Restrict a generic Class parameter to classes that implement Map












7















I'm trying to write a Map builder. One of the constructors will allow the client to specify the type of Map they wish to build



public class MapBuilder<K, V> {



    private Map<K, V> map;



    /**

     * Create a Map builder

     * @param mapType the type of Map to build. This type must support a default constructor

     * @throws Exception

     */

    public MapBuilder(Class<? extends Map<K, V>> mapType) throws Exception {

        map = mapType.newInstance();

    }



    // remaining implementation omitted

}


The intent is that it should be possible to construct instances of the builder with:



MapBuilder<Integer, String> builder = new MapBuilder<Integer, String>(LinkedHashMap.class);


or



MapBuilder<Integer, String> builder = new MapBuilder<Integer, String>(HashMap.class);


It seems that the type signature of the constructor argument doesn't currently support this, because the line above causes a "Cannot resolve constructor" compilation error.



How can I change my constructor so that it accepts classes that implement Map only?






java generics







share|improve this question


















  • 4





    Why not use a Supplier instead? That way, the users can pass LinkedHashMap::new and not be bothered with reflection.

    – RealSkeptic
    18 hours ago
















7















I'm trying to write a Map builder. One of the constructors will allow the client to specify the type of Map they wish to build



public class MapBuilder<K, V> {



    private Map<K, V> map;



    /**

     * Create a Map builder

     * @param mapType the type of Map to build. This type must support a default constructor

     * @throws Exception

     */

    public MapBuilder(Class<? extends Map<K, V>> mapType) throws Exception {

        map = mapType.newInstance();

    }



    // remaining implementation omitted

}


The intent is that it should be possible to construct instances of the builder with:



MapBuilder<Integer, String> builder = new MapBuilder<Integer, String>(LinkedHashMap.class);


or



MapBuilder<Integer, String> builder = new MapBuilder<Integer, String>(HashMap.class);


It seems that the type signature of the constructor argument doesn't currently support this, because the line above causes a "Cannot resolve constructor" compilation error.



How can I change my constructor so that it accepts classes that implement Map only?






java generics







share|improve this question


















  • 4





    Why not use a Supplier instead? That way, the users can pass LinkedHashMap::new and not be bothered with reflection.

    – RealSkeptic
    18 hours ago














7












7








7


1






I'm trying to write a Map builder. One of the constructors will allow the client to specify the type of Map they wish to build



public class MapBuilder<K, V> {



    private Map<K, V> map;



    /**

     * Create a Map builder

     * @param mapType the type of Map to build. This type must support a default constructor

     * @throws Exception

     */

    public MapBuilder(Class<? extends Map<K, V>> mapType) throws Exception {

        map = mapType.newInstance();

    }



    // remaining implementation omitted

}


The intent is that it should be possible to construct instances of the builder with:



MapBuilder<Integer, String> builder = new MapBuilder<Integer, String>(LinkedHashMap.class);


or



MapBuilder<Integer, String> builder = new MapBuilder<Integer, String>(HashMap.class);


It seems that the type signature of the constructor argument doesn't currently support this, because the line above causes a "Cannot resolve constructor" compilation error.



How can I change my constructor so that it accepts classes that implement Map only?






java generics







share|improve this question














I'm trying to write a Map builder. One of the constructors will allow the client to specify the type of Map they wish to build



public class MapBuilder<K, V> {



    private Map<K, V> map;



    /**

     * Create a Map builder

     * @param mapType the type of Map to build. This type must support a default constructor

     * @throws Exception

     */

    public MapBuilder(Class<? extends Map<K, V>> mapType) throws Exception {

        map = mapType.newInstance();

    }



    // remaining implementation omitted

}


The intent is that it should be possible to construct instances of the builder with:



MapBuilder<Integer, String> builder = new MapBuilder<Integer, String>(LinkedHashMap.class);


or



MapBuilder<Integer, String> builder = new MapBuilder<Integer, String>(HashMap.class);


It seems that the type signature of the constructor argument doesn't currently support this, because the line above causes a "Cannot resolve constructor" compilation error.



How can I change my constructor so that it accepts classes that implement Map only?






java generics




java generics






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 18 hours ago









DónalDónal

122k156485750




122k156485750








  • 4





    Why not use a Supplier instead? That way, the users can pass LinkedHashMap::new and not be bothered with reflection.

    – RealSkeptic
    18 hours ago














  • 4





    Why not use a Supplier instead? That way, the users can pass LinkedHashMap::new and not be bothered with reflection.

    – RealSkeptic
    18 hours ago








4




4





Why not use a Supplier instead? That way, the users can pass LinkedHashMap::new and not be bothered with reflection.

– RealSkeptic
18 hours ago





Why not use a Supplier instead? That way, the users can pass LinkedHashMap::new and not be bothered with reflection.

– RealSkeptic
18 hours ago












3 Answers
3






active

oldest

votes


















9














Use a Supplier instead of a Class:



public MapBuilder(Supplier<? extends Map<K, V>> supplier) {

    map = supplier.get();

}


Which then can be called like this:



MapBuilder<Integer, Integer> builder = new MapBuilder<>(LinkedHashMap::new);


This is also safer, because a Class<Map> could have no default constructor, which would throw an error (which is not very responsive code)






share|improve this answer



















  • 1





    You can get rid of the wildcard altogether: public MapBuilder(Supplier<Map<K, V>> supplier)

    – ernest_k
    18 hours ago








  • 1





    @ernest_k you can't. Without the wildcard it would fail if you provide an instance of for example class Sup implements Supplier<HashMap<Integer, Integer>>. It's a very rare case, but it can happen

    – Lino
    18 hours ago








  • 1





    Granted. It would still fail even if you use a lambda (Supplier<HashMap<String, Integer>> supp = HashMap::new; MapBuilder<String, Integer> mb = new MapBuilder<String, Integer>(supp); - but this is something I'd personally be happy to have no support for (that's an opinion). This is a good answer, btw.

    – ernest_k
    18 hours ago






  • 2





    @ernest_k Thank you :), I agree though that in a small project it would not make a difference (using the wildcard or not) but when writing an API which is used across different projects it is the way to go and makes the lifes of people using your API a lot easier

    – Lino
    17 hours ago



















2














The following will work:



public MapBuilder(Class<? extends Map> mapType) throws Exception {

    map = mapType.newInstance();

}





share|improve this answer



















  • 1





    Will produce an unchecked warning though

    – Lino
    18 hours ago



















2














The problem is that LinkedHashMap.class is 



Class<LinkedHashMap>


and not something like



Class<LinkedHashMap<Integer, String>>


These are also inconvertible types (so you can't cast it) and there's no way to get an instance of the latter.



What you can do is change the constructor to 



public MapBuilder(Class<? extends Map> mapType) throws Exception


Generics are erased at run-time so at run-time all Maps will behave just like Map<Object, Object> anyway. Therefore it doesn't matter that the class you're constructing from is using the raw type.




By the way, Class::newInstance is deprecated. Use 



mapType.getConstructor().newInstance()





share|improve this answer

























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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9














    Use a Supplier instead of a Class:



    public MapBuilder(Supplier<? extends Map<K, V>> supplier) {
    
    map = supplier.get();
    
}


    Which then can be called like this:



    MapBuilder<Integer, Integer> builder = new MapBuilder<>(LinkedHashMap::new);


    This is also safer, because a Class<Map> could have no default constructor, which would throw an error (which is not very responsive code)






    share|improve this answer



















    • 1





      You can get rid of the wildcard altogether: public MapBuilder(Supplier<Map<K, V>> supplier)

      – ernest_k
      18 hours ago








    • 1





      @ernest_k you can't. Without the wildcard it would fail if you provide an instance of for example class Sup implements Supplier<HashMap<Integer, Integer>>. It's a very rare case, but it can happen

      – Lino
      18 hours ago








    • 1





      Granted. It would still fail even if you use a lambda (Supplier<HashMap<String, Integer>> supp = HashMap::new; MapBuilder<String, Integer> mb = new MapBuilder<String, Integer>(supp); - but this is something I'd personally be happy to have no support for (that's an opinion). This is a good answer, btw.

      – ernest_k
      18 hours ago






    • 2





      @ernest_k Thank you :), I agree though that in a small project it would not make a difference (using the wildcard or not) but when writing an API which is used across different projects it is the way to go and makes the lifes of people using your API a lot easier

      – Lino
      17 hours ago
















    9














    Use a Supplier instead of a Class:



    public MapBuilder(Supplier<? extends Map<K, V>> supplier) {
    
    map = supplier.get();
    
}


    Which then can be called like this:



    MapBuilder<Integer, Integer> builder = new MapBuilder<>(LinkedHashMap::new);


    This is also safer, because a Class<Map> could have no default constructor, which would throw an error (which is not very responsive code)






    share|improve this answer



















    • 1





      You can get rid of the wildcard altogether: public MapBuilder(Supplier<Map<K, V>> supplier)

      – ernest_k
      18 hours ago








    • 1





      @ernest_k you can't. Without the wildcard it would fail if you provide an instance of for example class Sup implements Supplier<HashMap<Integer, Integer>>. It's a very rare case, but it can happen

      – Lino
      18 hours ago








    • 1





      Granted. It would still fail even if you use a lambda (Supplier<HashMap<String, Integer>> supp = HashMap::new; MapBuilder<String, Integer> mb = new MapBuilder<String, Integer>(supp); - but this is something I'd personally be happy to have no support for (that's an opinion). This is a good answer, btw.

      – ernest_k
      18 hours ago






    • 2





      @ernest_k Thank you :), I agree though that in a small project it would not make a difference (using the wildcard or not) but when writing an API which is used across different projects it is the way to go and makes the lifes of people using your API a lot easier

      – Lino
      17 hours ago














    9












    9








    9







    Use a Supplier instead of a Class:



    public MapBuilder(Supplier<? extends Map<K, V>> supplier) {
    
    map = supplier.get();
    
}


    Which then can be called like this:



    MapBuilder<Integer, Integer> builder = new MapBuilder<>(LinkedHashMap::new);


    This is also safer, because a Class<Map> could have no default constructor, which would throw an error (which is not very responsive code)






    share|improve this answer













    Use a Supplier instead of a Class:



    public MapBuilder(Supplier<? extends Map<K, V>> supplier) {
    
    map = supplier.get();
    
}


    Which then can be called like this:



    MapBuilder<Integer, Integer> builder = new MapBuilder<>(LinkedHashMap::new);


    This is also safer, because a Class<Map> could have no default constructor, which would throw an error (which is not very responsive code)







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 18 hours ago









    LinoLino

    9,10422039




    9,10422039








    • 1





      You can get rid of the wildcard altogether: public MapBuilder(Supplier<Map<K, V>> supplier)

      – ernest_k
      18 hours ago








    • 1





      @ernest_k you can't. Without the wildcard it would fail if you provide an instance of for example class Sup implements Supplier<HashMap<Integer, Integer>>. It's a very rare case, but it can happen

      – Lino
      18 hours ago








    • 1





      Granted. It would still fail even if you use a lambda (Supplier<HashMap<String, Integer>> supp = HashMap::new; MapBuilder<String, Integer> mb = new MapBuilder<String, Integer>(supp); - but this is something I'd personally be happy to have no support for (that's an opinion). This is a good answer, btw.

      – ernest_k
      18 hours ago






    • 2





      @ernest_k Thank you :), I agree though that in a small project it would not make a difference (using the wildcard or not) but when writing an API which is used across different projects it is the way to go and makes the lifes of people using your API a lot easier

      – Lino
      17 hours ago














    • 1





      You can get rid of the wildcard altogether: public MapBuilder(Supplier<Map<K, V>> supplier)

      – ernest_k
      18 hours ago








    • 1





      @ernest_k you can't. Without the wildcard it would fail if you provide an instance of for example class Sup implements Supplier<HashMap<Integer, Integer>>. It's a very rare case, but it can happen

      – Lino
      18 hours ago








    • 1





      Granted. It would still fail even if you use a lambda (Supplier<HashMap<String, Integer>> supp = HashMap::new; MapBuilder<String, Integer> mb = new MapBuilder<String, Integer>(supp); - but this is something I'd personally be happy to have no support for (that's an opinion). This is a good answer, btw.

      – ernest_k
      18 hours ago






    • 2





      @ernest_k Thank you :), I agree though that in a small project it would not make a difference (using the wildcard or not) but when writing an API which is used across different projects it is the way to go and makes the lifes of people using your API a lot easier

      – Lino
      17 hours ago








    1




    1





    You can get rid of the wildcard altogether: public MapBuilder(Supplier<Map<K, V>> supplier)

    – ernest_k
    18 hours ago







    You can get rid of the wildcard altogether: public MapBuilder(Supplier<Map<K, V>> supplier)

    – ernest_k
    18 hours ago






    1




    1





    @ernest_k you can't. Without the wildcard it would fail if you provide an instance of for example class Sup implements Supplier<HashMap<Integer, Integer>>. It's a very rare case, but it can happen

    – Lino
    18 hours ago







    @ernest_k you can't. Without the wildcard it would fail if you provide an instance of for example class Sup implements Supplier<HashMap<Integer, Integer>>. It's a very rare case, but it can happen

    – Lino
    18 hours ago






    1




    1





    Granted. It would still fail even if you use a lambda (Supplier<HashMap<String, Integer>> supp = HashMap::new; MapBuilder<String, Integer> mb = new MapBuilder<String, Integer>(supp); - but this is something I'd personally be happy to have no support for (that's an opinion). This is a good answer, btw.

    – ernest_k
    18 hours ago





    Granted. It would still fail even if you use a lambda (Supplier<HashMap<String, Integer>> supp = HashMap::new; MapBuilder<String, Integer> mb = new MapBuilder<String, Integer>(supp); - but this is something I'd personally be happy to have no support for (that's an opinion). This is a good answer, btw.

    – ernest_k
    18 hours ago




    2




    2





    @ernest_k Thank you :), I agree though that in a small project it would not make a difference (using the wildcard or not) but when writing an API which is used across different projects it is the way to go and makes the lifes of people using your API a lot easier

    – Lino
    17 hours ago





    @ernest_k Thank you :), I agree though that in a small project it would not make a difference (using the wildcard or not) but when writing an API which is used across different projects it is the way to go and makes the lifes of people using your API a lot easier

    – Lino
    17 hours ago













    2














    The following will work:



    public MapBuilder(Class<? extends Map> mapType) throws Exception {
    
    map = mapType.newInstance();
    
}





    share|improve this answer



















    • 1





      Will produce an unchecked warning though

      – Lino
      18 hours ago
















    2














    The following will work:



    public MapBuilder(Class<? extends Map> mapType) throws Exception {
    
    map = mapType.newInstance();
    
}





    share|improve this answer



















    • 1





      Will produce an unchecked warning though

      – Lino
      18 hours ago














    2












    2








    2







    The following will work:



    public MapBuilder(Class<? extends Map> mapType) throws Exception {
    
    map = mapType.newInstance();
    
}





    share|improve this answer













    The following will work:



    public MapBuilder(Class<? extends Map> mapType) throws Exception {
    
    map = mapType.newInstance();
    
}






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 18 hours ago









    EranEran

    285k37463552




    285k37463552








    • 1





      Will produce an unchecked warning though

      – Lino
      18 hours ago














    • 1





      Will produce an unchecked warning though

      – Lino
      18 hours ago








    1




    1





    Will produce an unchecked warning though

    – Lino
    18 hours ago





    Will produce an unchecked warning though

    – Lino
    18 hours ago











    2














    The problem is that LinkedHashMap.class is 



    Class<LinkedHashMap>


    and not something like



    Class<LinkedHashMap<Integer, String>>


    These are also inconvertible types (so you can't cast it) and there's no way to get an instance of the latter.



    What you can do is change the constructor to 



    public MapBuilder(Class<? extends Map> mapType) throws Exception


    Generics are erased at run-time so at run-time all Maps will behave just like Map<Object, Object> anyway. Therefore it doesn't matter that the class you're constructing from is using the raw type.




    By the way, Class::newInstance is deprecated. Use 



    mapType.getConstructor().newInstance()





    share|improve this answer






























      2














      The problem is that LinkedHashMap.class is 



      Class<LinkedHashMap>


      and not something like



      Class<LinkedHashMap<Integer, String>>


      These are also inconvertible types (so you can't cast it) and there's no way to get an instance of the latter.



      What you can do is change the constructor to 



      public MapBuilder(Class<? extends Map> mapType) throws Exception


      Generics are erased at run-time so at run-time all Maps will behave just like Map<Object, Object> anyway. Therefore it doesn't matter that the class you're constructing from is using the raw type.




      By the way, Class::newInstance is deprecated. Use 



      mapType.getConstructor().newInstance()





      share|improve this answer




























        2












        2








        2







        The problem is that LinkedHashMap.class is 



        Class<LinkedHashMap>


        and not something like



        Class<LinkedHashMap<Integer, String>>


        These are also inconvertible types (so you can't cast it) and there's no way to get an instance of the latter.



        What you can do is change the constructor to 



        public MapBuilder(Class<? extends Map> mapType) throws Exception


        Generics are erased at run-time so at run-time all Maps will behave just like Map<Object, Object> anyway. Therefore it doesn't matter that the class you're constructing from is using the raw type.




        By the way, Class::newInstance is deprecated. Use 



        mapType.getConstructor().newInstance()





        share|improve this answer















        The problem is that LinkedHashMap.class is 



        Class<LinkedHashMap>


        and not something like



        Class<LinkedHashMap<Integer, String>>


        These are also inconvertible types (so you can't cast it) and there's no way to get an instance of the latter.



        What you can do is change the constructor to 



        public MapBuilder(Class<? extends Map> mapType) throws Exception


        Generics are erased at run-time so at run-time all Maps will behave just like Map<Object, Object> anyway. Therefore it doesn't matter that the class you're constructing from is using the raw type.




        By the way, Class::newInstance is deprecated. Use 



        mapType.getConstructor().newInstance()






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 18 hours ago

























        answered 18 hours ago









        MichaelMichael

        20.4k83471




        20.4k83471






























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