Simple Java Puzzler

Comparing Integer objects can be tricky.

Posted on December 3, 2017

My first ever blog post was about Java operators I thought I knew. Ironically, after working with Java daily for the last couple of years, I still find things I had no clue about.

My friend Massimo asked this the other day. Given

Integer a = 100;
Integer b = 100;
Integer c = 200;
Integer d = 200;

boolean ab = (a == b);
boolean cd = (c == d);

System.out.println(ab + ", " + cd);

what is printed to stdout?

Using Integer instead of an int, each variable points to an object. == checks for object equality and will compare memory locations. As we’re not reassigning any variables, each references a different object and thus the output should be false, false, no?

Wrong.

The reasoning above is not wrong, but misses the most important piece of this puzzle. When Java encounters code like Integer a = 100; the compiler converts it into Integer a = Integer.valueOf(100); and since we all know what Integer.valueOf does, we never bother to read the docs.

Here’s the SE7 Javadoc:

Returns an Integer instance representing the specified int value. If a new Integer instance is not required, this method should generally be used in preference to the constructor Integer(int), as this method is likely to yield significantly better space and time performance by caching frequently requested values. This method will always cache values in the range -128 to 127, inclusive, and may cache other values outside of this range.

Last sentence spills the beans. Integer.valueOf will always cache values in the interval [-128,127]. Here’s the implementation:

public static Integer valueOf(int i) {
  assert IntegerCache.high >= 127;
  if (i >= IntegerCache.low && i <= IntegerCache.high)
    return IntegerCache.cache[i + (-IntegerCache.low)];
  return new Integer(i);
}

IntegerCache caches all Integer objects between -128 and the upper bound, which will be at least 127, but can be extended by using AutoBoxCacheMax VM property.

Returning to the puzzle, we now know

Integer a = 100;
Integer b = 100;
boolean ab = (a == b);

both a and b will point to the same object as non-primitive 100 is cached, meaning == will yield true.

Similarly, assuming AutoBoxCacheMax remains at default 127,

Integer c = 200;
Integer d = 200;
boolean cd = (c == d);

c and d will not be cached, as 200 is out of bounds and therefore a new Integer object will be assigned to each variable. == will then compare two distinct memory locations and yield false.

Putting the two together, we can now confidently say the output is true, false, which is - unsurprisingly now - the correct answer.




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