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












8















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
    2 hours 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.
    2 hours 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
    2 hours 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
    2 hours ago











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

    – Nathan Adams
    2 hours ago
















8















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
    2 hours 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.
    2 hours 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
    2 hours 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
    2 hours ago











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

    – Nathan Adams
    2 hours ago














8












8








8


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 2 hours ago









Nathan AdamsNathan Adams

1688




1688













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

    – Nathan Adams
    2 hours 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.
    2 hours 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
    2 hours 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
    2 hours ago











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

    – Nathan Adams
    2 hours ago



















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

    – Nathan Adams
    2 hours 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.
    2 hours 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
    2 hours 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
    2 hours ago











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

    – Nathan Adams
    2 hours ago

















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

– Nathan Adams
2 hours ago





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

– Nathan Adams
2 hours 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.
2 hours 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.
2 hours 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
2 hours 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
2 hours 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
2 hours 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
2 hours ago













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

– Nathan Adams
2 hours ago





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

– Nathan Adams
2 hours ago












2 Answers
2






active

oldest

votes


















7














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
    1 hour ago











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

    – Ole V.V.
    1 hour ago











  • What's interesting to me, and sorry if I missed you already mentioning/implying this, is if the constructor (or method) is generic, having the incorrect number of type arguments results in a compilation error—at least for me in OpenJDK 12. For example, if you have class Test { <T> Test() {} } then calling new <String, Long>Test() is an error. Yet when the constructor (or method) is not generic it lets you add type arguments to your hearts desire... maybe this should be considered a bug (if not in the compiler then in the specification)?

    – Slaw
    25 mins ago













  • Hmm... maybe not a bug. See stackoverflow.com/questions/28014853/…

    – Slaw
    12 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
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    active

    oldest

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    oldest

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    active

    oldest

    votes









    7














    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
      1 hour ago











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

      – Ole V.V.
      1 hour ago











    • What's interesting to me, and sorry if I missed you already mentioning/implying this, is if the constructor (or method) is generic, having the incorrect number of type arguments results in a compilation error—at least for me in OpenJDK 12. For example, if you have class Test { <T> Test() {} } then calling new <String, Long>Test() is an error. Yet when the constructor (or method) is not generic it lets you add type arguments to your hearts desire... maybe this should be considered a bug (if not in the compiler then in the specification)?

      – Slaw
      25 mins ago













    • Hmm... maybe not a bug. See stackoverflow.com/questions/28014853/…

      – Slaw
      12 mins ago
















    7














    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
      1 hour ago











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

      – Ole V.V.
      1 hour ago











    • What's interesting to me, and sorry if I missed you already mentioning/implying this, is if the constructor (or method) is generic, having the incorrect number of type arguments results in a compilation error—at least for me in OpenJDK 12. For example, if you have class Test { <T> Test() {} } then calling new <String, Long>Test() is an error. Yet when the constructor (or method) is not generic it lets you add type arguments to your hearts desire... maybe this should be considered a bug (if not in the compiler then in the specification)?

      – Slaw
      25 mins ago













    • Hmm... maybe not a bug. See stackoverflow.com/questions/28014853/…

      – Slaw
      12 mins ago














    7












    7








    7







    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 1 hour ago

























    answered 2 hours 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
      1 hour ago











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

      – Ole V.V.
      1 hour ago











    • What's interesting to me, and sorry if I missed you already mentioning/implying this, is if the constructor (or method) is generic, having the incorrect number of type arguments results in a compilation error—at least for me in OpenJDK 12. For example, if you have class Test { <T> Test() {} } then calling new <String, Long>Test() is an error. Yet when the constructor (or method) is not generic it lets you add type arguments to your hearts desire... maybe this should be considered a bug (if not in the compiler then in the specification)?

      – Slaw
      25 mins ago













    • Hmm... maybe not a bug. See stackoverflow.com/questions/28014853/…

      – Slaw
      12 mins ago



















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

      – jaspreet
      1 hour ago











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

      – Ole V.V.
      1 hour ago











    • What's interesting to me, and sorry if I missed you already mentioning/implying this, is if the constructor (or method) is generic, having the incorrect number of type arguments results in a compilation error—at least for me in OpenJDK 12. For example, if you have class Test { <T> Test() {} } then calling new <String, Long>Test() is an error. Yet when the constructor (or method) is not generic it lets you add type arguments to your hearts desire... maybe this should be considered a bug (if not in the compiler then in the specification)?

      – Slaw
      25 mins ago













    • Hmm... maybe not a bug. See stackoverflow.com/questions/28014853/…

      – Slaw
      12 mins ago

















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

    – jaspreet
    1 hour ago





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

    – jaspreet
    1 hour ago













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

    – Ole V.V.
    1 hour ago





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

    – Ole V.V.
    1 hour ago













    What's interesting to me, and sorry if I missed you already mentioning/implying this, is if the constructor (or method) is generic, having the incorrect number of type arguments results in a compilation error—at least for me in OpenJDK 12. For example, if you have class Test { <T> Test() {} } then calling new <String, Long>Test() is an error. Yet when the constructor (or method) is not generic it lets you add type arguments to your hearts desire... maybe this should be considered a bug (if not in the compiler then in the specification)?

    – Slaw
    25 mins ago







    What's interesting to me, and sorry if I missed you already mentioning/implying this, is if the constructor (or method) is generic, having the incorrect number of type arguments results in a compilation error—at least for me in OpenJDK 12. For example, if you have class Test { <T> Test() {} } then calling new <String, Long>Test() is an error. Yet when the constructor (or method) is not generic it lets you add type arguments to your hearts desire... maybe this should be considered a bug (if not in the compiler then in the specification)?

    – Slaw
    25 mins ago















    Hmm... maybe not a bug. See stackoverflow.com/questions/28014853/…

    – Slaw
    12 mins ago





    Hmm... maybe not a bug. See stackoverflow.com/questions/28014853/…

    – Slaw
    12 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|improve this answer




























      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|improve this answer


























        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|improve this answer













        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|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        GhostCatGhostCat

        95.1k1793156




        95.1k1793156






























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