How to cast an Object to an int
JavaObjectCastingIntegerJava Problem Overview
How can I cast an Object to an int in java?
Java Solutions
Solution 1 - Java
If you're sure that this object is an Integer
:
int i = (Integer) object;
Or, starting from Java 7, you can equivalently write:
int i = (int) object;
Beware, it can throw a ClassCastException
if your object isn't an Integer
and a NullPointerException
if your object is null
.
This way you assume that your Object is an Integer (the wrapped int) and you unbox it into an int.
int
is a primitive so it can't be stored as an Object
, the only way is to have an int
considered/boxed as an Integer
then stored as an Object
.
If your object is a String
, then you can use the Integer.valueOf()
method to convert it into a simple int :
int i = Integer.valueOf((String) object);
It can throw a NumberFormatException
if your object isn't really a String
with an integer as content.
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Solution 2 - Java
Scenario 1: simple case
If it's guaranteed that your object is an Integer
, this is the simple way:
int x = (Integer)yourObject;
Scenario 2: any numerical object
In Java Integer
, Long
, BigInteger
etc. all implement the Number
interface which has a method named intValue
. Any other custom types with a numerical aspect should also implement Number
(for example: Age implements Number
). So you can:
int x = ((Number)yourObject).intValue();
Scenario 3: parse numerical text
When you accept user input from command line (or text field etc.) you get it as a String
. In this case you can use Integer.parseInt(String string)
:
String input = someBuffer.readLine();
int x = Integer.parseInt(input);
If you get input as Object
, you can use (String)input
, or, if it can have an other textual type, input.toString()
:
int x = Integer.parseInt(input.toString());
Scenario 4: identity hash
In Java there are no pointers. However Object
has a pointer-like default implementation for hashCode()
, which is directly available via System.identityHashCode(Object o)
. So you can:
int x = System.identityHashCode(yourObject);
Note that this is not a real pointer value. Objects' memory address can be changed by the JVM while their identity hashes are keeping. Also, two living objects can have the same identity hash.
You can also use object.hashCode()
, but it can be type specific.
Scenario 5: unique index
In same cases you need a unique index for each object, like to auto incremented ID values in a database table (and unlike to identity hash which is not unique). A simple sample implementation for this:
class ObjectIndexer {
private int index = 0;
private Map<Object, Integer> map = new WeakHashMap<>();
// or:
// new WeakIdentityHashMap<>();
public int indexFor(Object object) {
if (map.containsKey(object)) {
return map.get(object);
} else {
index++;
map.put(object, index);
return index;
}
}
}
Usage:
ObjectIndexer indexer = new ObjectIndexer();
int x = indexer.indexFor(yourObject); // 1
int y = indexer.indexFor(new Object()); // 2
int z = indexer.indexFor(yourObject); // 1
Scenario 6: enum member
In Java enum members aren't integers but full featured objects (unlike C/C++, for example). Probably there is never a need to convert an enum object to int
, however Java automatically associates an index number to each enum member. This index can be accessed via Enum.ordinal()
, for example:
enum Foo { BAR, BAZ, QUX }
// ...
Object baz = Foo.BAZ;
int index = ((Enum)baz).ordinal(); // 1
Solution 3 - Java
Assuming the object is an Integer
object, then you can do this:
int i = ((Integer) obj).intValue();
If the object isn't an Integer
object, then you have to detect the type and convert it based on its type.
Solution 4 - Java
@Deprecated
public static int toInt(Object obj)
{
if (obj instanceof String)
{
return Integer.parseInt((String) obj);
} else if (obj instanceof Number)
{
return ((Number) obj).intValue();
} else
{
String toString = obj.toString();
if (toString.matches("-?\d+"))
{
return Integer.parseInt(toString);
}
throw new IllegalArgumentException("This Object doesn't represent an int");
}
}
As you can see, this isn't a very efficient way of doing it. You simply have to be sure of what kind of object you have. Then convert it to an int the right way.
Solution 5 - Java
You have to cast it to an Integer (int's wrapper class). You can then use Integer's intValue() method to obtain the inner int.
Solution 6 - Java
Answer:
int i = ( Integer ) yourObject;
If, your object is an integer already, it will run smoothly. ie:
Object yourObject = 1;
// cast here
or
Object yourObject = new Integer(1);
// cast here
etc.
If your object is anything else, you would need to convert it ( if possible ) to an int first:
String s = "1";
Object yourObject = Integer.parseInt(s);
// cast here
Or
String s = "1";
Object yourObject = Integer.valueOf( s );
// cast here
Solution 7 - Java
I use a one-liner when processing data from GSON:
int i = object != null ? Double.valueOf(object.toString()).intValue() : 0;
Solution 8 - Java
You can't. An int
is not an Object
.
Integer
is an Object
though, but I doubt that's what you mean.
Solution 9 - Java
Can't be done. An int
is not an object, it's a primitive type. You can cast it to Integer, then get the int.
Integer i = (Integer) o; // throws ClassCastException if o.getClass() != Integer.class
int num = i; //Java 1.5 or higher
Solution 10 - Java
If the Object
was originally been instantiated as an Integer
, then you can downcast it to an int
using the cast operator (Subtype)
.
Object object = new Integer(10);
int i = (Integer) object;
Note that this only works when you're using at least Java 1.5 with autoboxing feature, otherwise you have to declare i
as Integer
instead and then call intValue()
on it.
But if it initially wasn't created as an Integer
at all, then you can't downcast like that. It would result in a ClassCastException
with the original classname in the message. If the object's toString()
representation as obtained by String#valueOf()
denotes a syntactically valid integer number (e.g. digits only, if necessary with a minus sign in front), then you can use Integer#valueOf()
or new Integer()
for this.
Object object = "10";
int i = Integer.valueOf(String.valueOf(object));
###See also:
Solution 11 - Java
int i = (Integer) object; //Type is Integer.
int i = Integer.parseInt((String)object); //Type is String.
Solution 12 - Java
If you mean cast a String to int, use Integer.valueOf("123")
.
You can't cast most other Objects to int though, because they wont have an int value. E.g. an XmlDocument has no int value.
Solution 13 - Java
I guess you're wondering why C or C++ lets you manipulate an object pointer like a number, but you can't manipulate an object reference in Java the same way.
Object references in Java aren't like pointers in C or C++... Pointers basically are integers and you can manipulate them like any other int. References are intentionally a more concrete abstraction and cannot be manipulated the way pointers can.
Solution 14 - Java
int[] getAdminIDList(String tableName, String attributeName, int value) throws SQLException {
ArrayList list = null;
Statement statement = conn.createStatement();
ResultSet result = statement.executeQuery("SELECT admin_id FROM " + tableName + " WHERE " + attributeName + "='" + value + "'");
while (result.next()) {
list.add(result.getInt(1));
}
statement.close();
int id[] = new int[list.size()];
for (int i = 0; i < id.length; i++) {
try {
id[i] = ((Integer) list.get(i)).intValue();
} catch(NullPointerException ne) {
} catch(ClassCastException ch) {}
}
return id;
}
// enter code here
This code shows why ArrayList
is important and why we use it. Simply casting int
from Object
. May be its helpful.
Solution 15 - Java
For Example Object variable; hastaId
Object hastaId = session.getAttribute("hastaID");
For Example Cast an Object to an int,hastaID
int hastaID=Integer.parseInt(String.valueOf(hastaId));
Solution 16 - Java
Refer This code:
public class sample
{
public static void main(String[] args)
{
Object obj=new Object();
int a=10,b=0;
obj=a;
b=(int)obj;
System.out.println("Object="+obj+"\nB="+b);
}
}
Solution 17 - Java
so divide1=me.getValue()/2;
int divide1 = (Integer) me.getValue()/2;
Solution 18 - Java
Finally, the best implementation for your specification was found.
public int tellMyNumber(Object any) {
return 42;
}
Solution 19 - Java
first check with instanceof keyword . if true then cast it.