Question

try the following code to see a nullpointer error (if you don’t see the error because of the autograding, you can copy it into the pencil icon scratch area to run it without the grader). since s is null, indexof throws an nullpointer error for s. comment out the first if statement and run the program again. the second if statement avoids the error with shortcircuit evaluation. because s !

Answer 1

Using the knowledge of computational language in JAVA it is possible to describe  since s is null, indexof throws an nullpointer error for s. comment out the first if statement and run the program again. the second if statement avoids the error with shortcircuit evaluation.

Writting the code:

public class Printer {

   private String name;

   public void print() {

       printString(name);

   }

public class Example {

   public static void main(String[] args) {

       Object obj = null;

       obj.hashCode();

   }

}

   public static void main(String[] args) {

       Printer printer = new Printer();

       printer.print();

   }

}

   }

public class Example {

   public static void main(String[] args) {

       Object obj = null;

       obj.hashCode();

   }

}

   public static void main(String[] args) {

       Printer printer = new Printer("123");

       printer.print();

   }

}

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