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Error "illegal generic type for instanceof" when using local classes

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Error “illegal generic type for instanceof” when using local classes



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Data science time! April 2019 and salary with experience
Should we burninate the [wrap] tag?
The Ask Question Wizard is Live!“illegal generic type of instanceof” when using instanceof on an inner class type?How to refer to a class when both simple and fully-qualified names clashCreate instance of generic type in Java?Calling a static method on a generic type parameterCollections.emptyList() returns a List<Object>?How to get the type of T from a member of a generic class or method?“illegal generic type of instanceof” when using instanceof on an inner class type?Get generic type of class at runtimeHow to get a class instance of generics type THow to make a Java Generic method static?Google Gson - deserialize list<class> object? (generic type)Get “Illegal generic type for instanceof” error when coparison



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31















I have the following Java code that uses a local class.



import java.util.Arrays;

public class X<T>
void m()
class Z

for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof Z)




It does not compile with the following error message:



X.java:8: error: illegal generic type for instanceof
if (o instanceof Z)
^
1 error


I understand that the local class Z inherits the generic type signature of X<T>, being an inner class. The same kind of compilation error appears in this example, where Z is not local, but still inner:



import java.util.Arrays;

public class X<T>
class Z

void m()
for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof Z) // Compilation error




It can be worked around either by making Z non-inner / static:



import java.util.Arrays;

public class X<T>
static class Z

void m()
for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof Z) // Compiles now




Or by qualifying X.Z:



import java.util.Arrays;

public class X<T>
class Z

void m()
for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof X.Z) // Compiles now
if (o instanceof X<?>.Z) // Also





But how can I qualify a local class, or work around this limitation, without changing the local class itself?










share|improve this question
























  • I think this is a kind of compiler pathology. It's a strange way of telling you the type is not reifiable. It's even worse in Eclipse - it tells you "Use the form Z instead".

    – RealSkeptic
    15 hours ago






  • 1





    @RealSkeptic: Eclipse's compiler message is just "unlucky" here. It is usually a helpful extra information.

    – Lukas Eder
    15 hours ago






  • 2





    A corollary of this is that Z[] array = new Z[0]; is also illegal.

    – Andy Turner
    15 hours ago


















31















I have the following Java code that uses a local class.



import java.util.Arrays;

public class X<T>
void m()
class Z

for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof Z)




It does not compile with the following error message:



X.java:8: error: illegal generic type for instanceof
if (o instanceof Z)
^
1 error


I understand that the local class Z inherits the generic type signature of X<T>, being an inner class. The same kind of compilation error appears in this example, where Z is not local, but still inner:



import java.util.Arrays;

public class X<T>
class Z

void m()
for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof Z) // Compilation error




It can be worked around either by making Z non-inner / static:



import java.util.Arrays;

public class X<T>
static class Z

void m()
for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof Z) // Compiles now




Or by qualifying X.Z:



import java.util.Arrays;

public class X<T>
class Z

void m()
for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof X.Z) // Compiles now
if (o instanceof X<?>.Z) // Also





But how can I qualify a local class, or work around this limitation, without changing the local class itself?










share|improve this question
























  • I think this is a kind of compiler pathology. It's a strange way of telling you the type is not reifiable. It's even worse in Eclipse - it tells you "Use the form Z instead".

    – RealSkeptic
    15 hours ago






  • 1





    @RealSkeptic: Eclipse's compiler message is just "unlucky" here. It is usually a helpful extra information.

    – Lukas Eder
    15 hours ago






  • 2





    A corollary of this is that Z[] array = new Z[0]; is also illegal.

    – Andy Turner
    15 hours ago














31












31








31


1






I have the following Java code that uses a local class.



import java.util.Arrays;

public class X<T>
void m()
class Z

for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof Z)




It does not compile with the following error message:



X.java:8: error: illegal generic type for instanceof
if (o instanceof Z)
^
1 error


I understand that the local class Z inherits the generic type signature of X<T>, being an inner class. The same kind of compilation error appears in this example, where Z is not local, but still inner:



import java.util.Arrays;

public class X<T>
class Z

void m()
for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof Z) // Compilation error




It can be worked around either by making Z non-inner / static:



import java.util.Arrays;

public class X<T>
static class Z

void m()
for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof Z) // Compiles now




Or by qualifying X.Z:



import java.util.Arrays;

public class X<T>
class Z

void m()
for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof X.Z) // Compiles now
if (o instanceof X<?>.Z) // Also





But how can I qualify a local class, or work around this limitation, without changing the local class itself?










share|improve this question
















I have the following Java code that uses a local class.



import java.util.Arrays;

public class X<T>
void m()
class Z

for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof Z)




It does not compile with the following error message:



X.java:8: error: illegal generic type for instanceof
if (o instanceof Z)
^
1 error


I understand that the local class Z inherits the generic type signature of X<T>, being an inner class. The same kind of compilation error appears in this example, where Z is not local, but still inner:



import java.util.Arrays;

public class X<T>
class Z

void m()
for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof Z) // Compilation error




It can be worked around either by making Z non-inner / static:



import java.util.Arrays;

public class X<T>
static class Z

void m()
for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof Z) // Compiles now




Or by qualifying X.Z:



import java.util.Arrays;

public class X<T>
class Z

void m()
for (Object o : Arrays.asList(1, 2, 3))
if (o instanceof X.Z) // Compiles now
if (o instanceof X<?>.Z) // Also





But how can I qualify a local class, or work around this limitation, without changing the local class itself?







java generics local-class






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 56 mins ago









Peter Mortensen

13.9k1987114




13.9k1987114










asked 16 hours ago









Lukas EderLukas Eder

137k74448987




137k74448987












  • I think this is a kind of compiler pathology. It's a strange way of telling you the type is not reifiable. It's even worse in Eclipse - it tells you "Use the form Z instead".

    – RealSkeptic
    15 hours ago






  • 1





    @RealSkeptic: Eclipse's compiler message is just "unlucky" here. It is usually a helpful extra information.

    – Lukas Eder
    15 hours ago






  • 2





    A corollary of this is that Z[] array = new Z[0]; is also illegal.

    – Andy Turner
    15 hours ago


















  • I think this is a kind of compiler pathology. It's a strange way of telling you the type is not reifiable. It's even worse in Eclipse - it tells you "Use the form Z instead".

    – RealSkeptic
    15 hours ago






  • 1





    @RealSkeptic: Eclipse's compiler message is just "unlucky" here. It is usually a helpful extra information.

    – Lukas Eder
    15 hours ago






  • 2





    A corollary of this is that Z[] array = new Z[0]; is also illegal.

    – Andy Turner
    15 hours ago

















I think this is a kind of compiler pathology. It's a strange way of telling you the type is not reifiable. It's even worse in Eclipse - it tells you "Use the form Z instead".

– RealSkeptic
15 hours ago





I think this is a kind of compiler pathology. It's a strange way of telling you the type is not reifiable. It's even worse in Eclipse - it tells you "Use the form Z instead".

– RealSkeptic
15 hours ago




1




1





@RealSkeptic: Eclipse's compiler message is just "unlucky" here. It is usually a helpful extra information.

– Lukas Eder
15 hours ago





@RealSkeptic: Eclipse's compiler message is just "unlucky" here. It is usually a helpful extra information.

– Lukas Eder
15 hours ago




2




2





A corollary of this is that Z[] array = new Z[0]; is also illegal.

– Andy Turner
15 hours ago






A corollary of this is that Z[] array = new Z[0]; is also illegal.

– Andy Turner
15 hours ago













4 Answers
4






active

oldest

votes


















20














To me this seems to be an oversight or limitation in the Java language and I do not think it is possible.



The referenced type in an instanceof expression must be reifiable according to JLS 4.7, meaning that it must be expressed as a reifiable type by its fully qualified name. At the same time, JLS 6.7 states that local classes do not have a fully qualified name, they can therefore not be expressed as reifiable.



If you declare Z as generic, the instanceof operator treats Z as a raw type where all generic properties to it - in this case the enclosing class - are considered raw as well. (Similar to a generic methods of a raw type being considered as raw despite any generic signature. This is a measure to retain backwards compatiblity of type generification.) Since any raw type is reifiable, declaring Z to be generic will compile.






share|improve this answer

























  • It's true, but curiously, it does allow it if Z itself is generic, despite the fact that you have no way to qualify it.

    – RealSkeptic
    15 hours ago











  • "oversight" - You're here to pick a fight with the JLS designers, right? :)

    – Lukas Eder
    15 hours ago






  • 2





    Generics were patched into the language and there are some corner cases such as class literals that were not properly covered. Nothing is perfect, I think there is a chance for an oversight. ;)

    – Rafael Winterhalter
    15 hours ago











  • I extended my answer to explain why generification of Z does the trick.

    – Rafael Winterhalter
    15 hours ago











  • @LukasEder I think oversight is right. It's sort of like this question: these cases could have been designed around, but they haven't been; I make no judgment as to whether they should have been.

    – Andy Turner
    15 hours ago



















14














A possible workaround is to use reflection:



import java.util.Arrays;

public class X<T>
void m()
class Z

for (Object o : Arrays.asList(1, 2, 3))
if (Z.class.isInstance(o))







share|improve this answer






























    2














    Apparently, by making Z generic compilation succeeds. I expected that to require <T> as the type parameter, but you just have to make it generic, so anything will do



    import java.util.Arrays;

    public class X<T>
    void m()
    class Z<Anything>

    for (Object o : Arrays.asList(1, 2, 3))
    if (Z.class.isInstance(o))




    Proper solution would be qualify the local class, but I don't think you can. Either you refactor it to a private static class or that's probably the best you can get.






    share|improve this answer




















    • 2





      Once you make Z generic, you don't need to apply the reflection workaround anymore...

      – Lukas Eder
      15 hours ago


















    0














    This should work either. Using reflection too. But seems a valid solution.



    import java.util.Arrays;

    public class X<T>


    void m()

    class Z2


    for(Object o: Arrays.asList(1,2,3))
    if(Z2.class.isAssignableFrom(o.getClass()))












    share|improve this answer

























    • I didn't downvote, but I guess it's because there's a slightly better way to use reflection here, as I've mentioned in my own answer

      – Lukas Eder
      15 hours ago











    Your Answer






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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    20














    To me this seems to be an oversight or limitation in the Java language and I do not think it is possible.



    The referenced type in an instanceof expression must be reifiable according to JLS 4.7, meaning that it must be expressed as a reifiable type by its fully qualified name. At the same time, JLS 6.7 states that local classes do not have a fully qualified name, they can therefore not be expressed as reifiable.



    If you declare Z as generic, the instanceof operator treats Z as a raw type where all generic properties to it - in this case the enclosing class - are considered raw as well. (Similar to a generic methods of a raw type being considered as raw despite any generic signature. This is a measure to retain backwards compatiblity of type generification.) Since any raw type is reifiable, declaring Z to be generic will compile.






    share|improve this answer

























    • It's true, but curiously, it does allow it if Z itself is generic, despite the fact that you have no way to qualify it.

      – RealSkeptic
      15 hours ago











    • "oversight" - You're here to pick a fight with the JLS designers, right? :)

      – Lukas Eder
      15 hours ago






    • 2





      Generics were patched into the language and there are some corner cases such as class literals that were not properly covered. Nothing is perfect, I think there is a chance for an oversight. ;)

      – Rafael Winterhalter
      15 hours ago











    • I extended my answer to explain why generification of Z does the trick.

      – Rafael Winterhalter
      15 hours ago











    • @LukasEder I think oversight is right. It's sort of like this question: these cases could have been designed around, but they haven't been; I make no judgment as to whether they should have been.

      – Andy Turner
      15 hours ago
















    20














    To me this seems to be an oversight or limitation in the Java language and I do not think it is possible.



    The referenced type in an instanceof expression must be reifiable according to JLS 4.7, meaning that it must be expressed as a reifiable type by its fully qualified name. At the same time, JLS 6.7 states that local classes do not have a fully qualified name, they can therefore not be expressed as reifiable.



    If you declare Z as generic, the instanceof operator treats Z as a raw type where all generic properties to it - in this case the enclosing class - are considered raw as well. (Similar to a generic methods of a raw type being considered as raw despite any generic signature. This is a measure to retain backwards compatiblity of type generification.) Since any raw type is reifiable, declaring Z to be generic will compile.






    share|improve this answer

























    • It's true, but curiously, it does allow it if Z itself is generic, despite the fact that you have no way to qualify it.

      – RealSkeptic
      15 hours ago











    • "oversight" - You're here to pick a fight with the JLS designers, right? :)

      – Lukas Eder
      15 hours ago






    • 2





      Generics were patched into the language and there are some corner cases such as class literals that were not properly covered. Nothing is perfect, I think there is a chance for an oversight. ;)

      – Rafael Winterhalter
      15 hours ago











    • I extended my answer to explain why generification of Z does the trick.

      – Rafael Winterhalter
      15 hours ago











    • @LukasEder I think oversight is right. It's sort of like this question: these cases could have been designed around, but they haven't been; I make no judgment as to whether they should have been.

      – Andy Turner
      15 hours ago














    20












    20








    20







    To me this seems to be an oversight or limitation in the Java language and I do not think it is possible.



    The referenced type in an instanceof expression must be reifiable according to JLS 4.7, meaning that it must be expressed as a reifiable type by its fully qualified name. At the same time, JLS 6.7 states that local classes do not have a fully qualified name, they can therefore not be expressed as reifiable.



    If you declare Z as generic, the instanceof operator treats Z as a raw type where all generic properties to it - in this case the enclosing class - are considered raw as well. (Similar to a generic methods of a raw type being considered as raw despite any generic signature. This is a measure to retain backwards compatiblity of type generification.) Since any raw type is reifiable, declaring Z to be generic will compile.






    share|improve this answer















    To me this seems to be an oversight or limitation in the Java language and I do not think it is possible.



    The referenced type in an instanceof expression must be reifiable according to JLS 4.7, meaning that it must be expressed as a reifiable type by its fully qualified name. At the same time, JLS 6.7 states that local classes do not have a fully qualified name, they can therefore not be expressed as reifiable.



    If you declare Z as generic, the instanceof operator treats Z as a raw type where all generic properties to it - in this case the enclosing class - are considered raw as well. (Similar to a generic methods of a raw type being considered as raw despite any generic signature. This is a measure to retain backwards compatiblity of type generification.) Since any raw type is reifiable, declaring Z to be generic will compile.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 14 hours ago

























    answered 15 hours ago









    Rafael WinterhalterRafael Winterhalter

    28.5k1368149




    28.5k1368149












    • It's true, but curiously, it does allow it if Z itself is generic, despite the fact that you have no way to qualify it.

      – RealSkeptic
      15 hours ago











    • "oversight" - You're here to pick a fight with the JLS designers, right? :)

      – Lukas Eder
      15 hours ago






    • 2





      Generics were patched into the language and there are some corner cases such as class literals that were not properly covered. Nothing is perfect, I think there is a chance for an oversight. ;)

      – Rafael Winterhalter
      15 hours ago











    • I extended my answer to explain why generification of Z does the trick.

      – Rafael Winterhalter
      15 hours ago











    • @LukasEder I think oversight is right. It's sort of like this question: these cases could have been designed around, but they haven't been; I make no judgment as to whether they should have been.

      – Andy Turner
      15 hours ago


















    • It's true, but curiously, it does allow it if Z itself is generic, despite the fact that you have no way to qualify it.

      – RealSkeptic
      15 hours ago











    • "oversight" - You're here to pick a fight with the JLS designers, right? :)

      – Lukas Eder
      15 hours ago






    • 2





      Generics were patched into the language and there are some corner cases such as class literals that were not properly covered. Nothing is perfect, I think there is a chance for an oversight. ;)

      – Rafael Winterhalter
      15 hours ago











    • I extended my answer to explain why generification of Z does the trick.

      – Rafael Winterhalter
      15 hours ago











    • @LukasEder I think oversight is right. It's sort of like this question: these cases could have been designed around, but they haven't been; I make no judgment as to whether they should have been.

      – Andy Turner
      15 hours ago

















    It's true, but curiously, it does allow it if Z itself is generic, despite the fact that you have no way to qualify it.

    – RealSkeptic
    15 hours ago





    It's true, but curiously, it does allow it if Z itself is generic, despite the fact that you have no way to qualify it.

    – RealSkeptic
    15 hours ago













    "oversight" - You're here to pick a fight with the JLS designers, right? :)

    – Lukas Eder
    15 hours ago





    "oversight" - You're here to pick a fight with the JLS designers, right? :)

    – Lukas Eder
    15 hours ago




    2




    2





    Generics were patched into the language and there are some corner cases such as class literals that were not properly covered. Nothing is perfect, I think there is a chance for an oversight. ;)

    – Rafael Winterhalter
    15 hours ago





    Generics were patched into the language and there are some corner cases such as class literals that were not properly covered. Nothing is perfect, I think there is a chance for an oversight. ;)

    – Rafael Winterhalter
    15 hours ago













    I extended my answer to explain why generification of Z does the trick.

    – Rafael Winterhalter
    15 hours ago





    I extended my answer to explain why generification of Z does the trick.

    – Rafael Winterhalter
    15 hours ago













    @LukasEder I think oversight is right. It's sort of like this question: these cases could have been designed around, but they haven't been; I make no judgment as to whether they should have been.

    – Andy Turner
    15 hours ago






    @LukasEder I think oversight is right. It's sort of like this question: these cases could have been designed around, but they haven't been; I make no judgment as to whether they should have been.

    – Andy Turner
    15 hours ago














    14














    A possible workaround is to use reflection:



    import java.util.Arrays;

    public class X<T>
    void m()
    class Z

    for (Object o : Arrays.asList(1, 2, 3))
    if (Z.class.isInstance(o))







    share|improve this answer



























      14














      A possible workaround is to use reflection:



      import java.util.Arrays;

      public class X<T>
      void m()
      class Z

      for (Object o : Arrays.asList(1, 2, 3))
      if (Z.class.isInstance(o))







      share|improve this answer

























        14












        14








        14







        A possible workaround is to use reflection:



        import java.util.Arrays;

        public class X<T>
        void m()
        class Z

        for (Object o : Arrays.asList(1, 2, 3))
        if (Z.class.isInstance(o))







        share|improve this answer













        A possible workaround is to use reflection:



        import java.util.Arrays;

        public class X<T>
        void m()
        class Z

        for (Object o : Arrays.asList(1, 2, 3))
        if (Z.class.isInstance(o))








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 16 hours ago









        Lukas EderLukas Eder

        137k74448987




        137k74448987





















            2














            Apparently, by making Z generic compilation succeeds. I expected that to require <T> as the type parameter, but you just have to make it generic, so anything will do



            import java.util.Arrays;

            public class X<T>
            void m()
            class Z<Anything>

            for (Object o : Arrays.asList(1, 2, 3))
            if (Z.class.isInstance(o))




            Proper solution would be qualify the local class, but I don't think you can. Either you refactor it to a private static class or that's probably the best you can get.






            share|improve this answer




















            • 2





              Once you make Z generic, you don't need to apply the reflection workaround anymore...

              – Lukas Eder
              15 hours ago















            2














            Apparently, by making Z generic compilation succeeds. I expected that to require <T> as the type parameter, but you just have to make it generic, so anything will do



            import java.util.Arrays;

            public class X<T>
            void m()
            class Z<Anything>

            for (Object o : Arrays.asList(1, 2, 3))
            if (Z.class.isInstance(o))




            Proper solution would be qualify the local class, but I don't think you can. Either you refactor it to a private static class or that's probably the best you can get.






            share|improve this answer




















            • 2





              Once you make Z generic, you don't need to apply the reflection workaround anymore...

              – Lukas Eder
              15 hours ago













            2












            2








            2







            Apparently, by making Z generic compilation succeeds. I expected that to require <T> as the type parameter, but you just have to make it generic, so anything will do



            import java.util.Arrays;

            public class X<T>
            void m()
            class Z<Anything>

            for (Object o : Arrays.asList(1, 2, 3))
            if (Z.class.isInstance(o))




            Proper solution would be qualify the local class, but I don't think you can. Either you refactor it to a private static class or that's probably the best you can get.






            share|improve this answer















            Apparently, by making Z generic compilation succeeds. I expected that to require <T> as the type parameter, but you just have to make it generic, so anything will do



            import java.util.Arrays;

            public class X<T>
            void m()
            class Z<Anything>

            for (Object o : Arrays.asList(1, 2, 3))
            if (Z.class.isInstance(o))




            Proper solution would be qualify the local class, but I don't think you can. Either you refactor it to a private static class or that's probably the best you can get.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 15 hours ago

























            answered 15 hours ago









            Edoardo VacchiEdoardo Vacchi

            759715




            759715







            • 2





              Once you make Z generic, you don't need to apply the reflection workaround anymore...

              – Lukas Eder
              15 hours ago












            • 2





              Once you make Z generic, you don't need to apply the reflection workaround anymore...

              – Lukas Eder
              15 hours ago







            2




            2





            Once you make Z generic, you don't need to apply the reflection workaround anymore...

            – Lukas Eder
            15 hours ago





            Once you make Z generic, you don't need to apply the reflection workaround anymore...

            – Lukas Eder
            15 hours ago











            0














            This should work either. Using reflection too. But seems a valid solution.



            import java.util.Arrays;

            public class X<T>


            void m()

            class Z2


            for(Object o: Arrays.asList(1,2,3))
            if(Z2.class.isAssignableFrom(o.getClass()))












            share|improve this answer

























            • I didn't downvote, but I guess it's because there's a slightly better way to use reflection here, as I've mentioned in my own answer

              – Lukas Eder
              15 hours ago















            0














            This should work either. Using reflection too. But seems a valid solution.



            import java.util.Arrays;

            public class X<T>


            void m()

            class Z2


            for(Object o: Arrays.asList(1,2,3))
            if(Z2.class.isAssignableFrom(o.getClass()))












            share|improve this answer

























            • I didn't downvote, but I guess it's because there's a slightly better way to use reflection here, as I've mentioned in my own answer

              – Lukas Eder
              15 hours ago













            0












            0








            0







            This should work either. Using reflection too. But seems a valid solution.



            import java.util.Arrays;

            public class X<T>


            void m()

            class Z2


            for(Object o: Arrays.asList(1,2,3))
            if(Z2.class.isAssignableFrom(o.getClass()))












            share|improve this answer















            This should work either. Using reflection too. But seems a valid solution.



            import java.util.Arrays;

            public class X<T>


            void m()

            class Z2


            for(Object o: Arrays.asList(1,2,3))
            if(Z2.class.isAssignableFrom(o.getClass()))













            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 15 hours ago









            Mark Rotteveel

            62.2k1479123




            62.2k1479123










            answered 15 hours ago









            JWThewesJWThewes

            191




            191












            • I didn't downvote, but I guess it's because there's a slightly better way to use reflection here, as I've mentioned in my own answer

              – Lukas Eder
              15 hours ago

















            • I didn't downvote, but I guess it's because there's a slightly better way to use reflection here, as I've mentioned in my own answer

              – Lukas Eder
              15 hours ago
















            I didn't downvote, but I guess it's because there's a slightly better way to use reflection here, as I've mentioned in my own answer

            – Lukas Eder
            15 hours ago





            I didn't downvote, but I guess it's because there's a slightly better way to use reflection here, as I've mentioned in my own answer

            – Lukas Eder
            15 hours ago

















            draft saved

            draft discarded
















































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