CodeGym/Help with Java Tasks/What the validator does not like here?

Level 22

Seversk

20.01.2021
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18comments

What the validator does not like here?

Question about the task

Building a file

Java Core, Level 8, Lesson 11

Resolved

Actually, code is working pretty fine

Let's build a file from various pieces.
Read file names from the console.
Each file has a name: <someName>.partN.

For example, Lion.avi.part1, Lion.avi.part2, ..., Lion.avi.part37.

The file names are supplied in random order. The word "end" is used to stop reading in file names.
In the folder where all the files are located, create a file without the "part" suffix, i.e. without ".<partN>".

For example, Lion.avi.

Use a buffer to copy all the bytes from the partial files to the created file.
Copy the first in the proper order, first the first part, then the second, ..., finally - the last part.
Close the streams.

Requirements:

The program must read file names from the console until the word "end" is entered.
Create a stream to write to the file without the "part" suffix (".<partN>") in the folder with all the "part" files.
Copy all the bytes from the *.partN files to the new file.
You should use a buffer for the reading and writing.
The file streams must be closed.
Don't use static variables.

package com.codegym.task.task18.task1825; import java.io.*; import java.util.*; /* Building a file */ public class Solution { public static void main(String[] args) throws IOException { ArrayList<InputStream> streamList = new ArrayList<>(); ArrayList<Integer> buffer = new ArrayList<>(); Scanner sc = new Scanner(System.in); String fileName = null; int partIndex = 0; String newFileName = null; while (true) { try { fileName = sc.nextLine(); if (fileName.equals("end")) break; partIndex = fileName.indexOf("part"); String getNumber = fileName.substring(partIndex+4,fileName.length()); // cut off number from fileName int number = Integer.parseInt(getNumber); newFileName = fileName.substring(0, partIndex-1); // cut out new file name //System.out.println(number); streamList.add(number-1, new FileInputStream(fileName)); //part1 will go to 0 index } catch (FileNotFoundException e) { e.printStackTrace(); } } for (InputStream is : streamList) { while (is.available() > 0) { buffer.add(is.read()); } } //System.out.println(newFileName); OutputStream os = new FileOutputStream(newFileName); for (int b : buffer) { os.write(b); } for (InputStream is : streamList) { is.close(); } os.close(); } }

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Nouser

Level 36 , Germany

20 January 2021, 08:30

The file names are supplied in random order. And you do not sort the input data (take care, that's still a mean trap). CG probably doesn't accept a list as buffer. And you shouldn't do that. At first you save bytes into the buffer. The list uses the Wrapper class Integer for each byte. Here you lose performance and memory (even if the frist 127 values are cached). And then you have the list. It has an overhead, too. Plus it dynamically is getting increased in size here while reading data.

Dmitri

Level 22 , Seversk, Russia

20 January 2021, 10:55

1. I do care about the order of files by adding them by index - streamList.add(number-1, new FileInputStream(fileName)) - so .part1 will go to index 0, .part3 to index 2 and so on... 2. What is buffer then? Actually I can just read -> write directly from/to files thank you

Nouser

Level 36 , Germany

20 January 2021, 11:16

1. I see that this may work but please test this with parts from 0 - 12 (and output the sorted list) 2. BufferedOutputStream or a byte array

Dmitri

Level 22 , Seversk, Russia

20 January 2021, 11:21

1. .part0 is not intended since has no meaning 2. BufferedOutputStream is buffer here)))?

Nouser

Level 36 , Germany

20 January 2021, 11:28

useful

have you tried to add filenames in mixed order.... as far as I remember that shouldn't work

ArrayList<String> test = new ArrayList<>();
test.add(15, "Hello");

System.out.println(test);

BufferedOutputStream uses an internal buffer ;) or, as said you could use a byte array (like CG has shown you in one of the previous lessons).

Dmitri

Level 22 , Seversk, Russia

20 January 2021, 11:37

if just outputting with BufferedOutputStream ok for CG I shall try it since it is simple.

Nouser

Level 36 , Germany

20 January 2021, 11:47

I had a look at my old solution and it verified with both, bos and byte array.

Dmitri

Level 22 , Seversk, Russia

20 January 2021, 12:58

solution

I reworked the code and replaced ArrayList with TreeSet and removed List as buffer. The part with TreeSet worked fine, but BufferedOutputStream did not work, I mean did not pass, it worked very well.

public class Solution {
    public static void main(String[] args) throws IOException {
        TreeSet<String> streamSet = new TreeSet<>();
        Scanner sc = new Scanner(System.in);
        String fileName = null;
        int partIndex = 0;
        String newFileName = null;
        while (true) {
            fileName = sc.nextLine();
            if (fileName.equals("end")) break;
            partIndex = fileName.indexOf("part");
            newFileName = fileName.substring(0, partIndex-1); // cut out new file name
            streamSet.add(fileName); //TreeSet sorts file names
        }
        BufferedOutputStream os = new BufferedOutputStream(new FileOutputStream(newFileName));
        for (String fn : streamSet) { // read from input streams to the output stream
           BufferedInputStream is = new BufferedInputStream(new FileInputStream(fn));
            while (is.available() > 0) {
                os.write(is.read());
            }
            is.close();
        }
        os.close();
    }
}

Dmitri

Level 22 , Seversk, Russia

20 January 2021, 13:02

How do you make such pretty decoration of code?

Nouser

Level 36 , Germany

20 January 2021, 13:15

solution

what pretty decoration do you mean? Using the [code] tags? If so, just click [code] in the above menu and add your code between the open and close tags. And i have the code to write the files pretty similar here. I just used a BufferedInputStream, too.

Dmitri

Level 22 , Seversk, Russia

20 January 2021, 13:20

Oh, yes thank you!!

Guadalupe Gagnon

Level 37 , Tampa, United States

20 January 2021, 16:49

#1 using indexof("part") would not work. If the file were any of these the code would fail: party.part# partner.part# apart.part# You are going to want to rethink this. #2 While TreeSets do sort, when they are Sets of Strings they do so in alphabetical order and not numerical order. Take for example if you had some files such as: a.part1 a.part2 a.part10 a.part11 The TreeSet would sort them like this: a.part1 a.part10 a.part11 a.part2 which is in alphabetical order and not the order that the task needs.

Nouser

Level 36 , Germany

20 January 2021, 17:21

It seems the validator isn't as strict anymore as it was. When I used a treeset it didn't validate at first. I had to write a comparator for the actual use case. However that's a little bit to much for that level. So another collection would be better suited.

Guadalupe Gagnon

Level 37 , Tampa, United States

20 January 2021, 17:46

Right. TreeSets work fine as long as you change the default sorting algorithm so it is not sorting alphabetically.

Dmitri

Level 22 , Seversk, Russia

21 January 2021, 03:43

Thank you I see your points. These are really important arguments. Just that you know the code passed the validator.

Dmitri

Level 22 , Seversk, Russia

21 January 2021, 03:46

It would be probably safer to use :

partIndex = fileName.indexOf(".part")

Dmitri

Level 22 , Seversk, Russia

21 January 2021, 04:19

I reworked it a little bit to change for ".part" and use TreeMap instead of TreeSet:

public class Solution {
    public static void main(String[] args) throws IOException {
        SortedMap<Integer, String> streamMap = new TreeMap<>();
        Scanner sc = new Scanner(System.in);
        String fileName = null;
        int partIndex = 0;
        String newFileName = null;
        int index = 0;
        while (true) {
            fileName = sc.nextLine();
            if (fileName.equals("end")) break;
            partIndex = fileName.indexOf(".part");
            newFileName = fileName.substring(0, partIndex); // cut out new file name
            index = Integer.parseInt(fileName.substring(partIndex+5,fileName.length()));
            streamMap.put(index, fileName); //TreeMap sorts file names
        }
        BufferedOutputStream os = new BufferedOutputStream(new FileOutputStream(newFileName)); // try this as a buffer
        for (Integer key : streamMap.keySet()) { // read from input streams to buffer
            BufferedInputStream is = new BufferedInputStream(new FileInputStream(streamMap.get(key)));
            while (is.available() > 0) {
                os.write(is.read());
            }
            is.close();
        }
        os.close();
    }
}

Guadalupe Gagnon

Level 37 , Tampa, United States

21 January 2021, 14:10

=D That is funny that it passed, they must have really lightened up the validator checks. Anyways, that new code looks a lot better in that it handles more cases properly. Looks good.