How to define an explicit constructor for an url object in Java

Question

I am defining a url in processing (or java) with the line:

URL base = new URL("http://www.google.com/");

So I get the following error: Default constructor cannot handle exception type MalformedURLException thrown by implicit super constructor. Must define an explicit constructor. Which I assume is because if my url isn't valid, there's no corresponding catch to catch the error (because I'm declaring my url outside of try).

But, i want to define this url outside of try, because my main try block is in a loop and the url is static. So there's no need to do the definition of the url more than once (and because it's static i'm not afraid of any MalformedURLExceptions).

How do I do that? Is there really no other way than to define the url within a separate try block? (because that seems a bit too much for just a simple static url)

// import libraries
import java.net.*;
import java.io.*;

// url
URL base = new URL("http://www.google.com");

void setup() {
}

void draw() {
  try {
    // retrieve url
    String[] results = loadStrings(base);

    // print results
    println(results[0]);
  }  
  catch (MalformedURLException e) {
    e.printStackTrace();
  }
  catch (ConnectException e) {
    e.printStackTrace();
  }
  catch (IOException e) {
    e.printStackTrace();
  }

  // stop looping
  noLoop();

}

Collin · Accepted Answer · 2012-03-24 15:16:35Z

13

You just need to define a default constructor for your class which throws MalformedURLException:

class MyClass {
    private URL url = new URL("http://www.google.com");

    public MyClass() throws MalformedURLException {}
}

The error is happening because the constructor Java is generating for your class isn't able to throw anything, you must specify that it might.

Edit

Seeing your code, your other option is this (I assume setup() gets called before you use the URL)

URL url;

void setup() {
    try {
        url = new URL("http://www.google.com");
    } catch (MalformedURLException ex) {
        throw new RuntimeException(ex);
    }
}

edited Mar 24, 2012 at 15:16

answered Mar 24, 2012 at 14:19

Collin

12.3k3 gold badges48 silver badges62 bronze badges

Sign up to request clarification or add additional context in comments.

9 Comments

user1218247 Over a year ago

But this isn't a class that i've written myself, just a standard java class. Should i still define the default constructor like you suggested in that case? Or am i misunderstanding?

Collin Over a year ago

Then we need to see more code for where this error is happening. Can you post the code around where you're using this url?

user1218247 Over a year ago

Of course, added my code at the bottom of my question (this is processing code btw., might be a little different from standard java).

Collin Over a year ago

Whatever class your code that you posted is in needs a default constructor defined to throw MalformedURLException. I don't know processing that well, but I assume it's creating some type of implicit class for your code. This code needs some sort of constructor that can throw the error for the implicitly created URL.

user1218247 Over a year ago

So your class would do that right? When i use it i get the error "The field panel.url is not visible", and changing the url from private to public doesn't change the error.

|

Jordão · Accepted Answer · 2012-03-24 14:59:57Z

1

Since the URL constructor throws a checked exception you have to either catch it or rethrow it. You can create your own factory method that just wraps it in a RuntimeException:

public class URLFactory {
  public static URL create(String urlString) {
    try {
      return new URL(urlString);
    } catch (MalformedURLException e) {
      throw new RuntimeException(e);
    }
  }
}

Or maybe just this in your environment?

URL createURL(String urlString) {
  try {
    return new URL(urlString);
  } catch (MalformedURLException e) {
    throw new RuntimeException(e);
  }
}

And use it instead:

URL base = URLFactory.create("http://www.google.com");

edited Mar 24, 2012 at 14:59

answered Mar 24, 2012 at 14:18

Jordão

56.8k13 gold badges118 silver badges147 bronze badges

2 Comments

user1218247 Over a year ago

Ok. Could you elaborate a little? How do i do that?

user1218247 Over a year ago

Thanks for the suggestions, i added your runtimeexception to the code.

Collectives™ on Stack Overflow

How to define an explicit constructor for an url object in Java

2 Answers 2

9 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

9 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related