java.lang.ClassCastException: java.lang.Short cannot be cast to java.lang.Integer

Question

In my project I have an Enum like this :

public enum MyEnum {

FIRST(1),

SECOND(2);

private int value;

private MyEnum(int value) {
    this.value = value;
}

public int getValue() {
    return value;
}

public static MyEnum fromValue(int value) {
    for (MyEnum e : MyEnum.values()) {
        if (e.getValue() == value) {
            return e;
        }
    }
    return null;
}

And I have this code :

Map<String, Object> myMap = new HashMap<>();
// Fill my Map with data from database
myMap = namedParameterJdbcTemplate.queryForList(qry, paramsMap);

***if (Arrays.asList(MyEnum.values())
                .contains(MyEnum.fromValue((int) 
                myMap.get("myKey")) )))*** {
    // Do something
    }

I get the Exception

 java.lang.ClassCastException: java.lang.Short cannot be cast to java.lang.Integer** on this line : **if (Arrays.asList(MyEnum.values())
            .contains(MyEnum.fromValue((int)myMap.get("myKey")) )))

myMap is filled with data from database, knowing that it is an SQL Server Database and that the myKey returned from database is of type tinyint in the database.

Can you please tell me what I am doing wrong ? Thanks.

Regards.

By the way, EnumSet.allOf(MyEnum.class) is far more efficient (and easier to read) than Arrays.asList(MyEnum.values()). See the documentation for more information. — VGR
– VGR, Commented Jul 26, 2017 at 16:39

HatsuPointerKun · Accepted Answer · 2017-07-26 10:32:27Z

2

Here :

(int)myMap.get("myKey")

I guess that the get method returned a reference to a Short object, which can't be cast to Integer, since short doesn't inherit from Integer.

What you can do is casting to Short, then call the intValue method, like this :

((Short)myMap.get("myKey")).intValue()

Which will return a integer value

answered Jul 26, 2017 at 10:32

HatsuPointerKun

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1 Comment

niryo Over a year ago

thanks, it worked. But I had to choose the cleanest solution

davidxxx · Accepted Answer · 2017-07-26 17:44:25Z

1

java.lang.ClassCastException: java.lang.Short cannot be cast to java.lang.Integer

Here myMap.get("myKey") returns a Short instance but the returned declared object is Object as the map is declared Map<String, Object>.

It is as if you had written :

Object myValue = myMap.get("myKey")

Then you pass this Short object declared as Object to MyEnum.fromValue() that has as parameter an int.

MyEnum.fromValue((int)myValue));

The compiler tries so to cast the object to an Integer.
But it is not an Integer but a Short.
So the cast fails.

To solve your problem, you should first check the instance of your object.
If the object is not an instance of Short, you should throw an exception or ignore the value.
If it is, you should cast from Object to Short and pass this Short to fromValue() that will be automatically unboxed to short :

   Object myValue =  myMap.get("myKey");
   if (!(myValue instanceof Short)){          
      throw new YourRuntimeException("A Short is expected but get a " + myValue.getClass());
      // or ignore the value
   }

   if (Arrays.asList(MyEnum.values())
              .contains(MyEnum.fromValue(((short) myValue)) {
           ....
    }

edited Jul 26, 2017 at 17:44

answered Jul 26, 2017 at 10:43

davidxxx

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3 Comments

niryo Over a year ago

Thank you it works. However I've written this little part of code which works : Short i = new Short((short) 1); int j = (int) i; System.out.println(i); . It's the same thing and it works in this case, can you please tell me why ? I'm just trying to understand. Thank you in advance.

davidxxx Over a year ago

The reason is simple. Here int j = (int) i; the compiler unboxes i value from Short to short and then does a widening conversion from short to int. Besides the cast is useless for widening conversion. int j = i; is enough. It is required only for narrowing conversion. In your actual code, the Short instance is declared as Object. So the compiler doesn't try to unbox it but do directly a cast to the target cast. I edited to explain it and provide a still more effective solution.

niryo Over a year ago

Thank you very much, I understand now.

user4910279 · Accepted Answer · 2017-07-26 10:50:08Z

0

Try this.

(int)(short)myMap.get("myKey")

answered Jul 26, 2017 at 10:50

user4910279

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