using template parameters in java interface declaration












0















I come from C++ and I'm taking a Java class right now, studying design patterns more specifically. Last class, the professor gave us an example project to help us get started with the Prototype pattern and there was an interface declaration in the project which I didn't understand very well (and didn't ask the professor either :/)



package pattern.prototype.impl;

public interface IPrototype<T extends IPrototype> extends Cloneable {

//clone: Permite realizar una clonacion superficial del prototipo.
public T clone();

//deepClone: Permite realizar una clonación profunda del prototipo.
public T deepClone();
}


could anyone give me some kind of explanation regarding the use of parameter T in this context IPrototype<T extends IPrototype>. What is its purpose there? Is it necessary or just one way of doing it?



Thank you










share|improve this question























  • This called the "Curiously re-occurring template pattern".

    – flakes
    Nov 23 '18 at 0:18






  • 1





    Java does not have C++ template parameters. That is a Java generic type. It is being named T, and declared that T must implement IPrototype

    – Elliott Frisch
    Nov 23 '18 at 0:18






  • 1





    docs.oracle.com/javase/tutorial/java/generics/index.html

    – Kartik
    Nov 23 '18 at 0:21











  • Was also going to say that Java generic types are not at all the same as C++ templates. Important to keep the two distinct in one's mind, in spite of the similar syntax.

    – markspace
    Nov 23 '18 at 0:51
















0















I come from C++ and I'm taking a Java class right now, studying design patterns more specifically. Last class, the professor gave us an example project to help us get started with the Prototype pattern and there was an interface declaration in the project which I didn't understand very well (and didn't ask the professor either :/)



package pattern.prototype.impl;

public interface IPrototype<T extends IPrototype> extends Cloneable {

//clone: Permite realizar una clonacion superficial del prototipo.
public T clone();

//deepClone: Permite realizar una clonación profunda del prototipo.
public T deepClone();
}


could anyone give me some kind of explanation regarding the use of parameter T in this context IPrototype<T extends IPrototype>. What is its purpose there? Is it necessary or just one way of doing it?



Thank you










share|improve this question























  • This called the "Curiously re-occurring template pattern".

    – flakes
    Nov 23 '18 at 0:18






  • 1





    Java does not have C++ template parameters. That is a Java generic type. It is being named T, and declared that T must implement IPrototype

    – Elliott Frisch
    Nov 23 '18 at 0:18






  • 1





    docs.oracle.com/javase/tutorial/java/generics/index.html

    – Kartik
    Nov 23 '18 at 0:21











  • Was also going to say that Java generic types are not at all the same as C++ templates. Important to keep the two distinct in one's mind, in spite of the similar syntax.

    – markspace
    Nov 23 '18 at 0:51














0












0








0








I come from C++ and I'm taking a Java class right now, studying design patterns more specifically. Last class, the professor gave us an example project to help us get started with the Prototype pattern and there was an interface declaration in the project which I didn't understand very well (and didn't ask the professor either :/)



package pattern.prototype.impl;

public interface IPrototype<T extends IPrototype> extends Cloneable {

//clone: Permite realizar una clonacion superficial del prototipo.
public T clone();

//deepClone: Permite realizar una clonación profunda del prototipo.
public T deepClone();
}


could anyone give me some kind of explanation regarding the use of parameter T in this context IPrototype<T extends IPrototype>. What is its purpose there? Is it necessary or just one way of doing it?



Thank you










share|improve this question














I come from C++ and I'm taking a Java class right now, studying design patterns more specifically. Last class, the professor gave us an example project to help us get started with the Prototype pattern and there was an interface declaration in the project which I didn't understand very well (and didn't ask the professor either :/)



package pattern.prototype.impl;

public interface IPrototype<T extends IPrototype> extends Cloneable {

//clone: Permite realizar una clonacion superficial del prototipo.
public T clone();

//deepClone: Permite realizar una clonación profunda del prototipo.
public T deepClone();
}


could anyone give me some kind of explanation regarding the use of parameter T in this context IPrototype<T extends IPrototype>. What is its purpose there? Is it necessary or just one way of doing it?



Thank you







java templates interface prototype-pattern






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 23 '18 at 0:11









ScaramoucheScaramouche

1,9951829




1,9951829













  • This called the "Curiously re-occurring template pattern".

    – flakes
    Nov 23 '18 at 0:18






  • 1





    Java does not have C++ template parameters. That is a Java generic type. It is being named T, and declared that T must implement IPrototype

    – Elliott Frisch
    Nov 23 '18 at 0:18






  • 1





    docs.oracle.com/javase/tutorial/java/generics/index.html

    – Kartik
    Nov 23 '18 at 0:21











  • Was also going to say that Java generic types are not at all the same as C++ templates. Important to keep the two distinct in one's mind, in spite of the similar syntax.

    – markspace
    Nov 23 '18 at 0:51



















  • This called the "Curiously re-occurring template pattern".

    – flakes
    Nov 23 '18 at 0:18






  • 1





    Java does not have C++ template parameters. That is a Java generic type. It is being named T, and declared that T must implement IPrototype

    – Elliott Frisch
    Nov 23 '18 at 0:18






  • 1





    docs.oracle.com/javase/tutorial/java/generics/index.html

    – Kartik
    Nov 23 '18 at 0:21











  • Was also going to say that Java generic types are not at all the same as C++ templates. Important to keep the two distinct in one's mind, in spite of the similar syntax.

    – markspace
    Nov 23 '18 at 0:51

















This called the "Curiously re-occurring template pattern".

– flakes
Nov 23 '18 at 0:18





This called the "Curiously re-occurring template pattern".

– flakes
Nov 23 '18 at 0:18




1




1





Java does not have C++ template parameters. That is a Java generic type. It is being named T, and declared that T must implement IPrototype

– Elliott Frisch
Nov 23 '18 at 0:18





Java does not have C++ template parameters. That is a Java generic type. It is being named T, and declared that T must implement IPrototype

– Elliott Frisch
Nov 23 '18 at 0:18




1




1





docs.oracle.com/javase/tutorial/java/generics/index.html

– Kartik
Nov 23 '18 at 0:21





docs.oracle.com/javase/tutorial/java/generics/index.html

– Kartik
Nov 23 '18 at 0:21













Was also going to say that Java generic types are not at all the same as C++ templates. Important to keep the two distinct in one's mind, in spite of the similar syntax.

– markspace
Nov 23 '18 at 0:51





Was also going to say that Java generic types are not at all the same as C++ templates. Important to keep the two distinct in one's mind, in spite of the similar syntax.

– markspace
Nov 23 '18 at 0:51












4 Answers
4






active

oldest

votes


















1














This is called the "Curiously re-occurring template pattern". As the name suggests, it was discovered for code written in C++ which uses templates, however, the pattern works with Generics in Java as well.



Here I can implement the interface as follows:



public class ConcretePrototype implements IPrototype<ConcretePrototype > {
@Override
public ConcretePrototype clone() { ... }

@Override
public ConcretePrototype deepClone() { ... }
}


Notice the method signature of the overriden methods. The base interface IPrototype does not know about ConcretePrototype, however, by using CRTP we can enforce that ConcretePrototype returns values of its own type.






share|improve this answer
























  • from your answer, can you tell me if these two assumptions are correct, please? 1) <T extends IPrototype>'s only purpose is to allow us to declare public T clone(); using T. 2) if all the interface's methods were, for instance, void, we would not need <T extends IPrototype>.

    – Scaramouche
    Nov 23 '18 at 1:48













  • @Scaramouche 1) For the code provided yes. It could also be used for parameters in the method signatures. 2) If you don't use the type then you don't need a type parameter. A slightly different way of writing the methods could be, IPrototype clone();, but then you would have to potentially cast the result to ConcretePrototype, which may or may not cause you extra work. eg if you have a variable ConcretePrototype p = new ConcretePrototype(); then do you want to write ConcretePrototype pClone = p.clone(); or ConcretePrototype pClone = (ConcretePrototype) p.clone();

    – flakes
    Nov 23 '18 at 1:54











  • great, you answered those two questions AND the third question I was about to ask, thank you, finally, do you really think IPrototype<T extends IPrototype> should be changed to IPrototype<T extends IPrototype<T>> as markspace suggests?

    – Scaramouche
    Nov 23 '18 at 1:56













  • @Scaramouche no problem! And yes IPrototype<T extends IPrototype<T>> is a better way to write the class definition!

    – flakes
    Nov 23 '18 at 1:59











  • @Scaramouche Without the type parameter you could write class ConcretePrototype implements IPrototype<IPrototype> which is probably not what you want the authors of the child classes to do. This would lead to writing code that looks like the non-generic example ConcretePrototype pClone = (ConcretePrototype) p.clone(). It depends on how strict you want to be. And generally you want to complete the type parameters or at least define them as unbounded directly : interface IPrototype<T extends IPrototype<?>>

    – flakes
    Nov 23 '18 at 2:06





















1














It’s not a template parameter, but a Java generic type, and it stands for any class implementing the given interface. In the context of prototype pattern it’s not necessary, just one possible implementation.






share|improve this answer































    1














    T is type parameter also a generic type.



    T extends IPrototype : To declare a bounded type parameter,T can be any type that is subclass of IPrototype



    public T clone(); return type would be generic.



    public T deepClone(); return type would be generic(means can be any type)






    share|improve this answer































      1














      BTW, I'm pretty sure the syntax you are using is wrong, so it might be best to inform your instructor anyway.



      The problem is that IPrototype is used as a raw type here. The second time it's used on the line, it's just IPrototype with no type variable. That's a no-no in Java.



      As for what is going on, all it means is that the type parameter for IPrototype must be a type of IPrototype--that is, some subclass of IProtoType. Take a look at Java's Enum type, which uses exactly the same pattern: https://docs.oracle.com/javase/10/docs/api/java/lang/Enum.html



      public interface IPrototype<T extends IPrototype<T>> extends Cloneable {
      // ^^^ add this

      //clone: Permite realizar una clonacion superficial del prototipo.
      public T clone();

      //deepClone: Permite realizar una clonación profunda del prototipo.
      public T deepClone();
      }





      share|improve this answer

























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        4 Answers
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        active

        oldest

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        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1














        This is called the "Curiously re-occurring template pattern". As the name suggests, it was discovered for code written in C++ which uses templates, however, the pattern works with Generics in Java as well.



        Here I can implement the interface as follows:



        public class ConcretePrototype implements IPrototype<ConcretePrototype > {
        @Override
        public ConcretePrototype clone() { ... }

        @Override
        public ConcretePrototype deepClone() { ... }
        }


        Notice the method signature of the overriden methods. The base interface IPrototype does not know about ConcretePrototype, however, by using CRTP we can enforce that ConcretePrototype returns values of its own type.






        share|improve this answer
























        • from your answer, can you tell me if these two assumptions are correct, please? 1) <T extends IPrototype>'s only purpose is to allow us to declare public T clone(); using T. 2) if all the interface's methods were, for instance, void, we would not need <T extends IPrototype>.

          – Scaramouche
          Nov 23 '18 at 1:48













        • @Scaramouche 1) For the code provided yes. It could also be used for parameters in the method signatures. 2) If you don't use the type then you don't need a type parameter. A slightly different way of writing the methods could be, IPrototype clone();, but then you would have to potentially cast the result to ConcretePrototype, which may or may not cause you extra work. eg if you have a variable ConcretePrototype p = new ConcretePrototype(); then do you want to write ConcretePrototype pClone = p.clone(); or ConcretePrototype pClone = (ConcretePrototype) p.clone();

          – flakes
          Nov 23 '18 at 1:54











        • great, you answered those two questions AND the third question I was about to ask, thank you, finally, do you really think IPrototype<T extends IPrototype> should be changed to IPrototype<T extends IPrototype<T>> as markspace suggests?

          – Scaramouche
          Nov 23 '18 at 1:56













        • @Scaramouche no problem! And yes IPrototype<T extends IPrototype<T>> is a better way to write the class definition!

          – flakes
          Nov 23 '18 at 1:59











        • @Scaramouche Without the type parameter you could write class ConcretePrototype implements IPrototype<IPrototype> which is probably not what you want the authors of the child classes to do. This would lead to writing code that looks like the non-generic example ConcretePrototype pClone = (ConcretePrototype) p.clone(). It depends on how strict you want to be. And generally you want to complete the type parameters or at least define them as unbounded directly : interface IPrototype<T extends IPrototype<?>>

          – flakes
          Nov 23 '18 at 2:06


















        1














        This is called the "Curiously re-occurring template pattern". As the name suggests, it was discovered for code written in C++ which uses templates, however, the pattern works with Generics in Java as well.



        Here I can implement the interface as follows:



        public class ConcretePrototype implements IPrototype<ConcretePrototype > {
        @Override
        public ConcretePrototype clone() { ... }

        @Override
        public ConcretePrototype deepClone() { ... }
        }


        Notice the method signature of the overriden methods. The base interface IPrototype does not know about ConcretePrototype, however, by using CRTP we can enforce that ConcretePrototype returns values of its own type.






        share|improve this answer
























        • from your answer, can you tell me if these two assumptions are correct, please? 1) <T extends IPrototype>'s only purpose is to allow us to declare public T clone(); using T. 2) if all the interface's methods were, for instance, void, we would not need <T extends IPrototype>.

          – Scaramouche
          Nov 23 '18 at 1:48













        • @Scaramouche 1) For the code provided yes. It could also be used for parameters in the method signatures. 2) If you don't use the type then you don't need a type parameter. A slightly different way of writing the methods could be, IPrototype clone();, but then you would have to potentially cast the result to ConcretePrototype, which may or may not cause you extra work. eg if you have a variable ConcretePrototype p = new ConcretePrototype(); then do you want to write ConcretePrototype pClone = p.clone(); or ConcretePrototype pClone = (ConcretePrototype) p.clone();

          – flakes
          Nov 23 '18 at 1:54











        • great, you answered those two questions AND the third question I was about to ask, thank you, finally, do you really think IPrototype<T extends IPrototype> should be changed to IPrototype<T extends IPrototype<T>> as markspace suggests?

          – Scaramouche
          Nov 23 '18 at 1:56













        • @Scaramouche no problem! And yes IPrototype<T extends IPrototype<T>> is a better way to write the class definition!

          – flakes
          Nov 23 '18 at 1:59











        • @Scaramouche Without the type parameter you could write class ConcretePrototype implements IPrototype<IPrototype> which is probably not what you want the authors of the child classes to do. This would lead to writing code that looks like the non-generic example ConcretePrototype pClone = (ConcretePrototype) p.clone(). It depends on how strict you want to be. And generally you want to complete the type parameters or at least define them as unbounded directly : interface IPrototype<T extends IPrototype<?>>

          – flakes
          Nov 23 '18 at 2:06
















        1












        1








        1







        This is called the "Curiously re-occurring template pattern". As the name suggests, it was discovered for code written in C++ which uses templates, however, the pattern works with Generics in Java as well.



        Here I can implement the interface as follows:



        public class ConcretePrototype implements IPrototype<ConcretePrototype > {
        @Override
        public ConcretePrototype clone() { ... }

        @Override
        public ConcretePrototype deepClone() { ... }
        }


        Notice the method signature of the overriden methods. The base interface IPrototype does not know about ConcretePrototype, however, by using CRTP we can enforce that ConcretePrototype returns values of its own type.






        share|improve this answer













        This is called the "Curiously re-occurring template pattern". As the name suggests, it was discovered for code written in C++ which uses templates, however, the pattern works with Generics in Java as well.



        Here I can implement the interface as follows:



        public class ConcretePrototype implements IPrototype<ConcretePrototype > {
        @Override
        public ConcretePrototype clone() { ... }

        @Override
        public ConcretePrototype deepClone() { ... }
        }


        Notice the method signature of the overriden methods. The base interface IPrototype does not know about ConcretePrototype, however, by using CRTP we can enforce that ConcretePrototype returns values of its own type.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 23 '18 at 0:25









        flakesflakes

        6,56111850




        6,56111850













        • from your answer, can you tell me if these two assumptions are correct, please? 1) <T extends IPrototype>'s only purpose is to allow us to declare public T clone(); using T. 2) if all the interface's methods were, for instance, void, we would not need <T extends IPrototype>.

          – Scaramouche
          Nov 23 '18 at 1:48













        • @Scaramouche 1) For the code provided yes. It could also be used for parameters in the method signatures. 2) If you don't use the type then you don't need a type parameter. A slightly different way of writing the methods could be, IPrototype clone();, but then you would have to potentially cast the result to ConcretePrototype, which may or may not cause you extra work. eg if you have a variable ConcretePrototype p = new ConcretePrototype(); then do you want to write ConcretePrototype pClone = p.clone(); or ConcretePrototype pClone = (ConcretePrototype) p.clone();

          – flakes
          Nov 23 '18 at 1:54











        • great, you answered those two questions AND the third question I was about to ask, thank you, finally, do you really think IPrototype<T extends IPrototype> should be changed to IPrototype<T extends IPrototype<T>> as markspace suggests?

          – Scaramouche
          Nov 23 '18 at 1:56













        • @Scaramouche no problem! And yes IPrototype<T extends IPrototype<T>> is a better way to write the class definition!

          – flakes
          Nov 23 '18 at 1:59











        • @Scaramouche Without the type parameter you could write class ConcretePrototype implements IPrototype<IPrototype> which is probably not what you want the authors of the child classes to do. This would lead to writing code that looks like the non-generic example ConcretePrototype pClone = (ConcretePrototype) p.clone(). It depends on how strict you want to be. And generally you want to complete the type parameters or at least define them as unbounded directly : interface IPrototype<T extends IPrototype<?>>

          – flakes
          Nov 23 '18 at 2:06





















        • from your answer, can you tell me if these two assumptions are correct, please? 1) <T extends IPrototype>'s only purpose is to allow us to declare public T clone(); using T. 2) if all the interface's methods were, for instance, void, we would not need <T extends IPrototype>.

          – Scaramouche
          Nov 23 '18 at 1:48













        • @Scaramouche 1) For the code provided yes. It could also be used for parameters in the method signatures. 2) If you don't use the type then you don't need a type parameter. A slightly different way of writing the methods could be, IPrototype clone();, but then you would have to potentially cast the result to ConcretePrototype, which may or may not cause you extra work. eg if you have a variable ConcretePrototype p = new ConcretePrototype(); then do you want to write ConcretePrototype pClone = p.clone(); or ConcretePrototype pClone = (ConcretePrototype) p.clone();

          – flakes
          Nov 23 '18 at 1:54











        • great, you answered those two questions AND the third question I was about to ask, thank you, finally, do you really think IPrototype<T extends IPrototype> should be changed to IPrototype<T extends IPrototype<T>> as markspace suggests?

          – Scaramouche
          Nov 23 '18 at 1:56













        • @Scaramouche no problem! And yes IPrototype<T extends IPrototype<T>> is a better way to write the class definition!

          – flakes
          Nov 23 '18 at 1:59











        • @Scaramouche Without the type parameter you could write class ConcretePrototype implements IPrototype<IPrototype> which is probably not what you want the authors of the child classes to do. This would lead to writing code that looks like the non-generic example ConcretePrototype pClone = (ConcretePrototype) p.clone(). It depends on how strict you want to be. And generally you want to complete the type parameters or at least define them as unbounded directly : interface IPrototype<T extends IPrototype<?>>

          – flakes
          Nov 23 '18 at 2:06



















        from your answer, can you tell me if these two assumptions are correct, please? 1) <T extends IPrototype>'s only purpose is to allow us to declare public T clone(); using T. 2) if all the interface's methods were, for instance, void, we would not need <T extends IPrototype>.

        – Scaramouche
        Nov 23 '18 at 1:48







        from your answer, can you tell me if these two assumptions are correct, please? 1) <T extends IPrototype>'s only purpose is to allow us to declare public T clone(); using T. 2) if all the interface's methods were, for instance, void, we would not need <T extends IPrototype>.

        – Scaramouche
        Nov 23 '18 at 1:48















        @Scaramouche 1) For the code provided yes. It could also be used for parameters in the method signatures. 2) If you don't use the type then you don't need a type parameter. A slightly different way of writing the methods could be, IPrototype clone();, but then you would have to potentially cast the result to ConcretePrototype, which may or may not cause you extra work. eg if you have a variable ConcretePrototype p = new ConcretePrototype(); then do you want to write ConcretePrototype pClone = p.clone(); or ConcretePrototype pClone = (ConcretePrototype) p.clone();

        – flakes
        Nov 23 '18 at 1:54





        @Scaramouche 1) For the code provided yes. It could also be used for parameters in the method signatures. 2) If you don't use the type then you don't need a type parameter. A slightly different way of writing the methods could be, IPrototype clone();, but then you would have to potentially cast the result to ConcretePrototype, which may or may not cause you extra work. eg if you have a variable ConcretePrototype p = new ConcretePrototype(); then do you want to write ConcretePrototype pClone = p.clone(); or ConcretePrototype pClone = (ConcretePrototype) p.clone();

        – flakes
        Nov 23 '18 at 1:54













        great, you answered those two questions AND the third question I was about to ask, thank you, finally, do you really think IPrototype<T extends IPrototype> should be changed to IPrototype<T extends IPrototype<T>> as markspace suggests?

        – Scaramouche
        Nov 23 '18 at 1:56







        great, you answered those two questions AND the third question I was about to ask, thank you, finally, do you really think IPrototype<T extends IPrototype> should be changed to IPrototype<T extends IPrototype<T>> as markspace suggests?

        – Scaramouche
        Nov 23 '18 at 1:56















        @Scaramouche no problem! And yes IPrototype<T extends IPrototype<T>> is a better way to write the class definition!

        – flakes
        Nov 23 '18 at 1:59





        @Scaramouche no problem! And yes IPrototype<T extends IPrototype<T>> is a better way to write the class definition!

        – flakes
        Nov 23 '18 at 1:59













        @Scaramouche Without the type parameter you could write class ConcretePrototype implements IPrototype<IPrototype> which is probably not what you want the authors of the child classes to do. This would lead to writing code that looks like the non-generic example ConcretePrototype pClone = (ConcretePrototype) p.clone(). It depends on how strict you want to be. And generally you want to complete the type parameters or at least define them as unbounded directly : interface IPrototype<T extends IPrototype<?>>

        – flakes
        Nov 23 '18 at 2:06







        @Scaramouche Without the type parameter you could write class ConcretePrototype implements IPrototype<IPrototype> which is probably not what you want the authors of the child classes to do. This would lead to writing code that looks like the non-generic example ConcretePrototype pClone = (ConcretePrototype) p.clone(). It depends on how strict you want to be. And generally you want to complete the type parameters or at least define them as unbounded directly : interface IPrototype<T extends IPrototype<?>>

        – flakes
        Nov 23 '18 at 2:06















        1














        It’s not a template parameter, but a Java generic type, and it stands for any class implementing the given interface. In the context of prototype pattern it’s not necessary, just one possible implementation.






        share|improve this answer




























          1














          It’s not a template parameter, but a Java generic type, and it stands for any class implementing the given interface. In the context of prototype pattern it’s not necessary, just one possible implementation.






          share|improve this answer


























            1












            1








            1







            It’s not a template parameter, but a Java generic type, and it stands for any class implementing the given interface. In the context of prototype pattern it’s not necessary, just one possible implementation.






            share|improve this answer













            It’s not a template parameter, but a Java generic type, and it stands for any class implementing the given interface. In the context of prototype pattern it’s not necessary, just one possible implementation.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 23 '18 at 0:26









            NenadNenad

            1408




            1408























                1














                T is type parameter also a generic type.



                T extends IPrototype : To declare a bounded type parameter,T can be any type that is subclass of IPrototype



                public T clone(); return type would be generic.



                public T deepClone(); return type would be generic(means can be any type)






                share|improve this answer




























                  1














                  T is type parameter also a generic type.



                  T extends IPrototype : To declare a bounded type parameter,T can be any type that is subclass of IPrototype



                  public T clone(); return type would be generic.



                  public T deepClone(); return type would be generic(means can be any type)






                  share|improve this answer


























                    1












                    1








                    1







                    T is type parameter also a generic type.



                    T extends IPrototype : To declare a bounded type parameter,T can be any type that is subclass of IPrototype



                    public T clone(); return type would be generic.



                    public T deepClone(); return type would be generic(means can be any type)






                    share|improve this answer













                    T is type parameter also a generic type.



                    T extends IPrototype : To declare a bounded type parameter,T can be any type that is subclass of IPrototype



                    public T clone(); return type would be generic.



                    public T deepClone(); return type would be generic(means can be any type)







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Nov 23 '18 at 0:28









                    GauravRai1512GauravRai1512

                    58811




                    58811























                        1














                        BTW, I'm pretty sure the syntax you are using is wrong, so it might be best to inform your instructor anyway.



                        The problem is that IPrototype is used as a raw type here. The second time it's used on the line, it's just IPrototype with no type variable. That's a no-no in Java.



                        As for what is going on, all it means is that the type parameter for IPrototype must be a type of IPrototype--that is, some subclass of IProtoType. Take a look at Java's Enum type, which uses exactly the same pattern: https://docs.oracle.com/javase/10/docs/api/java/lang/Enum.html



                        public interface IPrototype<T extends IPrototype<T>> extends Cloneable {
                        // ^^^ add this

                        //clone: Permite realizar una clonacion superficial del prototipo.
                        public T clone();

                        //deepClone: Permite realizar una clonación profunda del prototipo.
                        public T deepClone();
                        }





                        share|improve this answer






























                          1














                          BTW, I'm pretty sure the syntax you are using is wrong, so it might be best to inform your instructor anyway.



                          The problem is that IPrototype is used as a raw type here. The second time it's used on the line, it's just IPrototype with no type variable. That's a no-no in Java.



                          As for what is going on, all it means is that the type parameter for IPrototype must be a type of IPrototype--that is, some subclass of IProtoType. Take a look at Java's Enum type, which uses exactly the same pattern: https://docs.oracle.com/javase/10/docs/api/java/lang/Enum.html



                          public interface IPrototype<T extends IPrototype<T>> extends Cloneable {
                          // ^^^ add this

                          //clone: Permite realizar una clonacion superficial del prototipo.
                          public T clone();

                          //deepClone: Permite realizar una clonación profunda del prototipo.
                          public T deepClone();
                          }





                          share|improve this answer




























                            1












                            1








                            1







                            BTW, I'm pretty sure the syntax you are using is wrong, so it might be best to inform your instructor anyway.



                            The problem is that IPrototype is used as a raw type here. The second time it's used on the line, it's just IPrototype with no type variable. That's a no-no in Java.



                            As for what is going on, all it means is that the type parameter for IPrototype must be a type of IPrototype--that is, some subclass of IProtoType. Take a look at Java's Enum type, which uses exactly the same pattern: https://docs.oracle.com/javase/10/docs/api/java/lang/Enum.html



                            public interface IPrototype<T extends IPrototype<T>> extends Cloneable {
                            // ^^^ add this

                            //clone: Permite realizar una clonacion superficial del prototipo.
                            public T clone();

                            //deepClone: Permite realizar una clonación profunda del prototipo.
                            public T deepClone();
                            }





                            share|improve this answer















                            BTW, I'm pretty sure the syntax you are using is wrong, so it might be best to inform your instructor anyway.



                            The problem is that IPrototype is used as a raw type here. The second time it's used on the line, it's just IPrototype with no type variable. That's a no-no in Java.



                            As for what is going on, all it means is that the type parameter for IPrototype must be a type of IPrototype--that is, some subclass of IProtoType. Take a look at Java's Enum type, which uses exactly the same pattern: https://docs.oracle.com/javase/10/docs/api/java/lang/Enum.html



                            public interface IPrototype<T extends IPrototype<T>> extends Cloneable {
                            // ^^^ add this

                            //clone: Permite realizar una clonacion superficial del prototipo.
                            public T clone();

                            //deepClone: Permite realizar una clonación profunda del prototipo.
                            public T deepClone();
                            }






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Nov 23 '18 at 1:02

























                            answered Nov 23 '18 at 0:56









                            markspacemarkspace

                            6,93821529




                            6,93821529






























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