Why does String.valueOf(null) throw a NullPointerException?

The issue is that String.valueOf method is overloaded:

Java Specification Language mandates that in these kind of cases, the most specific overload is chosen:

JLS 15.12.2.5 Choosing the Most Specific Method

If more than one member method is both accessible and applicable to a method invocation, it is necessary to choose one to provide the descriptor for the run-time method dispatch. The Java programming language uses the rule that the most specific method is chosen.

A char[] is-an Object, but not all Object is-a char[]. Therefore, char[] is more specific than Object, and as specified by the Java language, the String.valueOf(char[]) overload is chosen in this case.

String.valueOf(char[]) expects the array to be non-null, and since null is given in this case, it then throws NullPointerException.

The easy “fix” is to cast the null explicitly to Object as follows:

System.out.println(String.valueOf((Object) null));
// prints "null"

Related questions


Moral of the story

There are several important ones:

  • Effective Java 2nd Edition, Item 41: Use overloading judiciously
    • Just because you can overload, doesn’t mean you should every time
    • They can cause confusion (especially if the methods do wildly different things)
  • Using good IDE, you can check which overload is selected at compile time
    • With Eclipse, you can mouse-hover on the above expression and see that indeed, the valueOf(char[]) overload is selected!
  • Sometimes you want to explicitly cast null (examples to follow)

See also


On casting null

There are at least two situations where explicitly casting null to a specific reference type is necessary:

  • To select overloading (as given in above example)
  • To give null as a single argument to a vararg parameter

A simple example of the latter is the following:

static void vararg(Object... os) {
    System.out.println(os.length);
}

Then, we can have the following:

vararg(null, null, null); // prints "3"
vararg(null, null);       // prints "2"
vararg(null);             // throws NullPointerException!

vararg((Object) null);    // prints "1"

See also

Related questions

Leave a Comment