Specifying throws IOException

Question

Can anyone help me solve my problem. I am a beginner in Java programming. Previously when I did not declare throws IOException it gave me an error:

Exception in thread "main" java.lang.RuntimeException: Uncompilable source code - unreported exception java.io.IOException; must be caught or declared to be thrown

The program is shown below:

import java.io.*;

public class addition {
    public static void main(String array[])throws IOException
    {
        InputStreamReader i = new InputStreamReader(System.in);
        BufferedReader b = new BufferedReader(i);
        System.out.println("Enter first number : ");
        int a1 = Integer.parseInt(b.readLine());
          System.out.println("Enter second number : ");
        int a2 = Integer.parseInt(b.readLine());
        int sum = a1 + a2 ;
        System.out.println("addition"+sum);
    }

}

Well, that exception message kind of summarizes all you need to know `must be caught or declared to be thrownand` — Rodrigo López, May 14 '15 at 18:10
You need to declare all exceptions that your code does not handle in Java. So in order for your code to compile you either need to handle any IOException in your own try-catch or you have to specify that the function can throw an IOException. — shuttle87, May 14 '15 at 18:10
What's your question? It looks like you solved your problem? — Vivin Paliath, May 14 '15 at 18:12
Also check this: https://docs.oracle.com/javase/specs/jls/se7/html/jls-11.html#jls-11.2.3 — shuttle87, May 14 '15 at 18:17

score 0 · Accepted Answer · answered May 14 '15 at 18:19

The BufferedReader function readLine() throws an IOException if there is an I/O failure while attempting to read from the input stream. In Java, you must use try catch statements to handle exceptions should they arise:

import java.io.*;

public class addition {
public static void main(String array[])throws IOException
{
    InputStreamReader i = new InputStreamReader(System.in);
    BufferedReader b = new BufferedReader(i);
    System.out.println("Enter first number : ");

    // Attempt to read in user input.
    try {
        int a1 = Integer.parseInt(b.readLine());
        System.out.println("Enter second number : ");
        int a2 = Integer.parseInt(b.readLine());
        int sum = a1 + a2 ;
        System.out.println("addition"+sum);
    }

    // Should there be some problem reading in input, we handle it gracefully.
    catch (IOException e) {
        System.out.println("Error reading input from user. Exiting now...");
        System.exit(0);
    }
}
}

thank you very much to all to clear my concept for IOException. — , May 14 '15 at 18:24
this should be modified to no longer have `throws IOException` since it is being handled via the catch. — Russell Uhl, May 14 '15 at 19:14

Specifying throws IOException

1 Answers1