unreported exception MalformedURLException when creating a URL

Question

Excuse me for my lack of Java skills, but I am normally a C kind of person.. I am beginning some Android development and I want to simply make a GET request. However, I cannot even get a simple URL type to compile correctly. I keep getting this error:

HelloWorld.java:17: error: unreported exception MalformedURLException; must be caught or declared to be thrown
 URL url = new URL("http://www.google.com/");
 ^
1 error

When running this simple code:

 import java.io.BufferedReader;
 import java.io.IOException;
 import java.io.InputStream;
 import java.io.InputStreamReader;
 import java.io.OutputStream;
 import java.io.OutputStreamWriter;
 import java.io.Reader;
 import java.io.Writer;
 import java.net.HttpURLConnection;
 import java.net.ProtocolException;
 import java.net.URL;
 import java.net.URLConnection;

public class HelloWorld{

     public static void main(String []args){
        URL url = new URL("http://www.google.com/");

        System.out.println(url.toString());
     }
}

What am I doing wrong here?

Answer 1

For all checked exception, it becomes mandatory to handle them in your code. here are 2 ways to do that.

in general, you can either pass on the exception handling to caller of the declaring method using throws clause. or you can handle them there itself using try-catch[-finally] construct.

in your case, you either need to add throws clause to main() method as

public static void main(String []args) throws MalformedURLException{

or you need to surround URL declaration with try-catch block, like here:

try{
    URL url = new URL("http://www.google.com/");
    //more code goes here
}catch(MalformedURLException ex){
//do exception handling here
}