Check type of primitive field

JavaReflectionPrimitive

Java Problem Overview


I'm trying to determine the type of a field on an object. I don't know the type of the object when it is passed to me but I need to find fields which are longs. It is easy enough to distinguish the boxed Longs but the primitive long seems more difficult.

I can make sure that the objects passed to me only have Longs, not the primitives, but I'd rather not. So what I have is:

for (Field f : o.getClass().getDeclaredFields()) {
    Class<?> clazz = f.getType();
    if (clazz.equals(Long.class)) {
        // found one -- I don't get here for primitive longs
    }
}

A hacky way, which seems to work, is this:

for (Field f : o.getClass().getDeclaredFields()) {
    Class<?> clazz = f.getType();
    if (clazz.equals(Long.class) ||  clazz.getName().equals("long")) {
        // found one
    }
}

I'd really like a cleaner way to do this if there is one. If there is no better way then I think that requiring the objects I receive to only use Long (not long) would be a better API.

Any ideas?

Java Solutions


Solution 1 - Java

You're using the wrong constant to check for Long primitives - use Long.TYPE, each other primitive type can be found with a similarly named constant on the wrapper. eg: Byte.TYPE, Character.TYPE, etc.

Solution 2 - Java

o.getClass().getField("fieldName").getType().isPrimitive();

Solution 3 - Java

You can just use

boolean.class
byte.class
char.class
short.class
int.class
long.class
float.class
double.class
void.class

If you are using reflection, why do you care, why do this check at all. The get/set methods always use objects so you don't need to know if the field is a primitive type (unless you try to set a primitive type to the null value.)

In fact, for the method get() you don't need to know which type it is. You can do

// any number type is fine.
Number n = field.get(object);
long l = n.longValue();

If you are not sure if it is a Number type you can do

Object o = field.get(object); // will always be an Object or null.
if (o instanceof Number) {
     Number n = (Number) o;
     long l = n.longValue();

Solution 4 - Java

  • To detect fields with long type use long.class or Long.TYPE.

  • To detect fields with Long type use Long.class.

Example:

for (Field f : o.getClass().getDeclaredFields()) {
    Class<?> clazz = f.getType();
    // to detect both Long and long types
    if (Long.class.equals(clazz) || long.class.equals(clazz)) {
        // found one
    }
}

Notice:

Long.TYPE is static Constant member and is equivalent to long.class.

snippet code form Long Class

> /** > * The {@link Class} object that represents the primitive type {@code long}. > */ > @SuppressWarnings("unchecked") > public static final Class TYPE > = (Class) long[].class.getComponentType();

Also check for answer for Difference between Integer.class and Integer.TYPE question

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionmacbutchView Question on Stackoverflow
Solution 1 - JavamP.View Answer on Stackoverflow
Solution 2 - JavaDrooView Answer on Stackoverflow
Solution 3 - JavaPeter LawreyView Answer on Stackoverflow
Solution 4 - Javaahmed hamdyView Answer on Stackoverflow