 
 ;
 ;

 public static void main(String... args)
 System.out.println("OUTER : "+var);

After running above code i got below result

OUTER :

After viewing this I'm totally clueless here about what is going on ... where did INNER gone ??? why it's not printing ? is it not called ?

Great help if anyone can explain this behavior and fundamental behind it.

edited 51 mins ago

maspinu

7032922

asked 56 mins ago

RBS

434

add a commentÂ |Â

up vote
7
down vote

favorite

I have actually 2 questions.

QUESTION 1

Consider the below code.

public class Test
 private static final List<String> var = new ArrayList<String>() 
 add("A");
 add("B");
 System.out.println("INNER : "+var);
 ;

 public static void main(String... args)
 System.out.println("OUTER : "+var);

QUESTION 2

I made some changes in above code and modified to below one(just put add method within first bracket)

public class Test
 private static final List<String> var = new ArrayList<String>() 
 public boolean add(String e) 
 add("A");
 add("B");
 System.out.println("INNER : " + var); return true; 
 ;
 ;

 public static void main(String... args)
 System.out.println("OUTER : "+var);

After running above code i got below result

OUTER :

After viewing this I'm totally clueless here about what is going on ... where did INNER gone ??? why it's not printing ? is it not called ?

Great help if anyone can explain this behavior and fundamental behind it.

edited 51 mins ago

maspinu

7032922

asked 56 mins ago

RBS

434

I have actually 2 questions.

QUESTION 1

Consider the below code.

public class Test
 private static final List<String> var = new ArrayList<String>() 
 add("A");
 add("B");
 System.out.println("INNER : "+var);
 ;

 public static void main(String... args)
 System.out.println("OUTER : "+var);

QUESTION 2

I made some changes in above code and modified to below one(just put add method within first bracket)

public class Test
 private static final List<String> var = new ArrayList<String>() 
 public boolean add(String e) 
 add("A");
 add("B");
 System.out.println("INNER : " + var); return true; 
 ;
 ;

 public static void main(String... args)
 System.out.println("OUTER : "+var);

After running above code i got below result

OUTER :

After viewing this I'm totally clueless here about what is going on ... where did INNER gone ??? why it's not printing ? is it not called ?

Great help if anyone can explain this behavior and fundamental behind it.

java arraylist collections

edited 51 mins ago

maspinu

7032922

asked 56 mins ago

RBS

434

edited 51 mins ago

maspinu

7032922

asked 56 mins ago

RBS

434

edited 51 mins ago

maspinu

7032922

edited 51 mins ago

maspinu

7032922

edited 51 mins ago

maspinu

7032922

asked 56 mins ago

RBS

434

asked 56 mins ago

RBS

434

asked 56 mins ago

RBS

434

add a commentÂ |Â

5 Answers
5

active

oldest

votes

up vote
5
down vote

accepted

Because the initializer block of the anonymous class off ArrayList is executed before the instance reference of the anonymous class is assigned to var.

The code is not executed because you never call the add(String e) method. If you did, it would result in a StackOverflowError, since add("A") is a recursive call now.

answered 49 mins ago

Andreas

69.5k450107

if i call add() from main method how does it becomes recursive call(i check it as you mentioned it gives StackOverflowError)
â€“Â RBS
32 mins ago

2

@RBS The first statement inside add(String e) is add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), and StackOverflowError.
â€“Â Andreas
25 mins ago

add a commentÂ |Â

up vote
1
down vote

In your code snippet 1, you are using the double brace initialization to add items to the array list. It is basically an anonymous inner class with an instance initializer. Since you print var before the object is constructed it is null.

In your second snippet, you are creating an anonymous class of ArrayList providing an implementation of the add method. But no one calls the add method on the ArrayList object.

EDIT: Good point made by Andreas@ - Even if you call the add method, it will result in an infinite recursive call resulting in a StackOverflowError.

Reference:

Initialization of an ArrayList in one line

answered 49 mins ago

user7

8,12032038

add a commentÂ |Â

up vote
1
down vote

The object you created has not been assigned to var in the initialization procedure, try this:

private static final List<String> var = new ArrayList<String>() 
 add("A");
 add("B");
 System.out.println("THIS : " + this); // THIS : [A, B]
 System.out.println("INNER : " + var); // INNER : null
;

You just override add method of ArrayList, it's nerver been called. Which basicly equals:

static class CustomList<E> extends ArrayList<E> 

 @Override
 public boolean add(String e) 
 add("A");
 add("B");
 System.out.println("INNER : " + var);
 return true;
 


private static final List<String> var = new CustomList<>();

edited 39 mins ago

answered 47 mins ago

Ã¥ÂÂ™Ã¥Â…Â´Ã¦Â–ÂŒ

14.8k41542

add a commentÂ |Â

up vote
0
down vote

In the first code you print the variable before that its constructor has returned as you add the elements and print the list inside the instance initializer.
So it can only print null.

In the second code you override add() to add hardcoded elements but add() is never invoked on the created instance. So add() doesn't print anything and nothing is added in the List instance either.

To add elements in a List you generally want to invoke add() on the List reference such as :

List<String> list = new ArrayList<>();
list.add("element A");
list.add("element B");

You don't create an anonymous class for such a requirement.

If you want to use a simpler syntax you can still do :

List<String> list = new ArrayList<>(Arrays.asList("element A", "element B"));

edited 40 mins ago

answered 47 mins ago

davidxxx

58.2k54581

add a commentÂ |Â

up vote
-1
down vote

In you first code var is null because the return value of new operator is not assigned to var.
In the second code you override the add method and it is never called, to call that run this code:

public class Test
private static final List<String> var = new ArrayList<String>() 
 public boolean add(String e) 
 /*add("A");
 add("B");*/
 super.add(e);
 System.out.println("INNER : " + var); return true;
 ;
 
 add("A");
 add("B");
 
;

public static void main(String... args)
 System.out.println("OUTER : "+var);

answered 40 mins ago

Rahim Dastar

707515

add a commentÂ |Â

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
5
down vote

accepted

Because the initializer block of the anonymous class off ArrayList is executed before the instance reference of the anonymous class is assigned to var.

The code is not executed because you never call the add(String e) method. If you did, it would result in a StackOverflowError, since add("A") is a recursive call now.

answered 49 mins ago

Andreas

69.5k450107

if i call add() from main method how does it becomes recursive call(i check it as you mentioned it gives StackOverflowError)
â€“Â RBS
32 mins ago

2

@RBS The first statement inside add(String e) is add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), and StackOverflowError.
â€“Â Andreas
25 mins ago

add a commentÂ |Â

up vote
5
down vote

accepted

Because the initializer block of the anonymous class off ArrayList is executed before the instance reference of the anonymous class is assigned to var.

The code is not executed because you never call the add(String e) method. If you did, it would result in a StackOverflowError, since add("A") is a recursive call now.

answered 49 mins ago

Andreas

69.5k450107

if i call add() from main method how does it becomes recursive call(i check it as you mentioned it gives StackOverflowError)
â€“Â RBS
32 mins ago

2

@RBS The first statement inside add(String e) is add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), and StackOverflowError.
â€“Â Andreas
25 mins ago

add a commentÂ |Â

up vote
5
down vote

accepted

Because the initializer block of the anonymous class off ArrayList is executed before the instance reference of the anonymous class is assigned to var.

The code is not executed because you never call the add(String e) method. If you did, it would result in a StackOverflowError, since add("A") is a recursive call now.

answered 49 mins ago

Andreas

69.5k450107

Because the initializer block of the anonymous class off ArrayList is executed before the instance reference of the anonymous class is assigned to var.

The code is not executed because you never call the add(String e) method. If you did, it would result in a StackOverflowError, since add("A") is a recursive call now.

answered 49 mins ago

Andreas

69.5k450107

answered 49 mins ago

Andreas

69.5k450107

answered 49 mins ago

Andreas

69.5k450107

answered 49 mins ago

Andreas

69.5k450107

if i call add() from main method how does it becomes recursive call(i check it as you mentioned it gives StackOverflowError)
â€“Â RBS
32 mins ago

2

@RBS The first statement inside add(String e) is add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), and StackOverflowError.
â€“Â Andreas
25 mins ago

add a commentÂ |Â

if i call add() from main method how does it becomes recursive call(i check it as you mentioned it gives StackOverflowError)
â€“Â RBS
32 mins ago

2

@RBS The first statement inside add(String e) is add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), and StackOverflowError.
â€“Â Andreas
25 mins ago

if i call add() from main method how does it becomes recursive call(i check it as you mentioned it gives StackOverflowError)
â€“Â RBS
32 mins ago

@RBS The first statement inside add(String e) is add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), which is a call to itself, so it executes add("A"), and StackOverflowError.
â€“Â Andreas
25 mins ago

add a commentÂ |Â

up vote
1
down vote

In your second snippet, you are creating an anonymous class of ArrayList providing an implementation of the add method. But no one calls the add method on the ArrayList object.

EDIT: Good point made by Andreas@ - Even if you call the add method, it will result in an infinite recursive call resulting in a StackOverflowError.

Reference:

Initialization of an ArrayList in one line

answered 49 mins ago

user7

8,12032038

add a commentÂ |Â

up vote
1
down vote

In your second snippet, you are creating an anonymous class of ArrayList providing an implementation of the add method. But no one calls the add method on the ArrayList object.

EDIT: Good point made by Andreas@ - Even if you call the add method, it will result in an infinite recursive call resulting in a StackOverflowError.

Reference:

Initialization of an ArrayList in one line

answered 49 mins ago

user7

8,12032038

add a commentÂ |Â

up vote
1
down vote

In your second snippet, you are creating an anonymous class of ArrayList providing an implementation of the add method. But no one calls the add method on the ArrayList object.

EDIT: Good point made by Andreas@ - Even if you call the add method, it will result in an infinite recursive call resulting in a StackOverflowError.

Reference:

Initialization of an ArrayList in one line

answered 49 mins ago

user7

8,12032038

In your second snippet, you are creating an anonymous class of ArrayList providing an implementation of the add method. But no one calls the add method on the ArrayList object.

EDIT: Good point made by Andreas@ - Even if you call the add method, it will result in an infinite recursive call resulting in a StackOverflowError.

Reference:

Initialization of an ArrayList in one line

answered 49 mins ago

user7

8,12032038

answered 49 mins ago

user7

8,12032038

answered 49 mins ago

user7

8,12032038

answered 49 mins ago

user7

8,12032038

add a commentÂ |Â

up vote
1
down vote

The object you created has not been assigned to var in the initialization procedure, try this:

private static final List<String> var = new ArrayList<String>() 
 add("A");
 add("B");
 System.out.println("THIS : " + this); // THIS : [A, B]
 System.out.println("INNER : " + var); // INNER : null
;

You just override add method of ArrayList, it's nerver been called. Which basicly equals:

static class CustomList<E> extends ArrayList<E> 

 @Override
 public boolean add(String e) 
 add("A");
 add("B");
 System.out.println("INNER : " + var);
 return true;
 


private static final List<String> var = new CustomList<>();

edited 39 mins ago

answered 47 mins ago

Ã¥ÂÂ™Ã¥Â…Â´Ã¦Â–ÂŒ

14.8k41542

add a commentÂ |Â

up vote
1
down vote

The object you created has not been assigned to var in the initialization procedure, try this:

private static final List<String> var = new ArrayList<String>() 
 add("A");
 add("B");
 System.out.println("THIS : " + this); // THIS : [A, B]
 System.out.println("INNER : " + var); // INNER : null
;

You just override add method of ArrayList, it's nerver been called. Which basicly equals:

static class CustomList<E> extends ArrayList<E> 

 @Override
 public boolean add(String e) 
 add("A");
 add("B");
 System.out.println("INNER : " + var);
 return true;
 


private static final List<String> var = new CustomList<>();

edited 39 mins ago

answered 47 mins ago

Ã¥ÂÂ™Ã¥Â…Â´Ã¦Â–ÂŒ

14.8k41542

add a commentÂ |Â

up vote
1
down vote

The object you created has not been assigned to var in the initialization procedure, try this:

private static final List<String> var = new ArrayList<String>() 
 add("A");
 add("B");
 System.out.println("THIS : " + this); // THIS : [A, B]
 System.out.println("INNER : " + var); // INNER : null
;

You just override add method of ArrayList, it's nerver been called. Which basicly equals:

static class CustomList<E> extends ArrayList<E> 

 @Override
 public boolean add(String e) 
 add("A");
 add("B");
 System.out.println("INNER : " + var);
 return true;
 


private static final List<String> var = new CustomList<>();

edited 39 mins ago

answered 47 mins ago

Ã¥ÂÂ™Ã¥Â…Â´Ã¦Â–ÂŒ

14.8k41542

The object you created has not been assigned to var in the initialization procedure, try this:

private static final List<String> var = new ArrayList<String>() 
 add("A");
 add("B");
 System.out.println("THIS : " + this); // THIS : [A, B]
 System.out.println("INNER : " + var); // INNER : null
;

You just override add method of ArrayList, it's nerver been called. Which basicly equals:

static class CustomList<E> extends ArrayList<E> 

 @Override
 public boolean add(String e) 
 add("A");
 add("B");
 System.out.println("INNER : " + var);
 return true;
 


private static final List<String> var = new CustomList<>();

edited 39 mins ago

answered 47 mins ago

Ã¥ÂÂ™Ã¥Â…Â´Ã¦Â–ÂŒ

14.8k41542

edited 39 mins ago

answered 47 mins ago

Ã¥ÂÂ™Ã¥Â…Â´Ã¦Â–ÂŒ

14.8k41542

answered 47 mins ago

Ã¥ÂÂ™Ã¥Â…Â´Ã¦Â–ÂŒ

14.8k41542

answered 47 mins ago

Ã¥ÂÂ™Ã¥Â…Â´Ã¦Â–ÂŒ

14.8k41542

add a commentÂ |Â

up vote
0
down vote

In the first code you print the variable before that its constructor has returned as you add the elements and print the list inside the instance initializer.
So it can only print null.

To add elements in a List you generally want to invoke add() on the List reference such as :

List<String> list = new ArrayList<>();
list.add("element A");
list.add("element B");

You don't create an anonymous class for such a requirement.

If you want to use a simpler syntax you can still do :

List<String> list = new ArrayList<>(Arrays.asList("element A", "element B"));

edited 40 mins ago

answered 47 mins ago

davidxxx

58.2k54581

add a commentÂ |Â

up vote
0
down vote

In the first code you print the variable before that its constructor has returned as you add the elements and print the list inside the instance initializer.
So it can only print null.

To add elements in a List you generally want to invoke add() on the List reference such as :

List<String> list = new ArrayList<>();
list.add("element A");
list.add("element B");

You don't create an anonymous class for such a requirement.

If you want to use a simpler syntax you can still do :

List<String> list = new ArrayList<>(Arrays.asList("element A", "element B"));

edited 40 mins ago

answered 47 mins ago

davidxxx

58.2k54581

add a commentÂ |Â

up vote
0
down vote

In the first code you print the variable before that its constructor has returned as you add the elements and print the list inside the instance initializer.
So it can only print null.

To add elements in a List you generally want to invoke add() on the List reference such as :

List<String> list = new ArrayList<>();
list.add("element A");
list.add("element B");

You don't create an anonymous class for such a requirement.

If you want to use a simpler syntax you can still do :

List<String> list = new ArrayList<>(Arrays.asList("element A", "element B"));

edited 40 mins ago

answered 47 mins ago

davidxxx

58.2k54581

In the first code you print the variable before that its constructor has returned as you add the elements and print the list inside the instance initializer.
So it can only print null.

To add elements in a List you generally want to invoke add() on the List reference such as :

List<String> list = new ArrayList<>();
list.add("element A");
list.add("element B");

You don't create an anonymous class for such a requirement.

If you want to use a simpler syntax you can still do :

List<String> list = new ArrayList<>(Arrays.asList("element A", "element B"));

edited 40 mins ago

answered 47 mins ago

davidxxx

58.2k54581

edited 40 mins ago

answered 47 mins ago

davidxxx

58.2k54581

answered 47 mins ago

davidxxx

58.2k54581

answered 47 mins ago

davidxxx

58.2k54581

add a commentÂ |Â

up vote
-1
down vote

In you first code var is null because the return value of new operator is not assigned to var.
In the second code you override the add method and it is never called, to call that run this code:

public class Test
private static final List<String> var = new ArrayList<String>() 
 public boolean add(String e) 
 /*add("A");
 add("B");*/
 super.add(e);
 System.out.println("INNER : " + var); return true;
 ;
 
 add("A");
 add("B");
 
;

public static void main(String... args)
 System.out.println("OUTER : "+var);

answered 40 mins ago

Rahim Dastar

707515

add a commentÂ |Â

up vote
-1
down vote

In you first code var is null because the return value of new operator is not assigned to var.
In the second code you override the add method and it is never called, to call that run this code:

public class Test
private static final List<String> var = new ArrayList<String>() 
 public boolean add(String e) 
 /*add("A");
 add("B");*/
 super.add(e);
 System.out.println("INNER : " + var); return true;
 ;
 
 add("A");
 add("B");
 
;

public static void main(String... args)
 System.out.println("OUTER : "+var);

answered 40 mins ago

Rahim Dastar

707515

add a commentÂ |Â

up vote
-1
down vote

In you first code var is null because the return value of new operator is not assigned to var.
In the second code you override the add method and it is never called, to call that run this code:

public class Test
private static final List<String> var = new ArrayList<String>() 
 public boolean add(String e) 
 /*add("A");
 add("B");*/
 super.add(e);
 System.out.println("INNER : " + var); return true;
 ;
 
 add("A");
 add("B");
 
;

public static void main(String... args)
 System.out.println("OUTER : "+var);

answered 40 mins ago

Rahim Dastar

707515

In you first code var is null because the return value of new operator is not assigned to var.
In the second code you override the add method and it is never called, to call that run this code:

public class Test
private static final List<String> var = new ArrayList<String>() 
 public boolean add(String e) 
 /*add("A");
 add("B");*/
 super.add(e);
 System.out.println("INNER : " + var); return true;
 ;
 
 add("A");
 add("B");
 
;

public static void main(String... args)
 System.out.println("OUTER : "+var);

answered 40 mins ago

Rahim Dastar

707515

answered 40 mins ago

Rahim Dastar

707515

answered 40 mins ago

Rahim Dastar

707515

answered 40 mins ago

Rahim Dastar

707515

add a commentÂ |Â

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