Why does null reference print as “null”

Question

In println, here o.toString() throws NPE but o1, does not. Why?

public class RefTest {
    public static void main(String[] args) {
        Object o = null;
         


        

Answer 1

It might help showing you the bytecode. Take a look at the following javap output of your class:

> javap -classpath target\test-classes -c RefTest

Compiled from "RefTest.java"
public class RefTest extends java.lang.Object{
public RefTest();
  Code:
   0:   aload_0
   1:   invokespecial   #8; //Method java/lang/Object."<init>":()V
   4:   return

public static void main(java.lang.String[]);
  Code:
   0:   aconst_null
   1:   astore_1
   2:   aconst_null
   3:   astore_2
   4:   getstatic       #17; //Field java/lang/System.out:Ljava/io/PrintStream;
   7:   aload_1
   8:   invokevirtual   #23; //Method java/lang/Object.toString:()Ljava/lang/String;
   11:  invokevirtual   #27; //Method java/io/PrintStream.println:(Ljava/lang/String;)V
   14:  getstatic       #17; //Field java/lang/System.out:Ljava/io/PrintStream;
   17:  aload_2
   18:  invokevirtual   #33; //Method java/io/PrintStream.print:(Ljava/lang/Object;)V
   21:  return

}

Just looking at the main method, you can see the lines of interest are where Code is 8 and 33.

Code 8 shows the bytecode for you calling o.toString(). Here o is null and so any attempt on a method invocation on null results in a NullPointerException.

Code 18 shows your null object being passed as a parameter to the PrintStream.print() method. Looking at the source code for this method will show you why this does not result in the NPE:

public void print(Object obj) {
    write(String.valueOf(obj));
}

and String.valueOf() will do this with nulls:

public static String valueOf(Object obj) {
    return (obj == null) ? "null" : obj.toString();
}

So you can see there is a test there which deals with null, and prevents an NPE.

Answer 2

System.out.println(o.toString())

o.toString() is trying to dereference a null object to convert it to a string, before passing it to println.

System.out.print(o1);

The print being called is the print(Object) variant, which is itself checking that the object is not null before proceeding.

Answer 3

It's because print(Object) uses String.valueOf(Object) for the conversion (aside: after the conversion println(Object) would behave as though print(String) was called, print(Object) effectively uses write(int)). String.valueOf(Object) doesn't throw the NPE like o.toString() does and is instead defined to return "null" for a null parameter.