IO problems...

[MysticVegeta]

I am using Ready and the Program is not writing to the file. It read integers 1, 2, 3 from a text file. Also, Why cant i read them as ints?

code:

import java.io.*; //import java.lang.String; public class ioExceptions {     public static void main (String[] Args)     {         String nums;         try         {             BufferedReader inputStream = new BufferedReader (new FileReader ("Database.txt"));             BufferedWriter out = new BufferedWriter (new FileWriter ("Out.txt"));             for (int x = 1 ; x < 4 ; x++)             {                 nums = inputStream.readLine ();                 System.out.println (nums);                 out.write (nums + "2");             }         }         catch (IOException asdf)         {             System.out.println ("FILE NOT FOUND");         }     } }

Thanks a lot.

[Hikaru79]

The thing with BufferedWriters is that if you don't fill the buffer, it doesn't get written to the file and just stays on the stream.

Basically, at the end of the program, write these two lines:

Java:

out.flush(); out.close();

And it will work just fine

NOTE: Be sure you don't put those lines in the loop, or else it will close the stream before you're done writing.

[JackTruong]

Java:

int num = Integer.parseInt(inputStream.readLine());

[wtd]

Or if you're using Java 1.5/5.0/numberOfTheWeek:

Java:

import java.util.Scanner; ... Scanner inputStreamScanner = new Scanner(inputStream); ... int num = inputStreamScanner.nextInt();

[1of42]

Hikaru79 wrote:

The thing with BufferedWriters is that if you don't fill the buffer, it doesn't get written to the file and just stays on the stream.

Basically, at the end of the program, write these two lines:

Java:

out.flush(); out.close();

And it will work just fine

NOTE: Be sure you don't put those lines in the loop, or else it will close the stream before you're done writing.

The flush is unnecessary, close() flushes the stream.

[MysticVegeta]

JackTruong wrote:

Java:

int num = Integer.parseInt(inputStream.readLine());

Why need to parse when it already exists in Integer format in the text file :S

[wtd]

MysticVegeta wrote:

JackTruong wrote:

Java:

int num = Integer.parseInt(inputStream.readLine());

Why need to parse when it already exists in Integer format in the text file :S

Does the text file know that it's an int?

[MysticVegeta]

Apparently turing does.

[wtd]

No, it doesn't. It assumes that if you ask it to read an integer, the next thing will be an integer.

What if it's not?

Well, then you've read a bunch of data from that stream for nothing. Can you get that data back and treat it differently? Yes, but not easily.

If I read in a string, then parse it, if there's an error, I still have the string, and I can still do something else with it.

[MysticVegeta]

I guess you are right. ok after reading in the string "1 2 3" I will use the stringsplit to store it into an array of ints I dont suppose there is a shorter way to do that. Do you think?

[wtd]

Again, if we're talking about Java 1.5.0...

Java:

String line = inputStream.readLine(); Scanner lineScanner = new Scanner(line); ArrayList<Integer> numbers = new ArrayList<Integer>(); for (int i = 0; i < 3; ++i) {    numbers.add(lineScanner.nextInt()); }

[wtd]

Or, if you're not sure how many numbers there are in your line:

Java:

String line = inputStream.readLine(); Scanner lineScanner = new Scanner(line); ArrayList<Integer> numbers = new ArrayList<Integer>(); while (lineScanner.hasNextInt()) {    numbers.add(lineScanner.nextInt()); }