Weird behaviour when using Java ternary operator

Question

When I write my java code like this:

Map<String, Long> map = new HashMap<>()
Long number =null;
if(map == null)
    number = (long) 0;
else
    number = map.get("non-existent key");

the app runs as expected but when I do this:

Map<String, Long> map = new HashMap<>();
Long number= (map == null) ? (long)0 : map.get("non-existent key");

I get a NullPointerException on the second line. The debug pointer jumps from the second line to this method in the java.lang.Thread class:

 /**
     * Dispatch an uncaught exception to the handler. This method is
     * intended to be called only by the JVM.
     */
     private void dispatchUncaughtException(Throwable e) {
         getUncaughtExceptionHandler().uncaughtException(this, e);
     }

What is happening here? Both these code paths are exactly equivalent isn't it?

---

Edit

I am using Java 1.7 U25

Answer 1

They are not equivalent.

The type of this expression

(map == null) ? (long)0 : map.get("non-existent key");

is long because the true result has type long.

The reason this expression is of type long is from section §15.25 of the JLS:

If one of the second and third operands is of primitive type T, and the type of the other is the result of applying boxing conversion (§5.1.7) to T, then the type of the conditional expression is T.

When you lookup a non-existant key the map returns null. So, Java is attempting to unbox it to a long. But it's null. So it can't and you get a NullPointerException. You can fix this by saying:

Long number = (map == null) ? (Long)0L : map.get("non-existent key");

and then you'll be okay.

However, here,

if(map == null)
    number = (long) 0;
else
    number = map.get("non-existent key");

since number is declared as Long, that unboxing to a long never occurs.