Space between inputs doesn't give exception on array index? - java

I'm not sure what to title this question(if anyone has input on what to name the question, please let me know). My program asks the user for 5 int and 5 doubles. Then those numbers are put in an array and passes it to a method to get the average. My question is if I separate the user input by spaces and press enter(like so, 5 space 6...enter; it allows me to enter more than what is allowed in the array index. Why doesn't it give you a error? and how do you prevent this? Also any advice on how I write code would be helpful too!
Here is the code.
import java.util.*;
public class Lab7A_BRC{
public static void main(String[] args) {
System.out.println("\t\t Average arrays ");
Scanner input = new Scanner(System.in);
//array count varriable
int n = 5;
//array declaration
int [] list1 = new int[n];
double [] list2 = new double[n];
System.out.print("Enter 5 integer values. ");
for(int i = 0; i < n; i++) {
list1[i]= input.nextInt();
if(i == (n - 1)){
System.out.println("\t The average of the 5 integers is "
+ average(list1, n));
}
}
System.out.println("Enter 5 double values. ");
for (int i = 0; i < n; i++){
list2[i]= input.nextDouble();
if(i == (n-1)){
System.out.println("\t The average of the 5 doubles is "
+ average(list2, n));
}
}
}
public static int average(int[] array, int n){
int sum = 0;
for(int i = 0; i < n; i++){
int holdNumber = array[i];
sum += holdNumber;
}
int average = sum / n;
return average;
}
public static double average(double[] array, int n){
double sum = 0;
for(int i = 0; i < n ; i++){
double holdNumber = array[i];
sum += holdNumber;
}
double average = sum / n;
return average;
}
}

It doesn't give you an error because you only read the first 5 values, as stated in your for loop.
The first thing is you should decouple your input logic from your output logic, so you know for sure you're in your 5th number when you exit the for loop.
Then you can check if there's anything else than a blank string left, if there is then you can throw an exception stating it has too many numbers.
I've adapted the integer part, you can easily adapt the doubles logic.
Feel free to ask if you have any doubts.
The adapted code:
import java.util.Scanner;
public class Lab7A_BRC {
public static void main(String[] args) {
System.out.println("\t\t Average arrays ");
Scanner input = new Scanner(System.in);
//array count varriable
int n = 5;
//array declaration
int[] list1 = new int[n];
double[] list2 = new double[n];
System.out.print("Enter 5 integer values. ");
for (int i = 0; i < n; i++) {
list1[i] = input.nextInt();
}
if (!input.nextLine().equals("")) {
throw new RuntimeException("Too many numbers entered!");
}
System.out.println("\t The average of the 5 integers is "
+ average(list1, n));
System.out.println("Enter 5 double values. ");
for (int i = 0; i < n; i++) {
list2[i] = input.nextDouble();
if (i == (n - 1)) {
System.out.println("\t The average of the 5 doubles is "
+ average(list2, n));
}
}
}
public static int average(int[] array, int n) {
int sum = 0;
for (int i = 0; i < n; i++) {
int holdNumber = array[i];
sum += holdNumber;
}
int average = sum / n;
return average;
}
public static double average(double[] array, int n) {
double sum = 0;
for (int i = 0; i < n; i++) {
double holdNumber = array[i];
sum += holdNumber;
}
double average = sum / n;
return average;
}
}

I think you're confusing two different concepts.
One is the input, and another one is your variable.
Input is a buffer (read: block of data) managed by the shell and the Scanner. It can contain an arbitrary amount of data, you have nothing to do with it.
What happens in your code is that the scanner takes the buffer and parses (read: interprets) the next valid value from the buffer and transforms it into the right data type - until the "nth" element. So, because you're taking "n" elements (controlled by the for), it doesn't matter how much data is available in the input buffer, you always read a finite amount.
The only way the amount of data matters is when there's no more input for the scanner to read from, in which case it asks for more input.

The reason is that you are iterating till the n number that you defined in the beginning.
for(int i = 0; i < n; i++) {
list1[i]= input.nextInt();
So if you try to enter 1 1 1 1 1 124124 1241 you will see that the average is 1 because the rest is ignored and not added to the list. Because it doest not try nextInt() more than n given.

I am a beginner, so my answer might be wrong:), sorry for that. This code is working for me. Like #iajrz mentioned, you can do spaece or newline when try to use system.in.

Because the for loop does n iterations so you pick up only the first n integers of the input. If your input is 1 2 3 4 5 6 7 8 it will select only 1 2 3 4 5 (because in your code n=5). You can also insert multiple digits number separated by spaces, so input 15 0 00 0010 0 has average=5

Related

Counting number of digits in integer

My code is supposed to prompt the user for an integer and count the occurrences of each digit in the integer. But it outputs nothing. Please help and here's my code.
import java.util.Scanner;
public class NumberCounts {
public static void main(String [] args) {
Scanner scan = new Scanner(System.in);
int arr [] = new int [10];
int num;
int outcome;
System.out.println(" Enter a number: ");
num = scan.nextInt();
while ( num != 0){
outcome = num % 10;
arr[num % 10]+=1;
num /= num;
}
for ( int i = 0; i <= arr.length; i++){
System.out.println(i + " : " + arr[i]);
}
}
}
Inside your while loop, you are dividing num by itself, then assigning the result back to num. This will make num equal to 1 forever. Change it to
num /= 10;
which will remove the last digit.
Also, you already calculated the last digit and stored the result in outcome. You might as well use it:
arr[outcome]+=1;
Then, when you get to the loop to print the array, in the condition, always use < an array length to stop after processing the last index length - 1. This will prevent the ArrayIndexOutOfBoundsException that is waiting to show up.
for (int i = 0; i < arr.length; i++){

Generating random number sequences & sorting them out in java

Good day,
I am currently working on this program, which has to generate as many numbers as I enter (between 1-100) and the number has to be in the range of 0 - 1 000 000.
Then, the program must print them out in a random order and after that using the insertion sort, it must sort the randomly generated numbers.
I've been tackling this for about 7 hours now, searched online for answers, but haven't found anything yet. I was hoping to get a fix for my problem here!
What the program is supposed to do:
Person enters how many random numbers they want the program to generate (between 1-100)
Print out the amount of random numbers (random numbers must must be in the range of 0 - 1 000 000), the person entered before.
Sort the numbers using insertion sorting and print them out.
This is what I currently have:
It doesn't print out the sorted list.
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter the amount:");
int amount = scan.nextInt();
int []numbers = new int[amount];
Random rand = new Random();
System.out.println("Random order:");
int MAX = 1000000;
int MIN = 0;
for (int i = 0; i < amount; i++) {
numbers[i] = rand.nextInt(MAX - MIN + 1) + MIN;
System.out.print(numbers[i] + ", ");
}
System.out.println("");
System.out.print("From smallest to biggest:");
sort(numbers);
}
public static int[] sort(int[] list) {
int i, j, key, temp;
for(i = 1; i < list.length; i++) {
key = list[i];
j = i-1;
while (j >= 0 && list[j] > key) {
temp = list[j];
list[j] = list[j+1];
list[j+1] = temp;
j--;
}
System.out.print(list);
}
return list;
}
}
I think u forgot to call the sort method.

How to return the Max Value and Max Count from a user input

I'm trying to get both, the largest number and the largest occurring number, from a user input. The problem with my code is it only returns the first value of the array.
public class CountMax {
public static void main(String [] args) {
//Create scanner object
Scanner input = new Scanner(System.in);
//Obtain user input
System.out.println("Enter numbers: ");
int num = input.nextInt();
int array[] = new int[num];
//loop through array
int max = array[0];
int count = 1;
for (int i = 0; i < array.length; i++) {
array[i] = num;
if(array[i] > max) {
max = array[i];
count = 1;
} else if(array[i] == max) {
count++;
}
}
//output results
System.out.println("The largest number is " + max);
System.out.println("The occurrence count of the largest number is " + count);
}}
I know this post is old, but Wyatt Lowery's solution is incorrect, just in case someones stumbles upon it from Google just like I did. You cannot count the number of max values in an array in the same loop like that until you have found the max value.
Example using Wyatt's class: 2 is obviously an incorrect answer.
Enter numbers:
1, 2, 3, 4, 5, 5, 7
The largest number is 7
The occurrence count of the largest number is 2
I would do:
int max = array[0];
int sum = 0;
for(int i = 1; i < array.length; i++) {
if(array[i] > max) max = array[i];
}
for(int i = 0; i < array.length; i++) {
if(array[i]==max) sum++;
}
One problem I noticed:
int num = input.nextInt();
When you do this, it is only going to take the first int (Meaning, only 1 number) As well when you are creating your array int array[] = new int[num], you are creating an array with the SIZE of num, and not actually creating an array with the VALUES of num. (Even though num is only a single number) To actually create an array of numbers, do something like this:
System.out.pritnln("Enter in numbers:");
String[] array = input.nextLine().split(", ");
An example input would be: "13, 12, 14, 14". Then the contents of the array would be those terms (And would remove spaces & commas). Your program should look something like this when finished:
public class CountMax {
public static void main(String [] args) {
//Create scanner object
Scanner input = new Scanner(System.in);
//Obtain user input
System.out.println("Enter numbers: ");
String[] array = input.nextLine().split(", ");
//Loop through array
int max = Integer.parseInt(array[0]);
int count = 0;
for (int i = 0; i < array.length; i++) {
if(Integer.parseInt(array[i]) > max) {
max = Integer.parseInt(array[i]);
} else if(Integer.parseInt(array[i]) == max) {
count++;
}
}
//Output
System.out.println("The largest number is " + max);
System.out.println("The occurrence count of the largest number is " + count);
}
}
Hope this helped :-)
Think more carefully about each step you need to take.
Do you know how many numbers will be entered by the user?
Right now you are only taking in one number because you are not looping on the input
int num = input.nextInt();
int array[] = new int[num];
Here, you are creating an array the size of whatever number the user entered. This is a correct approach, more typical of C, if the user will tell you "I will enter 10 numbers" and then enters the 10 numbers. This is convenient because you will know to loop 10 times, and you will need to count a maximum of 10 different numbers.
If we don't know how many numbers will be entered you will need to loop until EOF.. something like
while(input.hasNext()) {
int currentInt = input.next();
...
}
Now you have to consider how you will be counting these items.
I hope this gives you some things to think about towards your solution..

Create an array based on a range of values given by the user

What I'm trying to do is create an array based on values given by the user. The user has to give the length of the array plus the max and min values. I've managed to get the program to the point where it does output the correct amount of values (the correct length), but it just keeps outputting the exact same number (which is the max and min added). Based on research I did I tried converting them to a string so that wouldn't happen, but it still isn't working correctly. I've tried a couple of different methods including: Integer.toString, String.valueOf, and creating a whole new string. Any help would be greatly appreciated. Here's the code so far:
public static void main(String[] args) {
//Create a Scanner object
Scanner input = new Scanner(System.in);
//Ask the user to enter the length of the array
System.out.println("Please enter the length of the array:");
int arraylength = input.nextInt();
//Ask the user to enter a max value
System.out.println("Please enter the max value:");
int max = input.nextInt();
//Ask the user to input the minimum value
System.out.println("Please enter the min value:");
int min = input.nextInt();
//Initialize the array based on the user's input
double [] userArray = new double[arraylength];
/**
*The program comes up with random numbers based on the length
*entered by the user. The numbers are limited to being between
*the minimum and maximum value.
*/
for (int i = min; i < userArray.length; i++) {
userArray[i] = Math.random() * max;
}
//This code is supposed to sort the array and print out all of the numbers in order,
//with minimum in the beginning and max in the end.
for (int i = 0; i < userArray.length; i++) {
selectionSort(userArray);
Integer.toString(min);
Integer.toString(max);
System.out.println(min + userArray[i] + max);
}
//This code uses the method average to find the average
average(userArray);
//Close Scanner
input.close();
}
public static double average(double[] data) {
double sum = 0;
for (int i = 0; i < data.length; i++) {
sum = sum + data[i];
}
double average = sum / data.length;
return average;
}
public static void selectionSort(double[] list) {
for (int i = 0; i < list.length - 1; i++) {
//Find the minimum in the list[i...list.length-1]
double currentMin = list[i];
int currentMinIndex = i;
for (int j = i + 1; j < list.length; j++) {
if (currentMin > list[j]) {
currentMin = list[j];
currentMinIndex = j;
}
}
//Swap list[i] with list[currentMinIndex] if necessary
if (currentMinIndex != i) {
list[currentMinIndex] = list[i];
list[i] = currentMin;
}
}
}
}
Now, after I added that bit with the average calculation, the program did work once, though it did not compute the average (so it just created an array with min and max at the ends, sorted). It appears to be a fluke, because this is the exact same code and it hasn't done that since. Though maybe the average code somehow affected the rest?
If I understood you problem correctly, you need to change this line
System.out.println(min + userArray[i] + max);
to this:
System.out.println(min.toString() + " " + userArray[i].toString() + " " + max.toString());
My guess is, that it is java you are using. That tag would be helpful, too.
Edit:
You only need to sort your array once, and these do nothing: Integer.toString(min);
So the print routine could look like this:
selectionSort(userArray);
for (int i = 0; i < userArray.length; i++) {
System.out.println(min.toString() + " " + userArray[i].toString() + " " + max.toString());
}

Editing exception e to catch letters not decimals

I'm writing a program that takes 10 floating point numbers as input. However, whenever I enter a decimal the program sends me an error. I question is: how would I edit my current try-catch exception to catch only letters, etc., and allow decimals to be entered (and then store them into an array). Also, regardless of this problem, my program is also outputting the average many times, and always saying that it is equal to 0.
Below is the program:
import java.util.Scanner;
public class Average {
public static void main(String[] args) {
new Average().average(new double[10]);
}
public double average(double[] number) {
Scanner scanner = new Scanner(System.in);
int x = 0;
double sum = 0;
double[] numberList = new double[10]; //array to hold all numbers
double[] largerList = new double[10]; //array to hold numbers greater than the average
int numberIndex = 0;
int largerIndex = 0;
System.out.printf("Please enter 10 floating-point numberes.\nIf more than 10 values are entered, the numbers following 10 are ignored.\nIf less than 10 numbers are entered, the program will wait for you to enter 10.\n");
for (int i = 0; i < 10; i++) {
try { //try catch exception to catch decimal inputs as well as more /less than 10 integers
x = scanner.nextInt();
sum += numberList[x]; //add up all inputs to find sum
} catch (Exception e) {
System.out.println("Invalid input! Please reenter 10 integer values.");
scanner = new Scanner(System.in);
i = -1;
numberIndex = 0;
largerIndex = 0;
numberList = new double[10];
largerList = new double[10];
continue;
}
}
for (int i = 0; i < number.length; i++) {
sum = sum + number[i];
double average = sum / number.length;
//return average;
if (x > average) {
largerList[largerIndex] = x; //add negative input to negativeList array
largerIndex = largerIndex + 1;
}
System.out.println("Average value of your input is: " + average);
System.out.println();
}
for (int i = 0; i < largerIndex; i++) {
System.out.println(largerList[i]);
}
return 0;
}
}
You're using the nextInt() function, which only returns ints. Ints cannot hold decimals. Consult the API and take a look at the nextFloat() and nextDouble() methods instead.
For your Average problem, Your print statement is into a for loop so it will be executed number.length times. Move the print statements outside the loop. You will also need the declaration of the average variable to be outside the loop. Also you should only need to the loop to calculate the sum, you don't need the calculate the average everytime
double average;
for(/*loop conditions*/)
{
sum = sum + number[i];
}
average = sum / number.length;
System.out.println("Average value of your input is: " + average);
System.out.println();

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