CODEWITHSUNDEEP
javacastingdictionary
Mark
Mark

Reputation: 69920

Java - Cast a Map

How can I cast a Map<Object,Object> to Map<String,String> in the cleanest way?

Is there a way to do that without iterating over the map?

Thanks

Upvotes: 27

Views: 47091

Answers (5)

Nelson Azevedo
Nelson Azevedo

Reputation: 341

 public  Map<String, String> castOnlyStringValues(final Map<String, Object> map) {
    return map.entrySet().stream()
              .filter(x -> String.class.isAssignableFrom(x.getValue().getClass()))
              .map(x -> (Entry<?, ?>) x).map(x -> (Entry<String, String>) x)
              .collect(toMap(Entry::getKey, Entry::getValue));
}

Upvotes: 0

Thomas Decaux
Thomas Decaux

Reputation: 22661

If you want something clean, you must convert "deeply"! keys and values.

Using Java 8 stream, this take only a few lines:

static public Map<String, String> toStringString(Map<? extends Object, ? extends Object> map) {
        return map
                .entrySet()
                .stream()
                .collect(Collectors.toMap(
                    e -> e.getKey().toString(),
                    e -> e.getValue().toString()
                ));
    }

Upvotes: 10

Voo
Voo

Reputation: 30216

I think it's a good idea to explain why the simple solution doesn't work and why you never, ever should use this.

Assume you could cast List<Object> to List<String> (the same applies to Map, just a simpler interface). What would you expect to happen from the following code:

List<Object> m = Something;
m.add("Looks good.");
m.add(42);
List<String> s = (List<String>)m; // uhuh, no we don't want that.
String myString = s.get(1); // huh exception here.

Now you CAN hack it indeed using Bohemians/Chris solution, but you basically destroy Java's type system. DON'T DO THAT. You don't want a List<String> to contain an Integer! Have fun debugging that later on - the additional code of looping through all variables will avoid lots of headaches and hardly is a performance problem.

If there's a reason to declare the Map as taking an Object instead of a String someone may add any object to it - usually you should be able to avoid this with a better generic.

Upvotes: 12

Chris J
Chris J

Reputation: 9252

The actual answer is:

Map<Object,Object> valueMap = ...;
@SuppressWarnings("unchecked")
Map<String,String> targetMap = (Map)valueMap;

Upvotes: 14

Bohemian
Bohemian

Reputation: 424983

Building on Chris J's answer, here's his "hack" code in action, showing it works:

public static void main(String[] args) {
    Map<Object, Object> valueMap = new HashMap<Object, Object>();
    valueMap.put("foo", "bar");
    @SuppressWarnings("unchecked")
    Map<String, String> targetMap = (Map<String, String>) ((Object) valueMap); // hack alert!
    for (Map.Entry<String, String> entry : targetMap.entrySet()) {
        String key = entry.getKey();
        String value = entry.getValue();
        System.out.println(key + "=" + value); // prints foo=bar :)
    }
}

Wow! No runtime exceptions :)

Upvotes: -1

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