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Thread: Arrays in java:

  1. #1
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    Default Arrays in java:

    A program that analyzes a set of numbers can be very useful. Create an Analysis application that prompts the user for numbers in the range 1 through 50, terminated by a sentinel, and then performs the following analysis on the numbers
    Determine the average number
    Determine the maximum number
    Determine the range (maximum - minimum)
    Displays a histogram that shows the numbers in each five-unit range.

    --- Update ---

    how do i go about solving this problem?
    this is what i have so far:
    import java.util.Scanner;

    public class Arrays {
    public static void main (String [] args){

    Scanner scan = new Scanner (System.in);
    int array [] = new int[0];
    int num;
    System.out.println("Enter a number between 1 - 50 inclusive OR -1 if u wish to end");
    array = num.nextInt();

    //loop with sentinel value
    while( num != -1)
    array[0]++;

    // maximum value
    int maximum;
    for( num=0; num<=50; num++){
    if (maximum< array[num]){
    maximum= array[num];
    }
    }

    // minimum value
    int minimum;
    for (num=0; num >=50; num++){
    if(minimum>array[num]){
    minimum=array [num];
    }
    }

    // average
    int average;
    int sum;
    for(num=0; num<=50;num++){
    sum=sum+array[num];
    }
    average = sum/50;

    int range;
    range = maximum-minimum;

    }

    }


  2. #2
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    Default Re: Arrays in java:

    I am not sure, but I think your array holds no values in it. And you try to compare the number with the values within the array, which do not exist.
    And besides that, array = num.nextInt(); gives error.

    To get input from user, you should do this (using your variable names);

    num = scan.nextInt();

    Once you define an array (int array [] = new int[0]; ) you can't change its length. It seems you are trying to do that as well.

  3. The Following User Says Thank You to beer-in-box For This Useful Post:

    puny prince (February 14th, 2013)

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