No Problems here
public class MyList<E extends Number> extends ArrayList<E> {
}
.
Unexcepted bound. What does this mean? And why is it wrong? Thanks for help.
public class MyList<E extends Number> extends ArrayList<E extends Number> {
}
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hasNoPatience
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Can you please explain when exactly you are getting this error ? Is this compilation error ? Or it's coming while instantiating this class or what ? – Tushar J Dudhatra Mar 02 '17 at 10:52
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compilation error – hasNoPatience Mar 02 '17 at 10:53
2 Answers
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class MyList<E extends Number> is OK because you declare a type parameter, so you have to give it a name (E) and you can optionally declare it as bounded (extends Number).
In extends ArrayList<E> instead you just have to "use" a type parameter: with "<E>" you refer to the parameter declared in your class, for which a bound is already given in its declaration. "<? extends Number>" (with ? in place of E) would also be accepted by the compiler (although it would be not what you want). Instead, "<E extends Number>" is taken as a type parameter declaration, so it is a compile error.
rrobby86
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public class MyList<E extends Number> extends ArrayList<E> {
private static final long serialVersionUID = -1025575227555594680L;
}
This should work with no compilation error and no warnings even. Let me know if you still have the same error.
Tushar J Dudhatra
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