How do I find out what type each object is in a ArrayList<Object>?
JavaGenericsReflectionArraylistJava Problem Overview
I have a ArrayList
Java Solutions
Solution 1 - Java
In C#:
Fixed with recommendation from Mike
ArrayList list = ...;
// List<object> list = ...;
foreach (object o in list) {
if (o is int) {
HandleInt((int)o);
}
else if (o is string) {
HandleString((string)o);
}
...
}
In Java:
ArrayList<Object> list = ...;
for (Object o : list) {
if (o instanceof Integer)) {
handleInt((Integer o).intValue());
}
else if (o instanceof String)) {
handleString((String)o);
}
...
}
Solution 2 - Java
You can use the getClass()
method, or you can use instanceof. For example
for (Object obj : list) {
if (obj instanceof String) {
...
}
}
or
for (Object obj : list) {
if (obj.getClass().equals(String.class)) {
...
}
}
Note that instanceof will match subclasses. For instance, of C
is a subclass of A
, then the following will be true:
C c = new C();
assert c instanceof A;
However, the following will be false:
C c = new C();
assert !c.getClass().equals(A.class)
Solution 3 - Java
for (Object object : list) {
System.out.println(object.getClass().getName());
}
Solution 4 - Java
You almost never want you use something like:
Object o = ...
if (o.getClass().equals(Foo.class)) {
...
}
because you aren't accounting for possible subclasses. You really want to use Class#isAssignableFrom:
Object o = ...
if (Foo.class.isAssignableFrom(o)) {
...
}
Solution 5 - Java
In Java just use the instanceof operator. This will also take care of subclasses.
ArrayList<Object> listOfObjects = new ArrayList<Object>();
for(Object obj: listOfObjects){
if(obj instanceof String){
}else if(obj instanceof Integer){
}etc...
}
Solution 6 - Java
import java.util.ArrayList;
/**
* @author potter
*
*/
public class storeAny {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
ArrayList<Object> anyTy=new ArrayList<Object>();
anyTy.add(new Integer(1));
anyTy.add(new String("Jesus"));
anyTy.add(new Double(12.88));
anyTy.add(new Double(12.89));
anyTy.add(new Double(12.84));
anyTy.add(new Double(12.82));
for (Object o : anyTy) {
if(o instanceof String){
System.out.println(o.toString());
} else if(o instanceof Integer) {
System.out.println(o.toString());
} else if(o instanceof Double) {
System.out.println(o.toString());
}
}
}
}
Solution 7 - Java
Just call .getClass()
on each Object
in a loop.
Unfortunately, Java doesn't have map()
. :)
Solution 8 - Java
Instanceof works if you don't depend on specific classes, but also keep in mind that you can have nulls in the list, so obj.getClass() will fail, but instanceof always returns false on null.
Solution 9 - Java
Since Java 8
mixedArrayList.forEach((o) -> {
String type = o.getClass().getSimpleName();
switch (type) {
case "String":
// treat as a String
break;
case "Integer":
// treat as an int
break;
case "Double":
// treat as a double
break;
...
default:
// whatever
}
});
Solution 10 - Java
instead of using object.getClass().getName()
you can use object.getClass().getSimpleName()
, because it returns a simple class name without a package name included.
for instance,
Object[] intArray = { 1 };
for (Object object : intArray) {
System.out.println(object.getClass().getName());
System.out.println(object.getClass().getSimpleName());
}
gives,
java.lang.Integer
Integer
Solution 11 - Java
You say "this is a piece of java code being written", from which I infer that there is still a chance that you could design it a different way.
Having an ArrayList
Solution 12 - Java
If you expect the data to be numeric in some form, and all you are interested in doing is converting the result to a numeric value, I would suggest:
for (Object o:list) {
Double.parseDouble(o.toString);
}