Wildcard (Java): Difference between revisions

Revision as of 09:43, 8 March 2016

Revision as of 09:54, 8 March 2016

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== Wildcard as parameter type ==

In the body of a generic unit, the (formal) type parameter is handled like its [[bounded quantification|upper bound]] (expressed with <code>'''extends'''</code>; <code>Object</code> if not constrained). If the return type of a method is the type parameter, the result (e.g. of type <code>?</code>) can be referenced by a variable of the type of the upper bound (or <code>Object</code>). In the other direction, the wildcard fits no other type, not even <code>Object</code>: If <code>?</code> has been applied as the formal type parameter of a method, no actual parameters can be passed to it. ItHowever, objects of the unknown type can be calledread onlyfrom bythe castinggeneric theobject wildcardand assigned to a variable of a supertype of the reference:upperbound.

<source lang="java5">

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}

...

Generic<?UpperBound> wildcardReferenceconcreteTypeReference = new Generic<UpperBound>();

Generic<?> wildcardReference = concreteTypeReference;

UpperBound ub = wildcardReference.read(); // Object would also be OK

wildcardReference.write(new Object()); // type error

wildcardReference.write(new UpperBound()); // type error

((Generic<UpperBound>)wildcardReference)concreteTypeReference.write(new UpperBound()); // OK

</source>