Java: Read array of integers from file

Question

Say i have a file called "input.txt" that has a bunch of positive integers in it:

and so on....(one integer per line)

I want to read this file and make it into an array. The first integer (in this case 6) tells the number of indexes or elements in the array, so 6 spots. The other numbers fill in the array starting at 0. So at index 0, the number is 5, at index 1 the number is 6, and so on.

Can someone please show me how to read this file and make it into an array called A and return the integers in each index as n?

this is what i have so far:

import java.io.*;
public class inputFile {
    public static jobScheduleRecursive(int[] A, int i)
    {
        try
    {
        FileReader filereader = new FileReader("input.txt");
        BufferedReader bufferedreader = new BufferedReader(filereader);
        String line = bufferedreader.readLine();
        //While we have read in a valid line
        while (line != null) {
            //Try to parse integer from the String line
            try {
                System.out.println(Integer.parseInt(line));
            } catch (NumberFormatException nfe) {
                System.err.println("Failed to parse integer from line:" + line);
                System.err.println(nfe.getMessage());
                System.exit(1);
            }
            line = bufferedreader.readLine();
        }
    }
    catch(FileNotFoundException filenotfoundexception)
    {
        System.out.println("File not found.");
    }
    catch(IOException ioexception)
    {
        System.out.println("File input error occured!");
        ioexception.printStackTrace();
    }
    return A;
}

I think i'm doing something completely wrong. please help.

You don't have to put the number of entries on the first line if you use a List structure to store the numbers. You can easily convert the List to an array (which as a fixed length) once you are done reading by calling list.toArray() — Adriaan Koster, Oct 09 '11 at 20:12

score 15 · Answer 1 · answered Oct 09 '11 at 18:40

15

Using a Scanner and the Scanner.nextInt() method, you can solve this in just a few lines:

Scanner s = new Scanner(new File("input.txt"));
int[] array = new int[s.nextInt()];
for (int i = 0; i < array.length; i++)
    array[i] = s.nextInt();

answered Oct 09 '11 at 18:40

aioobe

413,195
112
811
826

There may be faster solutions. I wouldn't explore those until the application is profiled and bottlenecks are identified. – aioobe Apr 25 '13 at 16:58
Arent you moving the scanner already one line below when getting the length for the size of the array? Then when you populate the array, it will start in line two of the file – Bartzilla Mar 22 '18 at 16:52
@Bartzilla, yes. That was what OP was after, wasn't it? – aioobe Mar 22 '18 at 17:07
That is indeed true! – Bartzilla Mar 22 '18 at 20:13

score 6 · Answer 2 · answered Oct 17 '15 at 01:17

6

Java 8+

int[] ints = Files.lines(Paths.get("input.txt"))
                  .mapToInt(Integer::parseInt).toArray();

answered Oct 17 '15 at 01:17

Andrew

36,676
11
141
113

You could also do Files.lines(Paths.get("input.txt")).mapToInt(Integer::parseInt).boxed(); to get a List of non-primitives. – Daniel Gerber Aug 16 '16 at 08:17

score 5 · Answer 3 · answered Oct 09 '11 at 18:46

I think you need this for ACM-like competitions:) I use following template:

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

public class Task {

    private BufferedReader input;
    private PrintWriter output;
    private StringTokenizer stoken;

    String fin = "input";
    String fout = "output";


    private void solve() { // some solving code...
        int n = nextInt();
        int[] mas = new int[n];
        for (int i = 0; i<n; i++){
            mas[i] = nextInt();
        }
    }



    Task() throws IOException {
        input = new BufferedReader(new FileReader(fin + ".txt"));
        output = new PrintWriter(new FileWriter(fout + ".txt"));

        solve();

        input.close();
        output.flush();
        output.close();
    }


    int nextInt() {
        return Integer.parseInt(nextToken());
    }


    long nextLong() {
        return Long.parseLong(nextToken());
    }


    double nextFloat() {
        return Float.parseFloat(nextToken());
    }


    double nextDouble() {
        return Double.parseDouble(nextToken());
    }


    String nextToken() {
        while ((stoken == null) || (!stoken.hasMoreTokens())) {
            try {
                String line = input.readLine();
                stoken = new StringTokenizer(line);
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return stoken.nextToken();
    }


    public static void main(String[] args) throws IOException {
        new Task();
    }

}

In solve() method you can see how to read one number N (length of the following number sequence) and after that in loop (0..N) I read integers from input (in this case input is a file).

score 2 · Answer 4 · answered Oct 09 '11 at 19:08

import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;

public class filee{
    public static void main(String[] args) throws FileNotFoundException {
        File f = new File("l.txt");
        Scanner b = new Scanner(f);
        int[] arr = new int[b.nextInt()];
            for(int i = 0; i < arr.length; i++){
                arr[i] = b.nextInt();
            }
        for (int o : arr){
            System.out.println(o);
        }
    }
}

score 1 · Answer 5 · answered Oct 17 '15 at 05:51

If file is a classpath resource:

int[] ints = Files
            .lines(Paths.get(ClassLoader.getSystemResource("input.txt")
                    .toURI())).mapToInt(Integer::parseInt).toArray();

Printing the content from file:

 Arrays.stream(ints).forEach(System.out::println);

Java: Read array of integers from file

5 Answers5

Linked