5 pts.
 Java quick sort, reading from a user input file into any array (to be sorted)
Hey, What's up y'all,

I am trying to write some code in Java that will read in the numbers from a file (one # on each line of .txt file) put them into an array, and then run quick sort on the array. Eclipse is showing some red that I am having trouble with. My errors are marked with comments, and what the error is, if anyone can help me get this to run, thanks everyone!

-Kyle

[code]import java.io.*;
import java.util.Scanner;
import java.io.BufferedReader;
import java.io.File;


public class Lab3 {

public static void main(String[] args) throws IOException{


System.out.print("Name of file with array: ");
Scanner readIn = new Scanner(System.in);
String input=readIn.nextLine();}
**testScan1(input);**  /*Eclipse says "return type for method is missing" trying to call method testScan below*/


**public testScan1(String filename)**  //"Return type is missing for method"

{
File file = new File(filename);
Scanner scan;
try{
 int [] array = new int[5];

scan = new Scanner( file );
}
catch ( java.io.FileNotFoundException e )
{
System.out.println( "couldn't open. file not found "  );
return;
}
while(file.hasNext())      //"underlies "hasNext" "Method hasNext() is undefined"
{
for( int i = 0; i <= file.length(); ++i)
{

**array**[i]=scan.next();    //"Array cannot be Resolved"


}
}

int partition(int arr[], int left, int right)
{
 int i=left; int j = right;
 int tmp;
 int pivot = arr[(left+right)/2];
 while (i<=j){
  while(arr[i]<pivot)
   i++;
  while (arr[j]>pivot)
   j--;
  if (i<=j){
   tmp=arr[i];
   arr[i]=arr[j];
   arr[j]=tmp;
   i++; j--;
  }
 }
 return i;
    }
    void quickSort(int arr[], int left, int right){
 int index = partition(arr, left, right);
 if (left<index-1);
 quickSort(arr, left, index-1);
 if (index<right)
  quickSort(arr, index, right);
    }
    }
[/code][/pre]						

Software/Hardware used:
Eclipse, java
ASKED: November 5, 2009  11:00 PM
UPDATED: November 10, 2009  11:18 PM

Answer Wiki:
Instead of using .hasNext, using some sort of delimiter is what I'd suggest. I'd also check into any .next methods, which are less prone to erroring out. Hope this helps! -Schmidtw
Last Wiki Answer Submitted:  November 10, 2009  11:18 pm  by  Schmidtw   11,205 pts.
All Answer Wiki Contributors:  Schmidtw   11,205 pts.
To see all answers submitted to the Answer Wiki: View Answer History.


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The error messages are pretty explanatory, and with some very basic knowledge about the language you would know what it is referring to.

My recommendation:

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