“new” Keyword In Java Lambda Method Reference [duplicate]

8

This question already has an answer here:

• 
  Reference to an instance method of a particular object
  
  6 answers
  
  

I've seen a lot of methods where a new class is instantiated in a lambda method reference but can't seem to understand why. When is the new keyword needed in a method reference?

For example, the following passes compilation:

UnaryOperator<String>stringToUpperCase = String::toUpperCase;

But this doesn't:

UnaryOperator<String>stringToUpperCase = new String()::toUpperCase; 

java lambda java-8 method-reference

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edited Nov 21 '18 at 13:50

Eran

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asked Nov 21 '18 at 12:46

Clatty Cake

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Nov 21 '18 at 16:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  a new String in upper case is still just a blank string, so s -> "" will do the same thing
  – Michael
  Nov 21 '18 at 12:49
  

  
  
  
  

add a comment | 

8

This question already has an answer here:

• 
  Reference to an instance method of a particular object
  
  6 answers
  
  

I've seen a lot of methods where a new class is instantiated in a lambda method reference but can't seem to understand why. When is the new keyword needed in a method reference?

For example, the following passes compilation:

UnaryOperator<String>stringToUpperCase = String::toUpperCase;

But this doesn't:

UnaryOperator<String>stringToUpperCase = new String()::toUpperCase; 

java lambda java-8 method-reference

share|improve this question

edited Nov 21 '18 at 13:50

Eran

280k37453539

asked Nov 21 '18 at 12:46

Clatty Cake

3293511

marked as duplicate by Federico Peralta Schaffner java
Users with the  java badge can single-handedly close java questions as duplicates and reopen them as needed.

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Nov 21 '18 at 16:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  a new String in upper case is still just a blank string, so s -> "" will do the same thing
  – Michael
  Nov 21 '18 at 12:49
  

  
  
  
  

add a comment | 

8

8

8

2

This question already has an answer here:

• 
  Reference to an instance method of a particular object
  
  6 answers
  
  

I've seen a lot of methods where a new class is instantiated in a lambda method reference but can't seem to understand why. When is the new keyword needed in a method reference?

For example, the following passes compilation:

UnaryOperator<String>stringToUpperCase = String::toUpperCase;

But this doesn't:

UnaryOperator<String>stringToUpperCase = new String()::toUpperCase; 

java lambda java-8 method-reference

share|improve this question

edited Nov 21 '18 at 13:50

Eran

280k37453539

asked Nov 21 '18 at 12:46

Clatty Cake

3293511

This question already has an answer here:

• 
  Reference to an instance method of a particular object
  
  6 answers
  
  

I've seen a lot of methods where a new class is instantiated in a lambda method reference but can't seem to understand why. When is the new keyword needed in a method reference?

For example, the following passes compilation:

UnaryOperator<String>stringToUpperCase = String::toUpperCase;

But this doesn't:

UnaryOperator<String>stringToUpperCase = new String()::toUpperCase; 

This question already has an answer here:

• 
  Reference to an instance method of a particular object
  
  6 answers
  
  

java lambda java-8 method-reference

java lambda java-8 method-reference

share|improve this question

edited Nov 21 '18 at 13:50

Eran

280k37453539

asked Nov 21 '18 at 12:46

Clatty Cake

3293511

share|improve this question

edited Nov 21 '18 at 13:50

Eran

280k37453539

asked Nov 21 '18 at 12:46

Clatty Cake

3293511

share|improve this question

share|improve this question

edited Nov 21 '18 at 13:50

Eran

280k37453539

edited Nov 21 '18 at 13:50

Eran

280k37453539

edited Nov 21 '18 at 13:50

Eran

280k37453539

280k37453539

asked Nov 21 '18 at 12:46

Clatty Cake

3293511

asked Nov 21 '18 at 12:46

Clatty Cake

3293511

asked Nov 21 '18 at 12:46

Clatty Cake

3293511

3293511

marked as duplicate by Federico Peralta Schaffner java
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Nov 21 '18 at 16:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Federico Peralta Schaffner java
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Nov 21 '18 at 16:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  a new String in upper case is still just a blank string, so s -> "" will do the same thing
  – Michael
  Nov 21 '18 at 12:49
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  a new String in upper case is still just a blank string, so s -> "" will do the same thing
  – Michael
  Nov 21 '18 at 12:49
  

  
  
  
  

4

4

a new String in upper case is still just a blank string, so s -> "" will do the same thing
– Michael
Nov 21 '18 at 12:49

a new String in upper case is still just a blank string, so s -> "" will do the same thing
– Michael
Nov 21 '18 at 12:49

add a comment | 

2 Answers
2

active

oldest

votes

16

String::toUpperCase is a method reference that can be applied to any String instance.

new String()::toUpperCase is a method reference that can be applied to a specific String instance (the instance created by new String()).

Since UnaryOperator<String> expects a method that takes a String and returns a String, String::toUpperCase fits (since you can apply it on a String and get the upper case version of that String).

On the other hand, new String()::toUpperCase doesn't fit UnaryOperator<String>, since it is executed on an already specified String, so you can't pass another String instance to it.

It can, however, by assigned to a Supplier<String>, since it simply supplies an empty String instance:

Supplier<String> emptyStringToUpperCase = new String()::toUpperCase; 

This is similar to:

Supplier<String> emptyStringToUpperCase = () -> new String().toUpperCase();

while this:

UnaryOperator<String> stringToUpperCase = String::toUpperCase;

is similar to:

UnaryOperator<String> stringToUpperCase = s -> s.toUpperCase();

share|improve this answer

edited Nov 21 '18 at 12:56

answered Nov 21 '18 at 12:51

Eran

280k37453539

add a comment | 

5

There are four kinds of method references as shown below and your type falls in the second category, but UnaryOperator<String> essentially needs to represent a method which accepts any String argument and returns a String. However, the non-working method reference that you have used is actually working on a particular String object (i.e. not any String object)

Refer: https://docs.oracle.com/javase/tutorial/java/javaOO/methodreferences.html

share|improve this answer

edited Nov 21 '18 at 13:11

answered Nov 21 '18 at 12:51

Ankur Chrungoo

62039

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Strictly speaking, new String()::toUpperCase is indeed a method reference, of the third kind (new String() is an object which has the toUpperCase method). It doesn't take arguments, but returns a String. It could be used as a Supplier<String>. But it is a very complicated way to say () -> "".
  – glglgl
  Nov 21 '18 at 12:53
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @glglgl Actually, the second type, right?
  – Ankur Chrungoo
  Nov 21 '18 at 12:57
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  With 0-based counting even the 1st :P
  – Max Vollmer
  Nov 21 '18 at 13:00
  
  
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

16

String::toUpperCase is a method reference that can be applied to any String instance.

new String()::toUpperCase is a method reference that can be applied to a specific String instance (the instance created by new String()).

Since UnaryOperator<String> expects a method that takes a String and returns a String, String::toUpperCase fits (since you can apply it on a String and get the upper case version of that String).

On the other hand, new String()::toUpperCase doesn't fit UnaryOperator<String>, since it is executed on an already specified String, so you can't pass another String instance to it.

It can, however, by assigned to a Supplier<String>, since it simply supplies an empty String instance:

Supplier<String> emptyStringToUpperCase = new String()::toUpperCase; 

This is similar to:

Supplier<String> emptyStringToUpperCase = () -> new String().toUpperCase();

while this:

UnaryOperator<String> stringToUpperCase = String::toUpperCase;

is similar to:

UnaryOperator<String> stringToUpperCase = s -> s.toUpperCase();

share|improve this answer

edited Nov 21 '18 at 12:56

answered Nov 21 '18 at 12:51

Eran

280k37453539

add a comment | 

16

String::toUpperCase is a method reference that can be applied to any String instance.

new String()::toUpperCase is a method reference that can be applied to a specific String instance (the instance created by new String()).

Since UnaryOperator<String> expects a method that takes a String and returns a String, String::toUpperCase fits (since you can apply it on a String and get the upper case version of that String).

On the other hand, new String()::toUpperCase doesn't fit UnaryOperator<String>, since it is executed on an already specified String, so you can't pass another String instance to it.

It can, however, by assigned to a Supplier<String>, since it simply supplies an empty String instance:

Supplier<String> emptyStringToUpperCase = new String()::toUpperCase; 

This is similar to:

Supplier<String> emptyStringToUpperCase = () -> new String().toUpperCase();

while this:

UnaryOperator<String> stringToUpperCase = String::toUpperCase;

is similar to:

UnaryOperator<String> stringToUpperCase = s -> s.toUpperCase();

share|improve this answer

edited Nov 21 '18 at 12:56

answered Nov 21 '18 at 12:51

Eran

280k37453539

add a comment | 

16

16

16

String::toUpperCase is a method reference that can be applied to any String instance.

new String()::toUpperCase is a method reference that can be applied to a specific String instance (the instance created by new String()).

Since UnaryOperator<String> expects a method that takes a String and returns a String, String::toUpperCase fits (since you can apply it on a String and get the upper case version of that String).

On the other hand, new String()::toUpperCase doesn't fit UnaryOperator<String>, since it is executed on an already specified String, so you can't pass another String instance to it.

It can, however, by assigned to a Supplier<String>, since it simply supplies an empty String instance:

Supplier<String> emptyStringToUpperCase = new String()::toUpperCase; 

This is similar to:

Supplier<String> emptyStringToUpperCase = () -> new String().toUpperCase();

while this:

UnaryOperator<String> stringToUpperCase = String::toUpperCase;

is similar to:

UnaryOperator<String> stringToUpperCase = s -> s.toUpperCase();

share|improve this answer

edited Nov 21 '18 at 12:56

answered Nov 21 '18 at 12:51

Eran

280k37453539

String::toUpperCase is a method reference that can be applied to any String instance.

new String()::toUpperCase is a method reference that can be applied to a specific String instance (the instance created by new String()).

Since UnaryOperator<String> expects a method that takes a String and returns a String, String::toUpperCase fits (since you can apply it on a String and get the upper case version of that String).

On the other hand, new String()::toUpperCase doesn't fit UnaryOperator<String>, since it is executed on an already specified String, so you can't pass another String instance to it.

It can, however, by assigned to a Supplier<String>, since it simply supplies an empty String instance:

Supplier<String> emptyStringToUpperCase = new String()::toUpperCase; 

This is similar to:

Supplier<String> emptyStringToUpperCase = () -> new String().toUpperCase();

while this:

UnaryOperator<String> stringToUpperCase = String::toUpperCase;

is similar to:

UnaryOperator<String> stringToUpperCase = s -> s.toUpperCase();

share|improve this answer

edited Nov 21 '18 at 12:56

answered Nov 21 '18 at 12:51

Eran

280k37453539

share|improve this answer

share|improve this answer

edited Nov 21 '18 at 12:56

edited Nov 21 '18 at 12:56

edited Nov 21 '18 at 12:56

answered Nov 21 '18 at 12:51

Eran

280k37453539

answered Nov 21 '18 at 12:51

Eran

280k37453539

answered Nov 21 '18 at 12:51

Eran

280k37453539

280k37453539

add a comment | 

add a comment | 

5

There are four kinds of method references as shown below and your type falls in the second category, but UnaryOperator<String> essentially needs to represent a method which accepts any String argument and returns a String. However, the non-working method reference that you have used is actually working on a particular String object (i.e. not any String object)

Refer: https://docs.oracle.com/javase/tutorial/java/javaOO/methodreferences.html

share|improve this answer

edited Nov 21 '18 at 13:11

answered Nov 21 '18 at 12:51

Ankur Chrungoo

62039

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Strictly speaking, new String()::toUpperCase is indeed a method reference, of the third kind (new String() is an object which has the toUpperCase method). It doesn't take arguments, but returns a String. It could be used as a Supplier<String>. But it is a very complicated way to say () -> "".
  – glglgl
  Nov 21 '18 at 12:53
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @glglgl Actually, the second type, right?
  – Ankur Chrungoo
  Nov 21 '18 at 12:57
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  With 0-based counting even the 1st :P
  – Max Vollmer
  Nov 21 '18 at 13:00
  
  
  

  
  
  
  

add a comment | 

5

There are four kinds of method references as shown below and your type falls in the second category, but UnaryOperator<String> essentially needs to represent a method which accepts any String argument and returns a String. However, the non-working method reference that you have used is actually working on a particular String object (i.e. not any String object)

Refer: https://docs.oracle.com/javase/tutorial/java/javaOO/methodreferences.html

share|improve this answer

edited Nov 21 '18 at 13:11

answered Nov 21 '18 at 12:51

Ankur Chrungoo

62039

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Strictly speaking, new String()::toUpperCase is indeed a method reference, of the third kind (new String() is an object which has the toUpperCase method). It doesn't take arguments, but returns a String. It could be used as a Supplier<String>. But it is a very complicated way to say () -> "".
  – glglgl
  Nov 21 '18 at 12:53
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @glglgl Actually, the second type, right?
  – Ankur Chrungoo
  Nov 21 '18 at 12:57
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  With 0-based counting even the 1st :P
  – Max Vollmer
  Nov 21 '18 at 13:00
  
  
  

  
  
  
  

add a comment | 

5

5

5

There are four kinds of method references as shown below and your type falls in the second category, but UnaryOperator<String> essentially needs to represent a method which accepts any String argument and returns a String. However, the non-working method reference that you have used is actually working on a particular String object (i.e. not any String object)

Refer: https://docs.oracle.com/javase/tutorial/java/javaOO/methodreferences.html

share|improve this answer

edited Nov 21 '18 at 13:11

answered Nov 21 '18 at 12:51

Ankur Chrungoo

62039

There are four kinds of method references as shown below and your type falls in the second category, but UnaryOperator<String> essentially needs to represent a method which accepts any String argument and returns a String. However, the non-working method reference that you have used is actually working on a particular String object (i.e. not any String object)

Refer: https://docs.oracle.com/javase/tutorial/java/javaOO/methodreferences.html

share|improve this answer

edited Nov 21 '18 at 13:11

answered Nov 21 '18 at 12:51

Ankur Chrungoo

62039

share|improve this answer

share|improve this answer

edited Nov 21 '18 at 13:11

edited Nov 21 '18 at 13:11

edited Nov 21 '18 at 13:11

answered Nov 21 '18 at 12:51

Ankur Chrungoo

62039

answered Nov 21 '18 at 12:51

Ankur Chrungoo

62039

answered Nov 21 '18 at 12:51

Ankur Chrungoo

62039

62039

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Strictly speaking, new String()::toUpperCase is indeed a method reference, of the third kind (new String() is an object which has the toUpperCase method). It doesn't take arguments, but returns a String. It could be used as a Supplier<String>. But it is a very complicated way to say () -> "".
  – glglgl
  Nov 21 '18 at 12:53
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @glglgl Actually, the second type, right?
  – Ankur Chrungoo
  Nov 21 '18 at 12:57
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  With 0-based counting even the 1st :P
  – Max Vollmer
  Nov 21 '18 at 13:00
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Strictly speaking, new String()::toUpperCase is indeed a method reference, of the third kind (new String() is an object which has the toUpperCase method). It doesn't take arguments, but returns a String. It could be used as a Supplier<String>. But it is a very complicated way to say () -> "".
  – glglgl
  Nov 21 '18 at 12:53
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @glglgl Actually, the second type, right?
  – Ankur Chrungoo
  Nov 21 '18 at 12:57
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  With 0-based counting even the 1st :P
  – Max Vollmer
  Nov 21 '18 at 13:00
  
  
  

  
  
  
  

1

1

Strictly speaking, new String()::toUpperCase is indeed a method reference, of the third kind (new String() is an object which has the toUpperCase method). It doesn't take arguments, but returns a String. It could be used as a Supplier<String>. But it is a very complicated way to say () -> "".
– glglgl
Nov 21 '18 at 12:53

Strictly speaking, new String()::toUpperCase is indeed a method reference, of the third kind (new String() is an object which has the toUpperCase method). It doesn't take arguments, but returns a String. It could be used as a Supplier<String>. But it is a very complicated way to say () -> "".
– glglgl
Nov 21 '18 at 12:53

3

3

@glglgl Actually, the second type, right?
– Ankur Chrungoo
Nov 21 '18 at 12:57

@glglgl Actually, the second type, right?
– Ankur Chrungoo
Nov 21 '18 at 12:57

With 0-based counting even the 1st :P
– Max Vollmer
Nov 21 '18 at 13:00

With 0-based counting even the 1st :P
– Max Vollmer
Nov 21 '18 at 13:00

add a comment | 

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