How to resolve the java.lang.NullPointerException
In Java, the java.lang.NullPointerException is thrown when a reference variable is accessed (or de-referenced) and is not pointing to any object. This error can be resolved by using a try-catch block or an if-else condition to check if a reference variable is null before dereferencing it.
In some cases, the compiler prevents this exception with the compile-time error “The variable might not have been initialized” when a null reference variable is passed as an argument of a method:
String s;
foo(s); // Compiler gives an error.
However, the compiler does not give this error when null is directly passed to a function; however, there is a higher chance of it throwing a NullPointerException:
foo(null); // Higher probability of an exception
Real-world example
Now let’s enhance our understanding with a real-world example
public class NullPointerExample {public static void main(String[] args) {String s = "example";// Calling methods with null argumentfoo(null);bar(null);}// Using a try-catch block:static void foo(String x){try {System.out.println("First character: " + x.charAt(0));} catch(NullPointerException e) {System.out.println("NullPointerException thrown!");}}// Using if-else condition:static void bar(String x){if(x != null)System.out.println("First character: " + x.charAt(0));elseSystem.out.println("NullPointerException thrown!");}}
Upon running this code, we will get the java.lang.NullPointerException error because we are calling the function with a null argument. Now let's resolve it.
public class NullPointerExcept {public static void main(String[] args) {String s = "example";// Calling methods with valid argumentfoo(s); // Resolved: Passing a valid string "example"bar(s); // Resolved: Passing a valid string "example"}// Using a try-catch block:static void foo(String x){if (x == null) {System.out.println("String is null!");return;}System.out.println("First character: " + x.charAt(0));}// Using if-else condition:static void bar(String x){if(x == null) {System.out.println("String is null!");return;}System.out.println("First character: " + x.charAt(0));}}
We just solved the issue by passing a valid string argument to the function.
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