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





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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
    17 hours ago






  • 1





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

    – Lukas Eder
    17 hours ago






  • 2





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

    – Andy Turner
    17 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
    17 hours ago






  • 1





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

    – Lukas Eder
    17 hours ago






  • 2





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

    – Andy Turner
    17 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 2 hours ago









Peter Mortensen

13.9k1987114




13.9k1987114










asked 17 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
    17 hours ago






  • 1





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

    – Lukas Eder
    17 hours ago






  • 2





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

    – Andy Turner
    17 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
    17 hours ago






  • 1





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

    – Lukas Eder
    17 hours ago






  • 2





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

    – Andy Turner
    17 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
17 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
17 hours ago




1




1





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

– Lukas Eder
17 hours ago





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

– Lukas Eder
17 hours ago




2




2





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

– Andy Turner
17 hours ago







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

– Andy Turner
17 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
    17 hours ago











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

    – Lukas Eder
    17 hours ago






  • 3





    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
    17 hours ago











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

    – Rafael Winterhalter
    17 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
    17 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
      17 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
      17 hours ago












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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
      17 hours ago











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

      – Lukas Eder
      17 hours ago






    • 3





      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
      17 hours ago











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

      – Rafael Winterhalter
      17 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
      17 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
      17 hours ago











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

      – Lukas Eder
      17 hours ago






    • 3





      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
      17 hours ago











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

      – Rafael Winterhalter
      17 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
      17 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 16 hours ago

























    answered 17 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
      17 hours ago











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

      – Lukas Eder
      17 hours ago






    • 3





      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
      17 hours ago











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

      – Rafael Winterhalter
      17 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
      17 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
      17 hours ago











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

      – Lukas Eder
      17 hours ago






    • 3





      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
      17 hours ago











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

      – Rafael Winterhalter
      17 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
      17 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
    17 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
    17 hours ago













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

    – Lukas Eder
    17 hours ago





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

    – Lukas Eder
    17 hours ago




    3




    3





    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
    17 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
    17 hours ago













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

    – Rafael Winterhalter
    17 hours ago





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

    – Rafael Winterhalter
    17 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
    17 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
    17 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 17 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
              17 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
              17 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 17 hours ago

























            answered 17 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
              17 hours ago














            • 2





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

              – Lukas Eder
              17 hours ago








            2




            2





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

            – Lukas Eder
            17 hours ago





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

            – Lukas Eder
            17 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
              17 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
              17 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 17 hours ago









            Mark Rotteveel

            62.2k1479123




            62.2k1479123










            answered 17 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
              17 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
              17 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
            17 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
            17 hours ago


















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