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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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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
add a comment |
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
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 thatZ[] array = new Z[0];
is also illegal.
– Andy Turner
15 hours ago
add a comment |
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
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
java generics local-class
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 thatZ[] array = new Z[0];
is also illegal.
– Andy Turner
15 hours ago
add a comment |
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 thatZ[] 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
add a comment |
4 Answers
4
active
oldest
votes
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.
It's true, but curiously, it does allow it ifZ
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 ofZ
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
|
show 1 more comment
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))
add a comment |
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.
2
Once you makeZ
generic, you don't need to apply the reflection workaround anymore...
– Lukas Eder
15 hours ago
add a comment |
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()))
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
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
It's true, but curiously, it does allow it ifZ
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 ofZ
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
|
show 1 more comment
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.
It's true, but curiously, it does allow it ifZ
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 ofZ
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
|
show 1 more comment
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.
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.
edited 14 hours ago
answered 15 hours ago
Rafael WinterhalterRafael Winterhalter
28.5k1368149
28.5k1368149
It's true, but curiously, it does allow it ifZ
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 ofZ
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
|
show 1 more comment
It's true, but curiously, it does allow it ifZ
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 ofZ
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
|
show 1 more comment
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))
add a comment |
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))
add a comment |
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))
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))
answered 16 hours ago
Lukas EderLukas Eder
137k74448987
137k74448987
add a comment |
add a comment |
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.
2
Once you makeZ
generic, you don't need to apply the reflection workaround anymore...
– Lukas Eder
15 hours ago
add a comment |
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.
2
Once you makeZ
generic, you don't need to apply the reflection workaround anymore...
– Lukas Eder
15 hours ago
add a comment |
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.
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.
edited 15 hours ago
answered 15 hours ago
Edoardo VacchiEdoardo Vacchi
759715
759715
2
Once you makeZ
generic, you don't need to apply the reflection workaround anymore...
– Lukas Eder
15 hours ago
add a comment |
2
Once you makeZ
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
add a comment |
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()))
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
add a comment |
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()))
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
add a comment |
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()))
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()))
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
add a comment |
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
add a comment |
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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