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up vote 12 down vote favorite

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

share|improve this question
Why are you testing for the impossible? You create the map yourself; if it is null, your runtime system is broken. Since this is obviously a shortened version of other work you are doing, I would suggest you ensure that objects are always initializedso you don't have to do the map == null checks. The keys are your problem. – Eric Jablow Jul 9 at 13:57
@Eric Jablow This is just a simplified version of my code. The map in question is actually being retrieved from another place where they are dynamically created on demand. That is why the check. Here I included the initialization so that it is clear to people who are answering that the if-true case is not being executed when my problem is occuring – radiantRazor Jul 9 at 14:02
I now understand. Probably the best thing would be to throw an exception immediately if someone were to pass a null map to your code. Or, if someone passes in a null to your code, set your map to an empty one instead. – Eric Jablow Jul 9 at 18:19

3 Answers

up vote 14 down vote accepted

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.

share|improve this answer
In that case, could the poster use the ternary operator by casting 0 to Long instead of long? – MathSquared11235 Jul 9 at 13:48
1  
Yes, he'd have to cast 0L to a Long though. – Jason Jul 9 at 13:52
2  
Woww, the JLS has so many hidden catches. My obvious follow up question would be why is the JLS defined to infer the type as the primitive one? If it was defined that the type should be inferred as the autoboxed type, then such nullpointer exceptions could have been avoided. I will try asking that as a separate question maybe – radiantRazor Jul 9 at 14:04
up vote 3 down vote

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

They are not equivalent; the ternary operator has a few caveats.

The if-true argument of the ternary operator, (long) 0, is of the primitive type long. Consequently, the if-false argument will be automatically unboxed from Long to long (as per JLS §15.25):

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.

However, this argument is null (since your map does not contain the string "non-existent key", meaning get() returns null), so a NullPointerException occurs during the unboxing process.

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But why is the type of the if-true value affecting the type of if-false value? Shouldn't that be determined by the type on the LHS? – radiantRazor Jul 9 at 13:48
1  
@radiantRazor In short, because that's how the JLS defines the ternary operator. – arshajii Jul 9 at 13:51
up vote 0 down vote

I commented above suggesting that he ensure map never be null, but that does not help with the ternary problem. As a practical matter, it's easier to let the system do the work for you. He could use Apache Commons Collections 4 and its DefaultedMap class.

import static org.apache.commons.collections4.map.DefaultedMap.defaultedMap;

Map<String, Long> map = ...;  // Ensure not null.
Map<String, Long> dMap = defaultedMap(map, 0L);

Google Guava doesn't have anything as easy as that, but one can wrap map with the Maps.transformValues() method.

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