Unreported exception java.lang.Exception; must be caught or declared to be thrown

Unreported exception java.lang.Exception; must be caught or declared to be thrown

public static byte[] m16h(byte[] m) throws Exception

The signature of your method indicates that an Exception is susceptible of being thrown.

This means that the exception either :

1. Must be handled by the caller

   try {
       System.out.println(xor(m16h(add(xor(xor(m16h(add(k1, m16h(add(k2, m16h(k3))))), k3), k2), k1)), k3));
   } catch (Exception e) {
       e.printStackTrace();
   }
   

2. Must be rethrowed by the caller

   public static void main(String[] args) throws Exception
   

In your main method, the call to m16h may lead to an exception being thrown. In this case, you have two choices:

• handle the exception yourself in the main method.

// in the main
try {
    System.out.println(xor(m16h(...));
} catch(Exception e) {
    // do something, e.g. print e.getMessage()
}

• indicate that the main method can throw an exception, by appending throws Exception to its declaration.

public static void main(String args[]) throws Exception

Unreported exception java.lang.Exception; must be caught or declared to be thrown

Surround the line where you call this method with a try/catch block as follows:

try {
  System.out.println(xor(m16h(add(xor(xor(m16h(add(k1, m16h(add(k2, m16h(k3))))), k3), k2), k1)), k3));
}catch (Exception e){
  System.out.println (e.getMessage());
}