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Java - What do constructor type arguments mean when placed *before* the type?


How to round a number to n decimal places in JavaWhat is the difference between public, protected, package-private and private in Java?How do I call one constructor from another in Java?When to use LinkedList over ArrayList in Java?What does 'synchronized' mean?What is the Java equivalent for LINQ?What is a daemon thread in Java?Java: when to use static methodsWhat are the -Xms and -Xmx parameters when starting JVM?What does “Could not find or load main class” mean?













7















I've recently come across this unusual (to me) Java syntax...here's an example of it:



List list = new <String, Long>ArrayList();


Notice the positioning of the <String, Long> type arguments...it's not after the type as normal but before. I don't mind admitting I've never seen this syntax before. Also note there are 2 type arguments when ArrayList only has 1.



Does the positioning of the type arguments have the same meaning as putting them after the type? If not, what does the different positioning mean?



Why is it legal to have 2 type arguments when ArrayList only has 1?



I've searched the usual places, eg. Angelika Langer and on here but can't find any mention of this syntax anywhere apart from the grammar rules in the Java grammar file on the ANTLR project.










share|improve this question






















  • Yeah so do I, I'm not asking how to create a list lol

    – Nathan Adams
    55 mins ago






  • 2





    A constructor may have type arguments that are placed there (this particular constructor hasn’t, so <String, Long> is just ignored). See Generics Constructor.

    – Ole V.V.
    55 mins ago







  • 1





    OK that makes sense, although it's weird that there's no compile error even though there's no type arguments on the constructor

    – Nathan Adams
    50 mins ago






  • 1





    No error, but you get a warning about raw types. Don't use raw types. Do use the diamond operator. List<String> list = new ArrayList<>();

    – Elliott Frisch
    48 mins ago











  • @OleV.V. if you wanna put your comment and link as an answer I'll accept it

    – Nathan Adams
    47 mins ago















7















I've recently come across this unusual (to me) Java syntax...here's an example of it:



List list = new <String, Long>ArrayList();


Notice the positioning of the <String, Long> type arguments...it's not after the type as normal but before. I don't mind admitting I've never seen this syntax before. Also note there are 2 type arguments when ArrayList only has 1.



Does the positioning of the type arguments have the same meaning as putting them after the type? If not, what does the different positioning mean?



Why is it legal to have 2 type arguments when ArrayList only has 1?



I've searched the usual places, eg. Angelika Langer and on here but can't find any mention of this syntax anywhere apart from the grammar rules in the Java grammar file on the ANTLR project.










share|improve this question






















  • Yeah so do I, I'm not asking how to create a list lol

    – Nathan Adams
    55 mins ago






  • 2





    A constructor may have type arguments that are placed there (this particular constructor hasn’t, so <String, Long> is just ignored). See Generics Constructor.

    – Ole V.V.
    55 mins ago







  • 1





    OK that makes sense, although it's weird that there's no compile error even though there's no type arguments on the constructor

    – Nathan Adams
    50 mins ago






  • 1





    No error, but you get a warning about raw types. Don't use raw types. Do use the diamond operator. List<String> list = new ArrayList<>();

    – Elliott Frisch
    48 mins ago











  • @OleV.V. if you wanna put your comment and link as an answer I'll accept it

    – Nathan Adams
    47 mins ago













7












7








7


1






I've recently come across this unusual (to me) Java syntax...here's an example of it:



List list = new <String, Long>ArrayList();


Notice the positioning of the <String, Long> type arguments...it's not after the type as normal but before. I don't mind admitting I've never seen this syntax before. Also note there are 2 type arguments when ArrayList only has 1.



Does the positioning of the type arguments have the same meaning as putting them after the type? If not, what does the different positioning mean?



Why is it legal to have 2 type arguments when ArrayList only has 1?



I've searched the usual places, eg. Angelika Langer and on here but can't find any mention of this syntax anywhere apart from the grammar rules in the Java grammar file on the ANTLR project.










share|improve this question














I've recently come across this unusual (to me) Java syntax...here's an example of it:



List list = new <String, Long>ArrayList();


Notice the positioning of the <String, Long> type arguments...it's not after the type as normal but before. I don't mind admitting I've never seen this syntax before. Also note there are 2 type arguments when ArrayList only has 1.



Does the positioning of the type arguments have the same meaning as putting them after the type? If not, what does the different positioning mean?



Why is it legal to have 2 type arguments when ArrayList only has 1?



I've searched the usual places, eg. Angelika Langer and on here but can't find any mention of this syntax anywhere apart from the grammar rules in the Java grammar file on the ANTLR project.







java grammar






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 59 mins ago









Nathan AdamsNathan Adams

1638




1638












  • Yeah so do I, I'm not asking how to create a list lol

    – Nathan Adams
    55 mins ago






  • 2





    A constructor may have type arguments that are placed there (this particular constructor hasn’t, so <String, Long> is just ignored). See Generics Constructor.

    – Ole V.V.
    55 mins ago







  • 1





    OK that makes sense, although it's weird that there's no compile error even though there's no type arguments on the constructor

    – Nathan Adams
    50 mins ago






  • 1





    No error, but you get a warning about raw types. Don't use raw types. Do use the diamond operator. List<String> list = new ArrayList<>();

    – Elliott Frisch
    48 mins ago











  • @OleV.V. if you wanna put your comment and link as an answer I'll accept it

    – Nathan Adams
    47 mins ago

















  • Yeah so do I, I'm not asking how to create a list lol

    – Nathan Adams
    55 mins ago






  • 2





    A constructor may have type arguments that are placed there (this particular constructor hasn’t, so <String, Long> is just ignored). See Generics Constructor.

    – Ole V.V.
    55 mins ago







  • 1





    OK that makes sense, although it's weird that there's no compile error even though there's no type arguments on the constructor

    – Nathan Adams
    50 mins ago






  • 1





    No error, but you get a warning about raw types. Don't use raw types. Do use the diamond operator. List<String> list = new ArrayList<>();

    – Elliott Frisch
    48 mins ago











  • @OleV.V. if you wanna put your comment and link as an answer I'll accept it

    – Nathan Adams
    47 mins ago
















Yeah so do I, I'm not asking how to create a list lol

– Nathan Adams
55 mins ago





Yeah so do I, I'm not asking how to create a list lol

– Nathan Adams
55 mins ago




2




2





A constructor may have type arguments that are placed there (this particular constructor hasn’t, so <String, Long> is just ignored). See Generics Constructor.

– Ole V.V.
55 mins ago






A constructor may have type arguments that are placed there (this particular constructor hasn’t, so <String, Long> is just ignored). See Generics Constructor.

– Ole V.V.
55 mins ago





1




1





OK that makes sense, although it's weird that there's no compile error even though there's no type arguments on the constructor

– Nathan Adams
50 mins ago





OK that makes sense, although it's weird that there's no compile error even though there's no type arguments on the constructor

– Nathan Adams
50 mins ago




1




1





No error, but you get a warning about raw types. Don't use raw types. Do use the diamond operator. List<String> list = new ArrayList<>();

– Elliott Frisch
48 mins ago





No error, but you get a warning about raw types. Don't use raw types. Do use the diamond operator. List<String> list = new ArrayList<>();

– Elliott Frisch
48 mins ago













@OleV.V. if you wanna put your comment and link as an answer I'll accept it

– Nathan Adams
47 mins ago





@OleV.V. if you wanna put your comment and link as an answer I'll accept it

– Nathan Adams
47 mins ago












1 Answer
1






active

oldest

votes


















5














This is unusual alright, but fully valid Java. A class may have a generic constructor, for example:



public class TypeWithGenericConstructor 

public <T> TypeWithGenericConstructor(T arg)
// TODO Auto-generated constructor stub





I suppose that more often than not we don’t need to make the type argument explicit. For example:



 new TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));


Now T is clearly LocalDate. However there may be cases where Java cannot infer (deduce) the type argument. Then we supply it explicitly using the syntax from your question:



 new <LocalDate>TypeWithGenericConstructor(null);


Of course we may also supply it even though it is not necessary if we think it helps readability or for whatever reason:



 new <LocalDate>TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));


In your question you seem to be calling the java.util.ArrayList constructor. That constructor is not generic (only the ArrayList class as a whole is, that’s something else). Why Java allows you to supply type arguments in the call when they are not used, I don’t know, but it does. My Eclipse gives me a warning:




Unused type arguments for the non generic constructor ArrayList() of
type ArrayList; it should not be parameterized with arguments




But it’s not an error, and the program runs fine (I additionally get warnings about missing type arguments for List and ArrayList, but that again is a different story).




Does the positioning of the type arguments have the same meaning as
putting them after the type? If not, what does the different
positioning mean?




No, it’s different. The usual type argument/s after the type (ArrayList<Integer>) are for the generic class. The type arguments before are for the * constructor*.



The two forms may also be combined:



 List<Integer> list = new <String, Long>ArrayList<Integer>();


I would consider this a bit more correct since we can now see that the list stores Integer objects (I’d still prefer to leave out the meaningless <String, Long>, of course).




Why is it legal to have 2 type arguments when ArrayList only has 1?




First, if you supply type arguments before the type, you should supply the correct number for the constructor, not for the class, so it hasn’t got anything to do with how many type arguments the ArrayList class has got. That really means that in this case you shouldn’t supply any since the constructor doesn’t take type arguments (it’s not generic). When you supply some anyway, they are ignored, which is why it doesn’t matter how many or how few you supply (I repeat, I don’t know why Java allows you to supply them meaninglessly).






share|improve this answer

























  • But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?

    – jaspreet
    24 mins ago











  • Thanks for asking, @jaspreet. Please see my edit.

    – Ole V.V.
    13 mins ago










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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5














This is unusual alright, but fully valid Java. A class may have a generic constructor, for example:



public class TypeWithGenericConstructor 

public <T> TypeWithGenericConstructor(T arg)
// TODO Auto-generated constructor stub





I suppose that more often than not we don’t need to make the type argument explicit. For example:



 new TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));


Now T is clearly LocalDate. However there may be cases where Java cannot infer (deduce) the type argument. Then we supply it explicitly using the syntax from your question:



 new <LocalDate>TypeWithGenericConstructor(null);


Of course we may also supply it even though it is not necessary if we think it helps readability or for whatever reason:



 new <LocalDate>TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));


In your question you seem to be calling the java.util.ArrayList constructor. That constructor is not generic (only the ArrayList class as a whole is, that’s something else). Why Java allows you to supply type arguments in the call when they are not used, I don’t know, but it does. My Eclipse gives me a warning:




Unused type arguments for the non generic constructor ArrayList() of
type ArrayList; it should not be parameterized with arguments




But it’s not an error, and the program runs fine (I additionally get warnings about missing type arguments for List and ArrayList, but that again is a different story).




Does the positioning of the type arguments have the same meaning as
putting them after the type? If not, what does the different
positioning mean?




No, it’s different. The usual type argument/s after the type (ArrayList<Integer>) are for the generic class. The type arguments before are for the * constructor*.



The two forms may also be combined:



 List<Integer> list = new <String, Long>ArrayList<Integer>();


I would consider this a bit more correct since we can now see that the list stores Integer objects (I’d still prefer to leave out the meaningless <String, Long>, of course).




Why is it legal to have 2 type arguments when ArrayList only has 1?




First, if you supply type arguments before the type, you should supply the correct number for the constructor, not for the class, so it hasn’t got anything to do with how many type arguments the ArrayList class has got. That really means that in this case you shouldn’t supply any since the constructor doesn’t take type arguments (it’s not generic). When you supply some anyway, they are ignored, which is why it doesn’t matter how many or how few you supply (I repeat, I don’t know why Java allows you to supply them meaninglessly).






share|improve this answer

























  • But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?

    – jaspreet
    24 mins ago











  • Thanks for asking, @jaspreet. Please see my edit.

    – Ole V.V.
    13 mins ago















5














This is unusual alright, but fully valid Java. A class may have a generic constructor, for example:



public class TypeWithGenericConstructor 

public <T> TypeWithGenericConstructor(T arg)
// TODO Auto-generated constructor stub





I suppose that more often than not we don’t need to make the type argument explicit. For example:



 new TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));


Now T is clearly LocalDate. However there may be cases where Java cannot infer (deduce) the type argument. Then we supply it explicitly using the syntax from your question:



 new <LocalDate>TypeWithGenericConstructor(null);


Of course we may also supply it even though it is not necessary if we think it helps readability or for whatever reason:



 new <LocalDate>TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));


In your question you seem to be calling the java.util.ArrayList constructor. That constructor is not generic (only the ArrayList class as a whole is, that’s something else). Why Java allows you to supply type arguments in the call when they are not used, I don’t know, but it does. My Eclipse gives me a warning:




Unused type arguments for the non generic constructor ArrayList() of
type ArrayList; it should not be parameterized with arguments




But it’s not an error, and the program runs fine (I additionally get warnings about missing type arguments for List and ArrayList, but that again is a different story).




Does the positioning of the type arguments have the same meaning as
putting them after the type? If not, what does the different
positioning mean?




No, it’s different. The usual type argument/s after the type (ArrayList<Integer>) are for the generic class. The type arguments before are for the * constructor*.



The two forms may also be combined:



 List<Integer> list = new <String, Long>ArrayList<Integer>();


I would consider this a bit more correct since we can now see that the list stores Integer objects (I’d still prefer to leave out the meaningless <String, Long>, of course).




Why is it legal to have 2 type arguments when ArrayList only has 1?




First, if you supply type arguments before the type, you should supply the correct number for the constructor, not for the class, so it hasn’t got anything to do with how many type arguments the ArrayList class has got. That really means that in this case you shouldn’t supply any since the constructor doesn’t take type arguments (it’s not generic). When you supply some anyway, they are ignored, which is why it doesn’t matter how many or how few you supply (I repeat, I don’t know why Java allows you to supply them meaninglessly).






share|improve this answer

























  • But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?

    – jaspreet
    24 mins ago











  • Thanks for asking, @jaspreet. Please see my edit.

    – Ole V.V.
    13 mins ago













5












5








5







This is unusual alright, but fully valid Java. A class may have a generic constructor, for example:



public class TypeWithGenericConstructor 

public <T> TypeWithGenericConstructor(T arg)
// TODO Auto-generated constructor stub





I suppose that more often than not we don’t need to make the type argument explicit. For example:



 new TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));


Now T is clearly LocalDate. However there may be cases where Java cannot infer (deduce) the type argument. Then we supply it explicitly using the syntax from your question:



 new <LocalDate>TypeWithGenericConstructor(null);


Of course we may also supply it even though it is not necessary if we think it helps readability or for whatever reason:



 new <LocalDate>TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));


In your question you seem to be calling the java.util.ArrayList constructor. That constructor is not generic (only the ArrayList class as a whole is, that’s something else). Why Java allows you to supply type arguments in the call when they are not used, I don’t know, but it does. My Eclipse gives me a warning:




Unused type arguments for the non generic constructor ArrayList() of
type ArrayList; it should not be parameterized with arguments




But it’s not an error, and the program runs fine (I additionally get warnings about missing type arguments for List and ArrayList, but that again is a different story).




Does the positioning of the type arguments have the same meaning as
putting them after the type? If not, what does the different
positioning mean?




No, it’s different. The usual type argument/s after the type (ArrayList<Integer>) are for the generic class. The type arguments before are for the * constructor*.



The two forms may also be combined:



 List<Integer> list = new <String, Long>ArrayList<Integer>();


I would consider this a bit more correct since we can now see that the list stores Integer objects (I’d still prefer to leave out the meaningless <String, Long>, of course).




Why is it legal to have 2 type arguments when ArrayList only has 1?




First, if you supply type arguments before the type, you should supply the correct number for the constructor, not for the class, so it hasn’t got anything to do with how many type arguments the ArrayList class has got. That really means that in this case you shouldn’t supply any since the constructor doesn’t take type arguments (it’s not generic). When you supply some anyway, they are ignored, which is why it doesn’t matter how many or how few you supply (I repeat, I don’t know why Java allows you to supply them meaninglessly).






share|improve this answer















This is unusual alright, but fully valid Java. A class may have a generic constructor, for example:



public class TypeWithGenericConstructor 

public <T> TypeWithGenericConstructor(T arg)
// TODO Auto-generated constructor stub





I suppose that more often than not we don’t need to make the type argument explicit. For example:



 new TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));


Now T is clearly LocalDate. However there may be cases where Java cannot infer (deduce) the type argument. Then we supply it explicitly using the syntax from your question:



 new <LocalDate>TypeWithGenericConstructor(null);


Of course we may also supply it even though it is not necessary if we think it helps readability or for whatever reason:



 new <LocalDate>TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));


In your question you seem to be calling the java.util.ArrayList constructor. That constructor is not generic (only the ArrayList class as a whole is, that’s something else). Why Java allows you to supply type arguments in the call when they are not used, I don’t know, but it does. My Eclipse gives me a warning:




Unused type arguments for the non generic constructor ArrayList() of
type ArrayList; it should not be parameterized with arguments




But it’s not an error, and the program runs fine (I additionally get warnings about missing type arguments for List and ArrayList, but that again is a different story).




Does the positioning of the type arguments have the same meaning as
putting them after the type? If not, what does the different
positioning mean?




No, it’s different. The usual type argument/s after the type (ArrayList<Integer>) are for the generic class. The type arguments before are for the * constructor*.



The two forms may also be combined:



 List<Integer> list = new <String, Long>ArrayList<Integer>();


I would consider this a bit more correct since we can now see that the list stores Integer objects (I’d still prefer to leave out the meaningless <String, Long>, of course).




Why is it legal to have 2 type arguments when ArrayList only has 1?




First, if you supply type arguments before the type, you should supply the correct number for the constructor, not for the class, so it hasn’t got anything to do with how many type arguments the ArrayList class has got. That really means that in this case you shouldn’t supply any since the constructor doesn’t take type arguments (it’s not generic). When you supply some anyway, they are ignored, which is why it doesn’t matter how many or how few you supply (I repeat, I don’t know why Java allows you to supply them meaninglessly).







share|improve this answer














share|improve this answer



share|improve this answer








edited 15 mins ago

























answered 40 mins ago









Ole V.V.Ole V.V.

31.2k63956




31.2k63956












  • But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?

    – jaspreet
    24 mins ago











  • Thanks for asking, @jaspreet. Please see my edit.

    – Ole V.V.
    13 mins ago

















  • But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?

    – jaspreet
    24 mins ago











  • Thanks for asking, @jaspreet. Please see my edit.

    – Ole V.V.
    13 mins ago
















But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?

– jaspreet
24 mins ago





But how it is allowing 2 arguments <String, Long>.The list will allow storing which type of data?

– jaspreet
24 mins ago













Thanks for asking, @jaspreet. Please see my edit.

– Ole V.V.
13 mins ago





Thanks for asking, @jaspreet. Please see my edit.

– Ole V.V.
13 mins ago



















draft saved

draft discarded
















































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