[SOLVED] Why the existing file is not overwritten?

[Akirien]

Hello.. I have this code. I'm wondering why the content of existing file is not overwritten.

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

public class Sorting{
	
	public static void main(String[]args) throws Exception{
		
		File file = new File("Sorting.txt");
		
		if (file.exists()){
			
			System.out.println("File already exists!");
			System.exit(0);
			
		}
		
		PrintWriter output = new PrintWriter(file);
		
		for (int i = 0; i < 100; i++){
			
			int x = (int) (Math.random() * 100);
			output.print(x);
			output.print(" ");
		}
		output.close();
		
		Scanner readFile = null;
		ArrayList<Integer> list = new ArrayList<Integer>();
		
		try {
			readFile = new Scanner(new File("Sorting.txt"));
		} catch (FileNotFoundException e) {
			e.printStackTrace();
		}
		
		while(readFile.hasNext()){
			if (readFile.hasNextInt()) list.add(readFile.nextInt());
			else readFile.next();
		}
		
		Collections.sort(list);
		
		for (Integer y: list) System.out.println(y);
		
		
	}
}

[Fubarable]

Your program exits if the file already exists.

[Akirien]

Your program exits if the file already exists.

Thank you for replying... How can I overwrite the content of the existing file?

[Akirien]

Got it already.. I removed the System.exit(0);

Thank you :)

[Fubarable]

Thank you for replying... How can I overwrite the content of the existing file?

By deleting the block of code that tests if the file exists and exits if so...

Did you write this code yourself??

[Akirien]

By deleting the block of code that tests if the file exists and exits if so...

Did you write this code yourself??

I referred from the book. And the sorting part, I modified from the web. Can I ask something. The System.exit(Integer) is for what? I know the program will exit. I mean the parameter.

[Fubarable]

I referred from the book. And the sorting part, I modified from the web. Can I ask something. The System.exit(Integer) is for what? I know the program will exit. I mean the parameter.

Good question, one that is well answered in the System entry in the API:

The argument serves as a status code; by convention, a nonzero status code indicates abnormal termination.

So if the program is exiting normally -- not due to some error or exception, return (0) in the parameter. Otherwise return a non-0 number, one that will make sense to you when you review your code 6 months from now.

[Akirien]

Good question, one that is well answered in the System entry in the API:


So if the program is exiting normally -- not due to some error or exception, return (0) in the parameter. Otherwise return a non-0 number, one that will make sense to you when you review your code 6 months from now.

Thank you for the explanation. A little bit confuse but trying to understand. Haha..