Re: generics:< ? >vs.< T >

Lew <>
Thu, 10 Mar 2011 01:01:53 -0500
Robert Klemme wrote:

Well, yes that explains why it's a syntax error. What I rather meant (and
should have phrased better): what is the reasoning to forbid "super" here and
allow only the other direction. On first glance both variants (A extends B and
B super A) seem equivalent. Am I missing something or is this just an
arbitrary restriction?

I'm sure I don't know. I suppose it does keep things simpler.

But for the type of thing that you cited upthread, just a <T> is usually
better than a bounded wildcard type. Unlike types themselves, where the
advice is to generalize the declared type, for generic type parameters it's
often better to specify.

In lieu of 'Map<? super String, V>', a 'Map<String, V>' would be easier to
work with. You have to weigh what the extra assertion buys you. Given that
'toString()' is ubiquitous, not much. But it might be worth it - that's the
decision the designer makes.

OTOH, you need the 'extends' in another example:

<sscce source="">
  package eegee;
  import java.util.Map;

  /** GenericFooferol. */
  public class GenericFooferol
      * Copy a {@code Map}.
      * @param <T> target value type.
      * @param <S> source value type, must be a subtype of {@code T}.
      * @param source {@code Map} to copy.
      * @param target {@code Map} into which to copy.
      * @return Map {@code <String, T>} target map.
     public static <T, S extends T> Map <String, T> copy(
         Map <?, S> source, Map <String, T> target )
         if ( source == null || target == null )
             throw new IllegalArgumentException( "copy(): null argument" );
         for ( Map.Entry <?, S> entry : source.entrySet() )
             target.put( entry.getKey().toString(), entry.getValue() );
         return target;

     private GenericFooferol(){}

Honi soit qui mal y pense.

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