Java Auto-Unboxing Gotcha. Beware!


What do you think that the following code snippet will print?

Object o = true ? new Integer(1) : new Double(2.0);
System.out.println(o);

Yes! It will print:

1.0

What? 1.0? But I have assigned an Integer to my o variable. Why does it print 1.0? It turns out that there is a subtle little specification section in the JLS’s §15.25, which specifies the ternary operator. Here’s what is applied to the above:

The type of a conditional expression is determined as follows:

  • […]
  • Otherwise, if the second and third operands have types that are convertible (§5.1.8) to numeric types, then there are several cases:
    • […]
    • Otherwise, binary numeric promotion (§5.6.2) is applied to the operand types, and the type of the conditional expression is the promoted type of the second and third operands.

      Note that binary numeric promotion performs value set conversion (§5.1.13) and may perform unboxing conversion (§5.1.8).

Binary numeric promotion may implicitly perform unboxing conversion! Eek! Who would have expected this? You can get a NullPointerException from auto-unboxing, if one of the operands is null, the following will fail

Integer i = new Integer(1);
if (i.equals(1))
    i = null;
Double d = new Double(2.0);
Object o = true ? i : d; // NullPointerException!
System.out.println(o);

Obviously (obviously !?) you can circumvent this problem by casting numeric types to non-numeric types, e.g. Object

Object o1 = true 
  ? (Object) new Integer(1) 
  : new Double(2.0);
System.out.println(o1);

The above will now print

1

Credits for discovery of this gotcha go to Paul Miner, who has explained this more in detail here on reddit.

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