Re: Converting a HashMap<String, Thing> into a sorted array

From:
Lew <lew@nowhere.com>
Newsgroups:
comp.lang.java.programmer
Date:
Sat, 21 Oct 2006 20:47:07 -0400
Message-ID:
<_7OdnatfQdoRIqfYnZ2dnUVZ_rednZ2d@comcast.com>

Daniel Dyer wrote:

There's a more concise way:
     public Thing[] convert(HashMap<String, Thing> things)
    {
        Thing[] tArray = things.values().toArray(new Thing[things.size()]);
        Arrays.sort(tArray);
        return tArray;
    }


On Sun, 22 Oct 2006 00:51:59 +0100, Lew <lew@nowhere.com> wrote:

Another way to express the toArray() call is
   Thing [] tArray = things.values().toArray( new Thing [0] );


Daniel Dyer wrote:

The version I posted will copy the contents into the array that is
passed in. With your version, a new array will be created since the
argument array is not big enough (unless the map happens to be empty).

http://java.sun.com/j2se/1.5.0/docs/api/java/util/Collection.html#toArray(T[])

"If the collection fits in the specified array, it is returned therein.
Otherwise, a new array is allocated with the runtime type of the
specified array and the size of this collection."

Dan.

--Daniel Dyer
http://www.uncommons.org


That's exactly right.

In
   Thing[] tArray = things.values().toArray(new Thing[things.size()]);

the "new" of the returned array is explicit and done by the client code.

In
   public static final Thing [] ARRAY_TEMPLATE = new Thing [0];
    ...
   Thing [] tArray = things.values().toArray( ARRAY_TEMPLATE );

the "new" of the returned array is hidden in the toArray() call.

 > This is not the neatest method in the API.

I guess the reason for the second idiom is convenience for the programmer.

- Lew

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