FileNotFoundException with FileInputStream even though .getResource(filename) finds it

Question

This is the first question I ask on stackoverflow, so if you notice something unpleasant/bad/inappropriate in my question, please be kind and point it out to me.

I have tried doing my school homework in Java, because I was tired of C++ and I had already done something in Python. However, I've had problems with reading from a binary file (which should contain one double and two float numbers, in order).

Specifically: .getResource(filename) finds the file, but when I open a FileInputStream(path) (inside public static Integer Leggere(Dati [] dato, String nomefile)), it throws FileNotFoundException.

This is my code:

import java.util.*;
import java.io.*;

public class Main{
    public static void main(String[] args) {
        Dati [] data  = new Dati[23500];
        int contatore = 0;
        for(; contatore < 23500; contatore++){
            data[contatore] = new Dati(0, 0, 0);
        }
        contatore = 0;

        String path = Dati.class.getClassLoader().getResource("valori.bin").getPath().toString();
        contatore = Leggere(data, path);
        Acquisire(data, contatore);
        Scrivere(data, "risultati.txt", contatore);
    }

    public static Integer Leggere(Dati [] dato, String nomefile){
        int j = 0;
        try{
            DataInputStream in = new DataInputStream(new FileInputStream(nomefile));
            while(in.available() > 0){
                dato[j].dato1 = in.readDouble();
                dato[j].dato2 = in.readFloat();
                dato[j].dato3 = in.readFloat();
                j++;
            }
            in.close();
        }
        catch(IOException e){
            System.out.println("Problemi nell'apertura del file");
            System.out.println(nomefile);
            System.exit(0);
        }

        return j;
    }

    public static void Scrivere(Dati [] dato, String nomefile, int count){
        PrintWriter output;
        try {
            output = new PrintWriter(nomefile);

            Integer j = 0;
            while(j < count){
                output.println(dato[j]);
                j++;
            }

            output.close();
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }


    }

    public static void Acquisire(Dati [] dato, int count){
        Scanner cinput = new Scanner(System.in);
        double c = 0.0;
        int j = 0;

        System.out.println("Inserisci un fattore di conversione: ");
        while(c == 0.0){
            c = cinput.nextDouble();
        }

        while(j < count){
            dato[j].dato1 *= c;
            dato[j].dato2 *= c;
            dato[j].dato3 *= c;
        }

        cinput.close();
    }
}

The program ends with two messages, which handle the exception. The second one is the path of the file as found by the getResource() method:

Problemi nell'apertura del file
/C:/Users/Sebastian/Desktop/Archivio/Scuola/5C%20a.s.%202016-2017/Compiti%20Estate%202016/Informatica/03%20-%20Conversione/bin/valori.bin

I know that it's a FileNotFoundException because it says so in debug mode.

How would you have done this piece of code? Do you know what the problem is? Could you post an example to fix the problem or an example of an alternative method to cope with it?

SOLUTION TO THE PROBLEM

I'm writing this so that people with a similar problem may find their solution.

"Malfunctioning" code

String path = Dati.class.getClassLoader().getResource("valori.bin").getPath().toString();
DataInputStream in = new DataInputStream(new FileInputStream(path));

What happened was that .getResource("valori.bin") managed to find the file (and it's path with .getPath()), but when I tried to open a FileInputStream(path), I got a FileNotFoundException.

Working code

InputStream stream = Dati.class.getClassLoader().getResourceAsStream("valori.bin");
DataInputStream in = new DataInputStream(stream);

This way does it. By doing so, I don't need to worry about path name because .getResourceAsStream() returns the usable Stream as is. I don't know exactly why the former code didn't work, but heuristics tells me to not worry much, and so be it!

Answer 1

You might want to take a look at the getResourceAsStream method in the class loader. It gives you an input stream that you can plug directly and not have to deal with full paths and those kinds of problems.