Does int.class equal Integer.class or Integer.TYPE in Java?

Question

Let's imagine one retrieves the declaring type of a Field using reflection.

Which of the following tests will correctly indicate whether one is dealing with an int or an Integer?

Field f = ...
Class<?> c = f.getDeclaringClass();
boolean isInteger;

isInteger = c.equals(Integer.class);
isInteger = c.equals(Integer.TYPE);
isInteger = c.equals(int.class);

isInteger = ( c == Integer.class);
isInteger = ( c == Integer.TYPE);
isInteger = ( c == int.class);

Donnettedonni

The issue with testing myself only is that there is a risk of missing a corner case. That's why I am asking the question. – Spermophyte

JVersty is correct. Implementation and specification are different things. He want to know about specification which are more to hard then just run a piece of code... +1 – Hexastyle

Answer 1

Based on Field.getType() (instead of f.getDeclaringClass()), I get the following:

Type: java.lang.Integer

equals(Integer.class): true
equals(int.class)    : false
equals(Integer.TYPE) : false
== (Integer.class)   : true
== (int.class)       : false
== (Integer.TYPE)    : false

Type: int

equals(Integer.class): false
equals(int.class)    : true
equals(Integer.TYPE) : true
== (Integer.class)   : false
== (int.class)       : true
== (Integer.TYPE)    : true

Type: java.lang.Object

equals(Integer.class): false
equals(int.class)    : false
equals(Integer.TYPE) : false
== (Integer.class)   : false
== (int.class)       : false
== (Integer.TYPE)    : false

Meaning the following is true:

Integer.TYPE.equals(int.class)
Integer.TYPE == int.class

Meaning if I want to find out whether I am dealing with an int or an Integer, I can use any of the following tests:

isInteger = c.equals(Integer.class) || c.equals(Integer.TYPE);
isInteger = c.equals(Integer.class) || c.equals(int.class);
isInteger = (c == Integer.class) || (c == Integer.TYPE);
isInteger = (c == Integer.class) || (c == int.class );

Is there a corner case I am missing? If yes, please comment.

Donnettedonni

And there are no corner cases missing, unless you count a field that is declared as Object and winds up storing an Integer – Yoshida

Oh yeah, and Integer extends Number, so there's another possible corner case if it matters. – Yoshida

Answer 2

int.class compiles down to Integer.TYPE. However, I think you are using Field.getDeclaringClass() incorrectly, as this returns the Class object representing the class that declares the field. What you would want to use is Field.getType().

Yoshida

It would be more accurate to say that int.class points to Integer.TYPE – Bangweulu

.class is a compile time construct. int.class gets compiled to a field access which loads Integer.TYPE. – Bangweulu

@Bangweulu Interesting, I'd never looked at the bytecode. Edited and thanks. – Yoshida