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Index out of bounds issue with number systems code

I am currently having an issue with a problem where my java index is out of bounds. The specific error is inside my binaryToDecimal method and is Index 32 out of bounds for length 32. I have tried to reorganize my code as in declaring the int binaryVal[] = new int[32]; as a global variable to fix the error but it does not work. How do I fix this issue?
import java.util.Scanner;
public class NumberConverter {
public static Scanner input = new Scanner(System.in);
static boolean success;
static String originalNum;
static int value = 0;
static int choice = 0;
static int intNum = 0;
static int remainder;
static char [] valueBinaryArray;
public static void main(String[] args) {
// TODO Auto-generated method stub
greeting();
getOriginalNumber();
getOutputType();
//checkValidInput();
if (value == 2 && choice == 3) {
decimalToHex();
}
else if (value == 3 && choice == 2) {
hexToDecimal();
}
else if (value == 2 && choice == 1) {
decimalToBinary();
}
else if (value == 1 && choice == 2) {
binaryToDecimal();
}
}
public static void greeting() {
System.out.println("Hello and welcome to the number systems converter");
System.out.println("Below you will be givin options as to what numbers you'd like to convert");
System.out.println("");
}
public static String getOriginalNumber() {
System.out.println("Enter a number:");
originalNum = input.next();
System.out.println("What type of number is this?");
System.out.println("");
System.out.println("1. Binary");
System.out.println("2. Decimal");
System.out.println("3. Hexadecimal");
value = input.nextInt();
return originalNum;
}
public static void getOutputType() {
System.out.println("Which number system would you like to convert to?");
System.out.println("");
System.out.println("1. Binary");
System.out.println("2. Decimal");
System.out.println("3. Hexadecimal");
choice = input.nextInt();
}
public static void checkValidInput() {
}
public static void decimalToHex() {
int intNum = Integer.valueOf(originalNum);
String str2 = "";
char hex[] = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
while (choice > 0) {
remainder = intNum % 16;
str2 = hex[remainder] + str2;
intNum = intNum/16;
}
}
public static void hexToDecimal() {
int intNum = Integer.valueOf(originalNum);
int counter = 0;
String hexVal = "";
int digit;
while (choice > 0) {
digit = intNum % 16;
switch (digit) {
case 1:
hexVal+="F"; break;
case 2:
hexVal+="E"; break;
case 3:
hexVal+="D"; break;
case 4:
hexVal+="C"; break;
case 5:
hexVal+="B"; break;
case 6:
hexVal+="A"; break;
default:
hexVal+=Integer.toString(digit);
}
intNum = intNum/16;
}
for (counter = hexVal.length()-1; counter>=0; counter--)
System.out.print(hexVal.charAt(counter));
}
public static void decimalToBinary() {
int intNum = Integer.valueOf(originalNum);
int counter = 0;
int binaryVal[] = new int[32];
while (choice > 0) {
binaryVal[counter++] = intNum % 2;
intNum = intNum/2;
}
for (int i = counter-1; i >= 0; i--) {
System.out.println(binaryVal[i]);
}
}
public static void binaryToDecimal() {
int intNum = Integer.valueOf(originalNum);
int counter = 0;
int binaryVal[] = new int[32];
while (choice > 0) {
binaryVal[counter++] = intNum % 2;
intNum = intNum/2;
}
for (int i = counter-1; i >= 0; i--) {
System.out.print(binaryVal[i]);
}
}
}

If you have an array with 32 items the first one is index 0 and the last one is index 31. Trying to access the item at position 32 goes out of the range of the array which is why you get an array index out of bounds exception.
To see the output I would suggest maybe make the array 33 big instead of 32. It's not a proper fix but you should be able to see what it is doing.
If it still goes out of bounds I'd have a closer look at your while loop end condition.

so what is causing your error is your condition in your for statement. Currently, you are just checking if choice >= 0 the problem with this check is that it allows the for statement to go beyond the length of the array you have specified (32). In order to fix this error, you simply need to add another check to make sure the for loop cannot go beyond the length of the array you have specified. This can be done by adding this statement into your for loop: counter < binaryVal.length. This should prevent the for loop from going beyond the length of your array and you shouldn't run into any errors. Your final code for your binaryToDecimal method should look like this:
public static void binaryToDecimal() {
int intNum = Integer.valueOf(originalNum);
int counter = 0;
int binaryVal[] = new int[32];
while (choice > 0 && counter < binaryVal.length) {
binaryVal[counter++] = intNum % 2;
intNum = intNum/2;
}
for (int i = counter-1; i >= 0; i--) {
System.out.print(binaryVal[i]);
}
}

Multiple of an Integer inside range

I am doing a school project and I kinda got blocked.
I am looking forward building a javascript that asks the user of a number between 1 and 20 and then finds and lists all the multiples of that number inside range 0 and 100.
Here is what it looks like at the moment:
public static void main(String[] args) {
Scanner lector = new Scanner(System.in);
System.out.println("*** Program start ***\n");
System.out.println("Insert number [integer, between 1 and 20]: ");
boolean okay = false;
while (!okay) {
int n1 = lector.nextInt();
lector.nextLine();
if (n1<1 || n1>20) {
System.out.print("Invalid number!\nplease try again [between 1 and 20]: ");
} else {
okay = true;
System.out.println("Number accepted!");
}
int i = 0;
while (i <= 100) {
System.out.println(i);
if ((n1%100) == 0) {
System.out.println(n1);
}
i = i + 1;
}
System.out.println("\n*** End ***");
}
}
}
I am obviously bad at math cause I can't get the formula to work.
Thank you in advance!

public static void main(String[] args) {
Scanner lector = new Scanner(System.in);
System.out.println("*** Program start ***\n");
System.out.println("Insert number [integer, between 1 and 20]: ");
boolean okay = false;
while (!okay) {
int n1 = lector.nextInt();
lector.nextLine();
if (n1<1 || n1>20) {
System.out.print("Invalid number!\nplease try again [between 1 and 20]: ");
} else {
okay = true;
System.out.println("Number accepted!");
}
int i = 0;
while (i <= 100) {
System.out.println(i);
if ((n1%i) == 0) {
System.out.println(i);
}
i = i + 1;
}
System.out.println("\n*** End ***");
}
}
}

All multiples that fall between 0 and 100? That's the kind of thing for-loops are made for. After performing your IO and reading the number n1, change that while loop to the following.
for (int i = n1; i >= 0 && i <= 100; i = i + n1) {
System.out.println(i);
}
In case you aren't familiar with for loops, this basically says, set i to the value of n1, and keep adding n1 to it, printing each value as you go. When it leaves the range of 0..100, it terminates. So for instance, if 8 is entered it goes 8, 16, 24, 32, ..., 80, 88, 96.
If you really want to use a while loop, then try:
int i = n1;
while(i >= 0 && i <= 100) {
System.out.println(i);
i = i + n1;
}

public static void main(String[] args) {
Scanner lector = new Scanner(System.in);
System.out.println("*** Program start ***\n");
System.out.println("Insert number [integer, between 1 and 20]: ");
boolean okay = false;
while (!okay) {
int n1 = lector.nextInt();
lector.nextLine();
if (n1<1 || n1>20) {
System.out.print("Invalid number!\nplease try again [between 1 and 20]: ");
} else {
okay = true;
System.out.println("Number accepted!");
}
int i = 0;
while (i <= 100) {
System.out.println(i);
if ((n1*i) <= 100) {
System.out.println(i);
}
i = i + 1;
}
System.out.println("\n*** End ***");
}
}
}
This should work. You were just finding if the number n1 was divisible by 100.

Error in finding a integers factor by a method

My program seems to work with numbers '25, 223, and 11" but i have no idea why it stops when i enter the numbers 10 or 100. Any ideas? Any help on this matter will be greatly appreciated-thank you in advance.
*****Source Code*****
import java.util.Scanner;
public class hw_5 {
public static void main(String[] args) {
Scanner inputReader = new Scanner(System.in);
int num;
boolean primeTest;
System.out.print("Enter an integer value: ");
num = inputReader.nextInt();
if(num != -1) {
primeTest = calcPrime(num);
if(!primeTest) {
System.out.println(num+" is not prime.");
printFactors(num);
}
}
}
public static boolean calcPrime(int num) {
for(int i = 2; i < num; i++) {
if(num % i == 0)
return true; // If number is divisible by any number, return true.
}
return false; // If loop exits (means, number was not divisible by any number), return false.
}
public static void printFactors(int num) {
int nFactors = 0;
for(int i = 2; i < num; i++) {
if(num % i == 0) {
System.out.println(num+" is divisible by "+i);
nFactors++;
}
}
System.out.println(num+" has "+nFactors+" factors");
}
}

import java.util.Scanner;
public class Test {
public static boolean calcPrime(int num) {
for(int i = 2; i < num; i++) {
if(num % i == 0)
return false; // If number is divisible by any number, return false.
}
return true; // If loop exits (means, number was not divisible by any number), return true.
}
public static void printFactors(int num) {
int nFactors = 0;
for(int i = 2; i < num; i++) {
if(num % i == 0) {
System.out.println(num+" is divisible by "+i);
nFactors++;
}
}
System.out.println(num+" has "+nFactors+" factors");
}
public static void main(String[] args) {
Scanner inputReader = new Scanner(System.in);
int num = 0;
boolean primeTest = false;
while(true) {
System.out.print("Enter an integer value: ");
num = inputReader.nextInt();
if(num == -1)
break;
primeTest = calcPrime(num);
if(primeTest)
System.out.println(num+" is prime.");
else {
System.out.println(num+" is not prime.");
printFactors(num);
}
}
}
}

Your prime test is wrong :
public static int calcPrime(int number) {
int primer = number % 2;
return primer;
}
It just tests if the number is odd.

Below method returns true if it is a prime number ,
boolean isPrime(int n) {
//check if n is a multiple of 2
if (n%2==0) return false;
//if not, then just check the odds
for(int i=3;i*i<=n;i+=2) {
if(n%i==0)
return false;
}
return true;
}
You could add the numbers to the list and get the their values on iterating back

Prime number required by user in Java

I want to display the prime number required by the user. For example, if the user wants 3rd prime, I will display 5. I have the following java code.
import java.util.Scanner;
public class Prime {
private static Scanner scanner;
public static void main(String args[]) {
//get input till which prime number to be printed
// System.out.println("Enter which prime number to be printed: ");
// scanner = new Scanner(System.in);
// int limit = scanner.nextInt();
int count = 0;
int number = 2;
//System.out.println("Printing prime number from 1 to " + limit);
while(count<=3)
{
if(isPrime(number)){
count++;
// System.out.println(count);
}
number++;
}
if(count == 3)
System.out.println("10001 prime is "+number);
}
public static boolean isPrime(int number){
for(int i=2; i<number; i++){
if(number%i == 0){
return false;
}
}
return true;
}
}
When i run it, I am not able to gett any output. Where am i going wrong?
PS: For time being, I am running the loop only until 3.

You have while(count <= 3) so when you exit the loop, count == 4.
Therefore, your if(count == 3) is never entered and nothing is printed.

Anyways the better solution would be
public void calcPrime(int inp) {
ArrayList<Integer> arr = new ArrayList<Integer>();
arr.add(2);
arr.add(3);
int counter = 4;
while(arr.size() < inp) {
if(counter % 2 != 0 && counter%3 != 0) {
int temp = 4;
while(temp*temp <= counter) {
if(counter % temp == 0)
break;
temp ++;
}
if(temp*temp > counter) {
arr.add(counter);
}
}
counter++;
}
System.out.println("finish" +arr.get(inp-1));
}
}

Here is corrected code that works.
public class Main {
public static void main(String args[]) {
//Returns fourth prime number
System.out.println(getPrimeNumber(4));
}
public static int getPrimeNumber(int order) {
int currentOrder = 1;
int currentNumber = 1;
while (currentOrder < order) {
currentNumber++;
if (isPrime(currentNumber)) currentOrder++;
}
return currentNumber;
}
public static boolean isPrime(int number) {
for (int i = 2; i < number; i++) {
if (number % i == 0) {
return false;
}
}
return true;
}
}
There was no need to start your counter at 0 and it mathematically wrong to start with number 2 ! Just begin with order and currentNumber initialized at 1 and if first prime number is what your user is looking for, no loop will be required !
Also, your loop condition after correct variable initialization was corrected from "<=" to "<".
That's all !

Java - Using for loops to read user input

In Java, I am having trouble running multiple loops using a single sequence of user-inputted integers. Individually, they run fine individually but together it prints out incorrect numbers.
I'm at a loss as to what is causing this problem.
Here is my code.
import java.util.Scanner;
public class SequenceTester
{
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
System.out.println("Enter a sequence of integers. " +
"Enter a non-integer to terminate");
int sequence = in.nextInt();
//Print One
int min = sequence;
while(in.hasNextInt())
{
int input = in.nextInt();
if(input < smallest)
{
smallest = input;
}
}
System.out.println(smallest);
//Print Two
int max = sequence;
while(in.hasNextInt())
{
int input = in.nextInt();
if(input > max)
{
max = input;
}
}
System.out.println(max);
//Print Three
int even = 0;
int odd = 0;
while(in.hasNextInt())
{
int input = in.nextInt();
if((input %2) == 0)
{
even++;
}
else
{
odd++;
}
}
System.out.println( even);
System.out.println(odd);
//Print Four
double total = 0;
int count = 0;
while (in.hasNextInt())
{
Int input = in.nextInt();
total = total + input;
count++;
}
double average = 0;
if (count > 0)
{
average = total / count;
}
System.out.println(average);
}
}

Your code is very fragmented, but it looks like you can achieve what you want with one loop since the loop condition is the same in all of them. This, of course, is only based on the vague description you gave us.
while(in.hasNextInt()) {
int input = in.nextInt();
if(condition1) {
//do stuff
} else if (condition2) {
//do other stuff
} else if (conditionN) {
//do other other stuff
} else {
//last of the stuff to do
}
}