Hello all, As i think i did what task wants but it didn't work

Question about the task

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I go trhow file by bites if it is char between 48 -57 i add it in to string and than check, if line finished by (13) i check is number even? yes add to arr. Than sort array and show it on display. All works, but task don't accepted it. Can you help please?

1. Read a file name from the console.
2. Read a set of numbers from the file.
3. Display only the even numbers, sorted in ascending order.

Example input:
5 8 -2 11 3 -5 2 10

Example output:
-2 2 8 10

Requirements:

The program must read data from the console.
The program must create a FileInputStream using the line read from the console.
The program should display data on the screen.
The program should display all of the even numbers read from the file, sorted in ascending order.
The program must close the input stream used to read the file (FileInputStream).

package com.codegym.task.task13.task1326; /* Sorting even numbers from a file */ import java.io.BufferedReader; import java.io.FileInputStream; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.Collections; public class Solution { public static void main(String[] args) throws Exception { //write your code here BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); String str = reader.readLine(); FileInputStream fileIn = new FileInputStream(str); String strSt = ""; int check = 0; int numIn = 0; int num = fileIn.read(); int county = 0; ArrayList<Integer> arr = new ArrayList<>(); while (fileIn.available() > 0) { if(county != 0) { num = fileIn.read(); }else{ county++; } if (num > 47 && num < 58 || num == 45 ) { if(num == 45){ strSt = "-"; }else { numIn = num - 48; strSt = strSt + "" + numIn; check = Integer.parseInt(strSt); } } else if (num == 13 && check%2==0) { arr.add(check); strSt = ""; check = 0; } else if(num == 10){ strSt = ""; check = 0; } } reader.close(); fileIn.close(); Collections.sort(arr); for(Integer iny : arr){ System.out.println(iny); } } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

2 December 2020, 16:01

solution

This code fails when reading the last integer from the file because it only checks each complete string when the byte '13' (new line) occurs. A file won't end with a new line. You are going to have to account for this. Some alternatives: Look up how to read all the bytes from a FileInputStream and turn them into a readable/workable String. Line 17: this takes the System.in (which is an input stream), passes that to an input stream reader, then that to a buffered reader. FileInputStream is an input stream. Maybe consider making a new buffered reader that uses the FileInputStream instead of System.in by swapping it at the correct place. You could then use just 1 line to: - read a line String from the file, parse it to an int, and add it to the list which would replace in full lines 28-51. You already know how to do this, just need to apply it.

Aleksandr

Level 22 , Moscow, Russian Federation

3 December 2020, 11:05

Hello, thank you for your comments, it should help, but i dont' understand somethings.

BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
FileInputStream file = new FileInputStream(reader.readLine());
ArrayList<Integer> arr = new ArrayList<>();
byte num = 0;
String str = "";

while(file.available()> 0) {
    num = (byte) file.read();
    if(((char)num + "").matches("[0-9-]")){
            str = str + "" + (char) num;
        }
    if (num == 13 || file.available() == 0) {
        if (Integer.parseInt(str) % 2 == 0) {
            arr.add(Integer.parseInt(str));
            str = "";
        }else {
            str = "";
        }
    }
}

reader.close();
file.close();

Collections.sort(arr);

for(Integer tret : arr){
    System.out.println(tret);
}

here all works too and i check last number without \n(enter) and with it, all works and sort works well, but task not accepted it.

Guadalupe Gagnon

Level 37 , Tampa, United States

3 December 2020, 14:09

useful

That is because you are not accounting for new line bytes, which would be byte # '10'. If the file does not have any byte 13 but only has byte 10, this code would fail.

Aleksandr

Level 22 , Moscow, Russian Federation

3 December 2020, 14:21

Now i got it, thank you!