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

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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
    1 hour 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.
    1 hour 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
    1 hour 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
    1 hour ago











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

    – Nathan Adams
    1 hour 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
    1 hour 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.
    1 hour 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
    1 hour 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
    1 hour ago











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

    – Nathan Adams
    1 hour 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 1 hour ago









Nathan AdamsNathan Adams

1638




1638












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

    – Nathan Adams
    1 hour 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.
    1 hour 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
    1 hour 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
    1 hour ago











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

    – Nathan Adams
    1 hour ago

















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

    – Nathan Adams
    1 hour 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.
    1 hour 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
    1 hour 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
    1 hour ago











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

    – Nathan Adams
    1 hour ago
















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

– Nathan Adams
1 hour ago





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

– Nathan Adams
1 hour 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.
1 hour 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.
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago













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

– Nathan Adams
1 hour ago





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

– Nathan Adams
1 hour ago












2 Answers
2






active

oldest

votes


















6














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
    48 mins ago











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

    – Ole V.V.
    37 mins ago


















0














Here:



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


You have a raw type, as you don't specify a generic type for your list object on the left hand side. Providing the generic type on the right hand side is mostly not required in the first place, as the compiler can defer the generic type.



A bit of guessing here: due to fact that list itself is a raw type, the type specification given to new is somehow ignored. The other answer nicely explains why you can have type parameters in that place. But because a raw type doesn't care about type parameters, the compiler doesn't bother checking that detail here.





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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6














    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
      48 mins ago











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

      – Ole V.V.
      37 mins ago















    6














    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
      48 mins ago











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

      – Ole V.V.
      37 mins ago













    6












    6








    6







    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 39 mins ago

























    answered 1 hour ago









    Ole V.V.Ole V.V.

    31.3k63956




    31.3k63956












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

      – jaspreet
      48 mins ago











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

      – Ole V.V.
      37 mins ago

















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

      – jaspreet
      48 mins ago











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

      – Ole V.V.
      37 mins ago
















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

    – jaspreet
    48 mins ago





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

    – jaspreet
    48 mins ago













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

    – Ole V.V.
    37 mins ago





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

    – Ole V.V.
    37 mins ago













    0














    Here:



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


    You have a raw type, as you don't specify a generic type for your list object on the left hand side. Providing the generic type on the right hand side is mostly not required in the first place, as the compiler can defer the generic type.



    A bit of guessing here: due to fact that list itself is a raw type, the type specification given to new is somehow ignored. The other answer nicely explains why you can have type parameters in that place. But because a raw type doesn't care about type parameters, the compiler doesn't bother checking that detail here.





    share



























      0














      Here:



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


      You have a raw type, as you don't specify a generic type for your list object on the left hand side. Providing the generic type on the right hand side is mostly not required in the first place, as the compiler can defer the generic type.



      A bit of guessing here: due to fact that list itself is a raw type, the type specification given to new is somehow ignored. The other answer nicely explains why you can have type parameters in that place. But because a raw type doesn't care about type parameters, the compiler doesn't bother checking that detail here.





      share

























        0












        0








        0







        Here:



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


        You have a raw type, as you don't specify a generic type for your list object on the left hand side. Providing the generic type on the right hand side is mostly not required in the first place, as the compiler can defer the generic type.



        A bit of guessing here: due to fact that list itself is a raw type, the type specification given to new is somehow ignored. The other answer nicely explains why you can have type parameters in that place. But because a raw type doesn't care about type parameters, the compiler doesn't bother checking that detail here.





        share













        Here:



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


        You have a raw type, as you don't specify a generic type for your list object on the left hand side. Providing the generic type on the right hand side is mostly not required in the first place, as the compiler can defer the generic type.



        A bit of guessing here: due to fact that list itself is a raw type, the type specification given to new is somehow ignored. The other answer nicely explains why you can have type parameters in that place. But because a raw type doesn't care about type parameters, the compiler doesn't bother checking that detail here.






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