Determining if an Object is of primitive type.

Question

I have an Object[] array, and I am trying to find the ones that are primitives. I've tried to use Class.isPrimitive(), but it seems I'm doing something wrong:

int i = 3;
Object o = i;

System.out.println(o.getClass().getName() + ", " +
                   o.getClass().isPrimitive());

prints java.lang.Integer, false.

Is there a right way or some alternative?

Answer 1

The types in an Object[] will never really be primitive - because you've got references! Here the type of i is int whereas the type of the object referenced by o is Integer (due to auto-boxing).

It sounds like you need to find out whether the type is "wrapper for primitive". I don't think there's anything built into the standard libraries for this, but it's easy to code up:

import java.util.*;

public class Test
{
    public static void main(String[] args)        
    {
        System.out.println(isWrapperType(String.class));
        System.out.println(isWrapperType(Integer.class));
    }

    private static final HashSet<Class<?>> WRAPPER_TYPES = getWrapperTypes();

    public static boolean isWrapperType(Class<?> clazz)
    {
        return WRAPPER_TYPES.contains(clazz);
    }

    private static HashSet<Class<?>> getWrapperTypes()
    {
        HashSet<Class<?>> ret = new HashSet<Class<?>>();
        ret.add(Boolean.class);
        ret.add(Character.class);
        ret.add(Byte.class);
        ret.add(Short.class);
        ret.add(Integer.class);
        ret.add(Long.class);
        ret.add(Float.class);
        ret.add(Double.class);
        ret.add(Void.class);
        return ret;
    }
}

Answer 2

Integer is not a primitive, Class.isPrimitive() is not lying.

Answer 3

Starting in Java 1.5 and up, there is a new feature called auto-boxing. The compiler does this itself. When it sees an opportunity, it converts a primitive type into its appropriate wrapper class.

What is probably happening here is when you declare

Object o = i;

The compiler will compile this statement as saying

Object o = new Integer(i);

This is auto-boxing. This would explain the output you are receiving. This page from the Java 1.5 spec explains auto-boxing more in detail.

Answer 4

I think this happens due to auto-boxing.

int i = 3;
Object o = i;
o.getClass().getName(); // prints Integer

You may implement a utility method that matches these specific boxing classes and gives you if a certain class is primitive.

public static boolean isWrapperType(Class<?> clazz) {
 return clazz.equals(Boolean.class) || 
  clazz.equals(Integer.class) ||
  clazz.equals(Character.class) ||
  clazz.equals(Byte.class) ||
  clazz.equals(Short.class) ||
  clazz.equals(Double.class) ||
  clazz.equals(Long.class) ||
  clazz.equals(Float.class);
}

Answer 5

As several people have already said, this is due to autoboxing.

You could create a utility method to check whether the object's class is Integer, Double, etc. But there is no way to know whether an object was created by autoboxing a primitive; once it's boxed, it looks just like an object created explicitly.

So unless you know for sure that your array will never contain a wrapper class without autoboxing, there is no real solution.

Answer 6

Just so you can see that is is possible for isPrimitive to return true (since you have enough answers showing you why it is false):

public class Main
{
    public static void main(final String[] argv)
    {
        final Class clazz;

        clazz = int.class;
        System.out.println(clazz.isPrimitive());
    }
}

This matters in reflection when a method takes in "int" rather than an "Integer".

This code works:

import java.lang.reflect.Method;

public class Main
{
    public static void main(final String[] argv)
        throws Exception
    {
        final Method method;

        method = Main.class.getDeclaredMethod("foo", int.class);
    }

    public static void foo(final int x)
    {
    }
}

This code failes (cannot find the method):

import java.lang.reflect.Method;

public class Main
{
    public static void main(final String[] argv)
        throws Exception
    {
        final Method method;

        method = Main.class.getDeclaredMethod("foo", Integer.class);
    }

    public static void foo(final int x)
    {
    }
}

Answer 7

The primitve wrapper types will not respond to this value. This is for class representation of primitives, though aside from reflection I can't think of too many uses for it offhand. So, for example

System.out.println(Integer.class.isPrimitive());

prints "false", but

public static void main (String args[]) throws Exception
{
 Method m = Junk.class.getMethod( "a",null);
 System.out.println( m.getReturnType().isPrimitive());
}

public static int a()
{
 return 1;
}

prints "true"

Answer 8

For those who like terse code.

private static final Set<Class> WRAPPER_TYPES = new HashSet(Arrays.asList(
    Boolean.class, Character.class, Byte.class, Short.class, Integer.class, Long.class, Float.class, Double.class, Void.class));
public static boolean isWrapperType(Class clazz) {
    return WRAPPER_TYPES.contains(clazz);
}

Answer 9

Get ahold of BeanUtils from Spring http://static.springsource.org/spring/docs/3.0.x/javadoc-api/

Probably the apache variation (commons beans) has similar funtionality.

Answer 10

commons-lang ClassUtils has wrapperToPrimitive(clazz) method, which will return the primitive correspondence. So you can:

boolean isPrimitiveOrWrapped = 
    clazz.isPrimitive() || ClassUtils.wrapperToPrimitive(clazz) != null;