How to Handle Exceptions in Java's AES Encryption Method?

Exception Handling in Java Code

Consider the following Java code snippet:

import java.awt.*;
import javax.swing.*;
import java.awt.event.*;
import javax.crypto.*;
import javax.crypto.spec.*;
import java.security.*;
import java.io.*;


public class EncryptURL extends JApplet implements ActionListener {

    // ...

    public void encrypt(String toEncrypt) throws Exception {
        try {
            String plaintext = toEncrypt;
            String key = "01234567890abcde";
            String iv = "fedcba9876543210";

            SecretKeySpec keyspec = new SecretKeySpec(key.getBytes(), "AES");
            IvParameterSpec ivspec = new IvParameterSpec(iv.getBytes());

            Cipher cipher = Cipher.getInstance("AES/CBC/NoPadding");
            cipher.init(Cipher.ENCRYPT_MODE, keyspec, ivspec);
            byte[] encrypted = cipher.doFinal(toEncrypt.getBytes());

            return encrypted;
        }
        catch(Exception e) {
            return null; // Always must return something
        }
    }


    public void actionPerformed(ActionEvent e) {
        if (e.getActionCommand() == "Submit") {

            // ...

            String concatURL =
                "user=" + subUserName + "&f=" + subFName +
                "&l=" + subLName + "&em=" + subEmail +
                "&p=" + subPhone + "&h=" + subHeartbeatID +
                "&re=" + subRegionCode + "&ret=" + subRetRegionCode;

            // ...

            try {
                byte[] encrypted = encrypt(concatURL);
                String encryptedString = bytesToHex(encrypted);
                // ...
            } catch (Exception exc) {
                // TODO: handle exception
            }
        }
    }

    // ...

}

The following questions and answers address the problems faced while compiling this code snippet.

Question 1: Why does the code snippet throw an "Exception; must be caught or declared to be thrown" exception when trying to compile it?

Answer:

This exception occurs because the encrypt method is declared to throw an Exception but does not handle it in its try-catch block. To resolve this, a catch block has been added to the encrypt method to handle the Exception.

Question 2: Why does the code snippet throw a "missing return statement" exception for line 109?

Answer:

The encrypt method has a declared return type of byte[], which means it must return a value of that type in all cases. The original code did not have a return statement in the catch block, resulting in an exception. A return statement has been added to the catch block to return null in case of an exception.

Lessons Learned:

1. A method with a return type must always return an object of that type in all possible scenarios.
2. All checked exceptions must always be handled.

The above is the detailed content of How to Handle Exceptions in Java's AES Encryption Method?. For more information, please follow other related articles on the PHP Chinese website!