Why are these == but not `equals()`?

I'm a bit confused about the way Java treats == and equals() when it comes to int, Integer and other types of numbers. For example:

Integer X = 9000;
int x = 9000;
Short Y = 9000;
short y = 9000;
List<Boolean> results = new ArrayList<Boolean>();
// results.add(X == Y); DOES NOT COMPILE        1)
results.add(Y == 9000);                      // 2)
results.add(X == y);                         // 3)
results.add(X.equals(x));                    // 4)
results.add(X.equals(Y));                    // 5)
results.add(X.equals(y));                    // 6)
System.out.println(results);

outputs (maybe you should make your guess first):

[true, true, true, false, false]

1. That X == Y does not compile is to be expected, being different objects.
2. I'm a little surprised that Y == 9 is true, given that 9 is by default an int, and given that 1) didn't even compile. Note that you can't put an int into a method expecting a Short, yet here they are equal.
3. This is surprising for the same reason as two, but it seems worse.
4. Not surprising, as x is autoboxed to and Integer.
5. Not surprising, as objects in different classes should not be equal().
6. What?? X == y is true but X.equals(y) is false? Shouldn't == always be stricter than equals()?

I'd appreciate it if anyone can help me make sense of this. For what reason do == and equals() behave this way?

Edit: I have changed 9 to 9000 to show that this behavior is not related to the any unusual ways that the integers from -128 to 127 behave.

2^nd Edit: OK, if you think you understand this stuff, you should consider the following, just to make sure:

Integer X = 9000;
Integer Z = 9000;
short y = 9000;
List<Boolean> results = new ArrayList<Boolean>();
results.add(X == Z);                      // 1)
results.add(X == y);                      // 2)
results.add(X.equals(Z));                 // 3)
results.add(X.equals(y));                 // 4)
System.out.println(results);

outputs:

[false, true, true, false]

The reason, as best as I understand it:

1. Different instance, so different.
2. X unboxed, then same value, so equal.
3. Same value, so equal.
4. y cannot be boxed to an Integer so cannot be equal.

Solution 1:

(small) Integer instances are cached, so the invariant x == y is holded for small instances (actually -127 +128, depends on JVM):

Integer a = 10;
Integer b = 10;

assert(a == b); // ok, same instance reused

a = 1024;
b = 1024;

assert(a == b); // fail, not the same instance....
assert(a.equals(b)); // but same _value_

EDIT

4) and 5) yield false because equals check types: X is an Integer whereas Y is a Short. This is the java.lang.Integer#equals method:

public boolean equals(Object obj) {
    if (obj instanceof Integer) {
        return value == ((Integer)obj).intValue();
    }

    return false;
}

Solution 2:

The reason for

X == y

being true has to do with binary numeric promotion. When at least one operand to the equality operator is convertible to a numeric type, the numeric equality operator is used. First, the first operand is unboxed. Then, both operands are converted to int.

While

X.equals(y)

is a normal function call. As has been mentioned, y will be autoboxed to a Short object. Integer.equals always returns false if the argument is not an Integer instance. This can be easily seen by inspecting the implementation.

One could argue that this is a design flaw.

Solution 3:

The morale of the story:

Autoboxing/unboxing is confusing, as is type promotion. Together, they make for good riddles but horrendous code.

In practice, it seldom makes sense to use numeric types smaller than int, and I'm almost inclined to configure my eclipse compiler to flag all autoboxing and -unboxing as an error.