“无法解决” Java扫描仪

Question

I've recently been learning Java and decided as a little task to understand user input, I would create a times tables generator, for a user entered number. Here's the code:

import java.util.Scanner;

public class Tables {

public static void main( String[] args) {

    int IFactor, num, ans;  

    Scanner Input = new Scanner(System.in);
    try {
        System.out.println("Please enter a number to be the factor: ");
        String SFactor = Input.next();
        IFactor = Integer.parseInt(SFactor); 

        num = 1;

        while (num < 11) {   
            ans = num * IFactor;
            System.out.println(num + " * " + IFactor + " = " + ans);
            num++; 
        }

    }
    finally {
        in.close();
    }

}

}

I originally had an error when I was declaring the Scanner 'input' with Eclipse stating that there was a resource leak and that it wasn't closed. I did a bit of research and saw that inserting a try{ } and a finally { } with 'in.close();' would solve the problem. However, this wasn't the case as I now have the error: 'in cannot be resolved'.

Any help would be much appreciated! Thanks!

Answer 1

in is not assigned to anything. You would have to close your scanner called Input.

 try{
// code
}
catch(Exception ex)
{
// Exception handling
}
finally{
        if(Input!=null){ 
         Input.close();
        }
}

Answer 2

Your variable name is Input and you are trying to do in.close(). It should be:

finally {
   Input.close();
}

Answer 3

try with resources is the modern / recommended way to close AutoCloseable resources like Scanner . Eg

try (Scanner Input = new Scanner(System.in)) {
    // do stuff with Input
}

and skip the finally block. Input will be closed when at the end of the try block or earlier if an exception is thrown. And you don't have to worry about it.

Check Item 7 in Effective Java for reasons to avoid finally block

Answer 4

Try

finally {
    Input.close();
}

instead. Please note that with java, variable names begin with a lowercase letter generally (and Classes with an upperface) - so it would be better to rename that variable to `input? as well.

Answer 5

The problem is

finally {
    in.close();
}

You can try this code which uses try-with-resources. It is more java8, and you do not need the finally clause:

try(Scanner Input = new Scanner(System.in);)
    {
        System.out.println("Please enter a number to be the factor: ");
        String SFactor = Input.next();
        IFactor = Integer.parseInt(SFactor); 

        num = 1;

        while (num < 11) {   
            ans = num * IFactor;
            System.out.println(num + " * " + IFactor + " = " + ans);
            num++; 
        }

    }