Java for-loop is not executed

Question

The first for-loop you see does not execute and I'm not sure why. It is completely ignored, I tried it in a separate method and I tried it in the main method but something seems to be ignoring but I'm not sure how to get it to run, it simply goes to the next method run in the main method.

package math;
import java.util.Scanner;


public class mathAverageValue {

    static int numOfVals;
    static double total;
    static double average;
    static double[] arr = new double[numOfVals];
    static String boole;

    public static void input() {
        Scanner s = new Scanner(System.in);
        System.out.println("How many values will be averaged ? : ");
        numOfVals = s.nextInt();
        for(int i=0; i<arr.length; i++){
                System.out.print("Enter Element No."+(i+1)+": ");
                arr[i] = s.nextDouble();
        }
    }


    public static void process() {
        for (int i=0; i < arr.length; i++) {
            total = total + arr[i];
        }

        average = total / arr.length;   
    }

    public static void output() {

        System.out.println("Your average is : " + average);

        System.out.println("Would you like to average again? Y or N : ");
        Scanner i = new Scanner(System.in);
        boole = i.next();

        if ("Y".equals(boole)) {
            input();
            output();
        }
    }

    public static void main(String[] args) {
        input();
        output();
    }

}

Easy static int numOfVals; means it is 0; then static double[] arr = new double[numOfVals]; is static double[] arr = new double[0]; - arr.length is 0 and the loop isn't entered. — Elliott Frisch
– Elliott Frisch, Commented Mar 3, 2018 at 21:31
replace for(int i=0; i<arr.length; i++) with for(int i=0; i<numOfVals ; i++) — Cassian
– Cassian, Commented Mar 3, 2018 at 21:34

Abhijit Pritam Dutta · Accepted Answer · 2018-03-03 21:39:05Z

2

Assign some value to static int numOfVals. Java by default assign 0 to it. Hence your for loop will never run. Also modify your array declaration like below:-

 static double arr = new double[numOfVals];

edited Mar 3, 2018 at 21:39

answered Mar 3, 2018 at 21:32

Abhijit Pritam Dutta

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3 Comments

markspace Over a year ago

After assigning a value to numOfVals (the OP does this), then the array should be created arr = new double[numOfVals]; The problem is the OP does the array creation and then assigns a value to numOfVals in the wrong order.

Abhijit Pritam Dutta Over a year ago

@markspace Good Catch. Modified the answer. Thanks.

swills044 Over a year ago

Thankyou, i'm new to java so this helps !

markspace · Accepted Answer · 2018-03-03 21:36:35Z

0

The problem is that you have assigned a value to numOfVals and then created the array in the wrong order.

public static void input() {
    Scanner s = new Scanner(System.in);
    System.out.println("How many values will be averaged ? : ");
    numOfVals = s.nextInt();
    arr = new double[numOfVals];       // <-- PUT THIS HERE
    for(int i=0; i<arr.length; i++){
            System.out.print("Enter Element No."+(i+1)+": ");
            arr[i] = s.nextDouble();
    }
}

answered Mar 3, 2018 at 21:36

markspace

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Comments

John · Accepted Answer · 2018-03-03 21:34:30Z

0

It is ignored because it is a zero length array:

static int numOfVals;  // This implicitly equals 0.
static double total;
static double average;
static double[] arr = new double[numOfVals]; // so this has no elements.

hence

    for(int i=0; i<arr.length; i++){   //arr.length is 0
            System.out.print("Enter Element No."+(i+1)+": ");
            arr[i] = s.nextDouble();
    }

doesn't iterate

answered Mar 3, 2018 at 21:34

John

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Comments

User · Accepted Answer · 2018-03-03 21:35:25Z

0

According to java primitive data types initialization, all types have a default value. In your case, static int numOfVals will be assigned with 0. This is the reason why the for loop is ignored. see https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html

answered Mar 3, 2018 at 21:35

User

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