HashMap to Array: throws java.lang.ArrayStoreException

Question

I am supposed to convert hashes into arrays, but it throws a java.lang.ArrayStoreException at me. I'm taking it one step at a time, trying to see how I can structure it, but it won't run.

Hashes:

{
    objectOne : {
        attributeOne : 1,
        attributeTwo : 2
    },
    objectTwo : {
        attributeOne : 3,
        attributeTwo : 4
    }
}

into Arrays with structure:

[
    {
        name : 'objectOne',
        attributes : {
            attributeOne : 1,
            attributeTwo : 2
        }
    }, {
        name : 'objectTwo',
        attributes : {
            attributeOne : 3,
            attributeTwo : 4
        }
    }
]

My code throws an:

java.lang.ArrayStoreException: java.lang.String
    at java.util.AbstractCollection.toArray(AbstractCollection.java:171)
    at hashToArray2.main(hashToArray2.java:36)

I've tried to change the parameters to:

Object[] keys = map.keySet().toArray(new Object[map.size()][]));
Object[] values = map.values().toArray(new Object[map.size()][]));

But it yields the same problem.

Code:

public class hashToArray2{
    public static void main (String[] args) throws ClassNotFoundException{
        Class.forName("hashToArray2");
        System.out.println("hashToArray class successfully loaded");

        //Creating object1 + input values
        HashMap<String, Integer> obj1 = new HashMap<String, Integer>();
        obj1.put("attributeOne", 1);
        obj1.put("attributeTwo", 2);

        //Creating object2 + input values
        HashMap<String, Integer> obj2 = new HashMap<String, Integer>();
        obj2.put("attributeOne", 3);
        obj2.put("attributeTwo", 4);

        //Combining obj1+2
        //HashMap<String, Integer> map = new HashMap<String, Integer>();
        //map.putAll(obj1);
        //map.putAll(obj2);

        //to array
        Object[][] arr1 = new String[obj1.size()][2];
        Object[][] arr2 = new String[obj2.size()][2];
        //obj1
        Object[] keys1 = obj1.keySet().toArray(new Object[obj1.size()][]);
        Object[] values1 = obj1.values().toArray(new Object[obj1.size()][]);

        for(int i = 0; i < arr1.length; i++){
            arr1[i][0] = keys1[i];
            arr1[i][1] = values1[i];
        }

        for(int j = 0; j < arr1.length; j++){
            for(int k = 0; k < arr1[j].length; k++){
                System.out.println(arr1[j][k]);
            }
        }

        //obj2
        Object[] keys2 = obj2.keySet().toArray(new Object[obj2.size()][]);
        Object[] values2 = obj2.values().toArray(new Object[obj2.size()][]);

        for(int a = 0; a < arr2.length; a++){
            arr2[a][0] = keys2[a];
            arr2[a][1] = values2[a];
        }

        for(int b = 0; b < arr2.length; b++){
            for(int c = 0; c < arr2[b].length; c++){
                System.out.println(arr2[b][c]);
            }
        }
    }
}

JB Nizet · Accepted Answer · 2015-07-26 13:54:43Z

1

The exception is caused by this line of code:

Object[] keys1 = obj1.keySet().toArray(new Object[obj1.size()][]);

So, what is it doing: it calls toArray(T[]) on a Set<String>. toArray() takes an array as argument, and stores all the values of the set (String instances, in this case), in that array. But you're passing new Object[obj1.size()][] as argument, i.e. an array of Object[], i.e. an array of arrays of objects. A String is not an Object[], hence the error.

I don't really know what you're trying to do, given that your description really doesn't match with the code. What I know is that your code is begging for custom classes with attributes, instead of maps and arrays.

answered Jul 26, 2015 at 13:54

JB Nizet

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Comments

Ahmed Khalaf · Accepted Answer · 2015-07-26 14:03:03Z

0

I don't like to just post code, but I think this is what you want.

List<Map> list = new ArrayList<>();
for(Map.Entry<String, Integer> entry : hashes.entrySet()) {
    Map newMap = new HashMap();
    newMap.put("name", entry.getKey());
    newMap.put("attributes", entry.getValue());

    list.add(newMap);
}

answered Jul 26, 2015 at 14:03

Ahmed Khalaf

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1 Comment

impulsiveey Over a year ago

Thanks for that! I actually just re-wrote the whole thing, but your code helped. :)

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