


Why Does My Java Code Throw an 'Exception; must be caught or declared to be thrown' Error?
Exception: "Exception; must be caught or declared to be thrown" while Compiling Java Code
Background:
When encountering the error "Exception; must be caught or declared to be thrown," it indicates that an exception has been raised during program execution, but the programmer has not handled or declared it to be thrown.
The Issue:
In the provided code snippet:
byte[] encrypted = cipher.doFinal(toEncrypt.getBytes()); return encrypted;
The encrypt method is missing an exception declaration in its method signature and fails to handle any exceptions that may occur within the try block.
Solution:
To resolve the issue, modify the encrypt method signature to declare the exception it might throw:
public static byte[] encrypt(String toEncrypt) throws Exception { // ... code within the try block ... return encrypted; }
Additionally, in the actionPerformed method:
public void actionPerformed(ActionEvent e) { // ... code ... try { byte[] encrypted = encrypt(concatURL); // ... code ... } catch (Exception exc) { // ... handle the exception ... } }
Ensure that all checked exceptions thrown by called methods are handled or propagated by throwing them again. In this case, the encrypt method must handle or declare any exceptions it might encounter.
Additional Considerations:
- Always declare exceptions in method signatures to provide clear information about possible exceptions.
- Handle exceptions appropriately by catching them and handling them based on the specific requirements.
- Ensure that methods consistently return values or exceptions as their return types, even in case of exceptions.
The above is the detailed content of Why Does My Java Code Throw an 'Exception; must be caught or declared to be thrown' Error?. For more information, please follow other related articles on the PHP Chinese website!

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