What exactly does comparing Integers with == do?

Question

EDIT: OK, OK, I misread. I\'m not comparing an int to an Integer. Duly noted.

My SCJP book says:

When == is used to compare a primitive to a

Answer 1

Integer i1 = 1;
Integer i2 = new Integer(1);
System.out.println(i1 == i2);

When you assign 1 to i1 that value is boxed, creating an Integer object. The comparison then compares the two object references. The references are unequal, so the comparison fails.

Integer i1 = 100;
Integer i2 = 100;
System.out.println(i1 != i2);

Because these are initialized with compile-time constants the compiler can and does intern them and makes both point to the same Integer object.

(Note that I changed the values from 1000 to 100. As @NullUserException points out, only small integers are interned.)

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Here's a really interesting test. See if you can figure this out. Why does the first program print true, but the second one false? Using your knowledge of boxing and compiler time analysis you should be able to figure this out:

// Prints "true".
int i1 = 1;
Integer i2 = new Integer(i1);
System.out.println(i1 == i2);

// Prints "false".
int i1 = 0;
Integer i2 = new Integer(i1);
i1 += 1;
System.out.println(i1 == i2);

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If you understand the above, try to predict what this program prints:

int i1 = 0;
i1 += 1;
Integer i2 = new Integer(i1);
System.out.println(i1 == i2);

(After you guess, run it and see!)