CodeGym/Help with Java Tasks/Any help on this one would be welcome, doesnt Validate so...

Level 22

Bangkok

07.05.2019
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Any help on this one would be welcome, doesnt Validate somehow.

Question about the task

Resolved

Checked with multiple files, seems to work on all occasions. I decided not to use the while (available >0) loop as it is not needed when writing everything in the array at once. The requirement failures seem to indicate the validation process is expecting something. Would be good to know if my code is bad or the validation is too strict. Thank you!

Read 3 file names from the console: file1, file2, file3.
Split file1 as follows:
Write the first half of its bytes to file2,and write the second half of its bytes to file3.
If the number of bytes in file1 is odd, then file2 should contain the larger half.
Close the streams.

Requirements:

The program should read a file name three times from the console.
Use FileInputStream to read from a file, and use FileOutputStream to write to files.
The first half of the bytes in the first file must be written to the second file.
The second half of the bytes in the first file must be written to the third file.
The FileInputStream and FileOutputStream must be closed.

package com.codegym.task.task18.task1808; /* Splitting a file */ import java.io.*; public class Solution { public static void main(String[] args) throws IOException { //Requirements 1,2 --> start initialization using File streams BufferedReader r = new BufferedReader(new InputStreamReader(System.in)); String file1 = r.readLine(); String file2 = r.readLine(); String file3 = r.readLine(); FileInputStream in = new FileInputStream(file1); FileOutputStream out1 = new FileOutputStream(file2); FileOutputStream out2 = new FileOutputStream(file3); int half; byte[] buffer1; byte[] buffer2; // end init // Preparing buffers for input if (in.available() % 2 != 0) { half = (in.available() / 2) + 1; buffer1 = new byte[half]; buffer2 = new byte[half - 1]; } else { half = in.available() / 2; buffer1 = new byte[half]; buffer2 = new byte[half]; } // 3,4 reading and writing halves of bytes boolean moveOn = false; while (!moveOn) { in.read(buffer1); if (buffer1[half - 1] != 0) ; moveOn = true; } in.read(buffer2); out1.write(buffer1); out2.write(buffer2); // end 3,4 //5 close streams in.close(); out1.close(); out2.close(); } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

7 May 2019, 15:43

Just a neat tip: You know that when you divide ints in Java that the remainder gets lopped off, effectively rounding down every time. You can use this to split the byte arrays accoringly without having to determine if the available bytes are odd/even. Take a look:

byte[] buffer1 = new byte[in.available()/2], buffer2;
in.read(buffer1);
in.read((buffer2 = new byte[in.available()])); //in.available returns the unread portion,
// which is exactly half and +1 if the original order was odd

This effectively can replace lines 21-44 Now the problem with your code is that you coded for the first half to be the larger of the 2 when the conditions say that the second half should be the larger of the 2. The easiest fix is to write the first half to output2 and the second half to output1, this would effectively correct the issue. **Edit** The tasks does not say that file 2 must be larger as i said. It specifically says that "If the number of bytes in file1 is odd, then file2 should contain the larger half. So you (and I) both coded that one file should be larger if the total bytes is odd. It should first determine if the total bytes is odd or not, then if that is true figure out if the first half bytes is odd or not, if that is true then the second half should be larger by 1, if that is false then the first half should be larger by 1, so this should be the code:

half = (in.available() / 2);

if (in.available() % 2 != 0)
   if(half % 2 == 0) half++;

in.read((buffer1 = new byte[half]);
in.read((buffer2 = new byte[in.available()]));

Roy

Level 22 , Bangkok, Thailand

7 May 2019, 16:32

Hey Guadalupe, first of all, thanks for the tip. Thats indeed correct and makes the code more compact. i dont need to initialize buffer2 upfront. However, when initializing buffer 1 you need to know if its odd or not and add 1, as this should be the larger of the two. The solution you suggested, however, is not working. **If the number of bytes in file1 is odd, then file2 should contain the larger half** There are 3 files and file2 is the first one of the written files, so that should be the larger one, which is also validating correct as you can see. Strangely enough it doesnt validate on the in/outputstreams. I have no idea what causes it, perhaps i might try checking it again tomorrow after a good night sleep, but would be great if one of the admins could check out the reason behind it.

Guadalupe Gagnon

Level 37 , Tampa, United States

7 May 2019, 16:32

check my edit, I read through too fast and 'assumed' i knew what was going on. It happens to us all.

Roy

Level 22 , Bangkok, Thailand

7 May 2019, 16:42

indeed, it happens to me all the time with these tasks. It should first determine if the total bytes is odd or not, then if that is true By this you mean if in.available is odd (true) figure out if the first half bytes is odd or not, When in.available is even, you know that both halves are always even too, so no need to check after. When in.available is odd, both halves with always be odd as well and only in that case you need to make the first written file(file2) larger by 1. if that is true then the second half should be larger by 1, if that is false then the first half should be larger by 1, so this should be the code: Can you send me your solution through pm please, so i will compare it tomorrow with mine

Guadalupe Gagnon

Level 37 , Tampa, United States

7 May 2019, 16:45

ok ... edit edit edit edit. I just read your comment to my first comment, I made the edit first and was all discombobulated. So, now that you explained if file1, the input file, is odd then file 2, the first output file, should contain the larger half. So this is my final solution:

half = (in.available() / 2);  //gets half

// adds one if the input is odd
if (in.available() % 2 != 0)
   half++;

// writes data
in.read((buffer1 = new byte[half]);
in.read((buffer2 = new byte[in.available()]));

sorry for all the edits. I don't have the ability to test verification, and I don't have my code that I used to pass this originally. ill send the total code that I came up with.

Roy

Level 22 , Bangkok, Thailand

7 May 2019, 16:53

Hmm, what you just wrote is just an improved version of my code. The output will be still the same after. My updated version looks like this now:

public class Solution {
    public static void main(String[] args) throws IOException {
        //Requirements 1,2  --> start initialization using File streams
        BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
        String file1 = r.readLine();
        String file2 = r.readLine();
        String file3 = r.readLine();

        FileInputStream in = new FileInputStream(file1);
        FileOutputStream out1 = new FileOutputStream(file2);
        FileOutputStream out2 = new FileOutputStream(file3);
        int half = (in.available() / 2);  //gets half
        byte[] buffer1;
        byte[] buffer2;
        // end init

        // Preparing buffers for input
        if (in.available() % 2 != 0)
            half++;
        buffer1 = new byte[half];

        // 3,4 reading and writing halves of bytes
        boolean moveOn = false;
        while (!moveOn) {
            in.read(buffer1);
            if (buffer1[half - 1] != 0) ;
            moveOn = true;
        }
        in.read((buffer2 = new byte[in.available()]));
        out1.write(buffer1);
        out2.write(buffer2);
        // end 3,4

        //5 close streams
        in.close();
        out1.close();
        out2.close();
    }
}

Guadalupe Gagnon

Level 37 , Tampa, United States

7 May 2019, 17:01

solution

Try this:

//boolean moveOn = false;
//while (!moveOn) {
in.read(buffer1);
//if (buffer1[half - 1] != 0) ;
//moveOn = true;
//}

Roy

Level 22 , Bangkok, Thailand

7 May 2019, 17:06

Yay! now i seriously ask myself why i used this loop in the first place....i think at first i thought it was written byte by byte every loop run, but later figured this wasnt the case and just never changed it anymore. Anyway, this solved the issue. Thanks a lot Guadalupe!

Guadalupe Gagnon

Level 37 , Tampa, United States

7 May 2019, 17:11

I kind of wish I had just started with that suggestion instead of interpreting the task requirements in all sorts of wrong ways.

Roy

Level 22 , Bangkok, Thailand

7 May 2019, 17:14

haha well it should hv validated just fine even with this unnecessary code, so its not something you look at right away.