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I'm trying to make my program throw an IllegalArgumentException if the input is not a number

I'm trying to make it to were is will throw my exception if the input is not a number and i cant figure it out, Could someone help me get on the right track
import java.util.Scanner;
class calculations
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
int number;
int total = 0;
try
{
} catch ( IllegalArgumentException e)
{
//error
}
while (true)
{
number = scan.nextInt();
if (number == 0)
{
break;
}
total += number;
}
System.out.println("Total is " + total);
}
}

You should use hasNextInt, which will allow you to check if the next token in the stream can be parsed as an int.
if (! scanner.hasNextInt()) throw new IllegalArgumentException()

Repeating a method until conditions are met

how can i get the following code to repeat input() until a numeric is entered and at the same time tell the user what type of of variable was entered,be it string,double or integer and if conditions are met prints out a success message?
package returnin;
import java.util.*;
public class trycatch {
public static void main(String[]args){
String chck=input();
String passed =check(chck);
System.out.println("If you see this message it means that you passed the test");
}
static String input(){
Scanner sc= new Scanner(System.in);
System.out.println("Enter a value");
String var=sc.nextLine();
return var;
}
static String check(String a){
double d = Double.valueOf(a);
if (d==(int)d){
System.out.println( "integer "+(int) d);
}
else {
System.out.println(" double "+d);
}
return a;
}
}

Here's a commented example:
package returnin;
import java.util.*;
public class trycatch {
public static void main(String[] args) {
// Don't recreate Scanner inside input method.
Scanner sc = new Scanner(System.in);
// Read once
String chck = input(sc);
// Loop until check is okay
while (!check(chck)) {
// read next
chck = input(sc);
}
System.out.println("If you see this message it means that you passed the test");
}
static String input(Scanner sc) {
System.out.println("Enter a value");
return sc.nextLine();
}
static boolean check(String a) {
try {
// Try parsing as an Integer
Integer.parseInt(a);
System.out.println("You entered an Integer");
return true;
} catch (NumberFormatException nfe) {
// Not an Integer
}
try {
// Try parsing as a long
Long.parseLong(a);
System.out.println("You entered a Long");
return true;
} catch (NumberFormatException nfe) {
// Not an Integer
}
try {
// Try parsing as a double
Double.parseDouble(a);
System.out.println("You entered a Double");
return true;
} catch (NumberFormatException nfe) {
// Not a Double
}
System.out.println("You entered a String.");
return false;
}
}

Catching IllegalArgumentException?

I am having a little bit of a problem here. I am trying to figure out how to catch the IllegalArgumentException. For my program, if the user enters a negative integer, the program should catch the IllegalArgumentException and ask the user if he/she wants to try again. But when the exception is thrown, it doesn't give that option. It just terminates. I tried to use the try and catch method but it doesn't work for me. How do I catch this particular exception to continue to run instead of terminating?
public static void main(String[] args) throws IllegalArgumentException
{
String keepGoing = "y";
Scanner scan = new Scanner(System.in);
while(keepGoing.equals("y") || keepGoing.equals("Y"))
{
System.out.println("Enter an integer: ");
int val = scan.nextInt();
if (val < 0)
{
throw new IllegalArgumentException
("value must be non-negative");
}
System.out.println("Factorial (" + val + ") = "+ MathUtils.factorial(val));
System.out.println("Another factorial? (y/n)");
keepGoing = scan.next();
}
}
}
and
public class MathUtils
{
public static int factorial(int n)
{
int fac = 1;
for(int i = n; i > 0; i--)
{
fac *= i;
}
return fac;
}
}

You need to add the try catch block inside the loop to continue the working for the loop. Once it hits the illegal argument exception catch it in catch block and ask if the user wants to continue
import java.util.Scanner;
public class Test {
public static void main(String[] args)
{
String keepGoing = "y";
populate(keepGoing);
}
static void populate( String keepGoing){
Scanner scan = new Scanner(System.in);
while(keepGoing.equalsIgnoreCase("y")){
try{
System.out.println("Enter an integer: ");
int val = scan.nextInt();
if (val < 0)
{
throw new IllegalArgumentException
("value must be non-negative");
}
System.out.println("Factorial (" + val + ") = "+ MathUtils.factorial(val));
System.out.println("Another factorial? (y/n)");
keepGoing = scan.next();
}
catch(IllegalArgumentException i){
System.out.println("Negative encouneterd. Want to Continue");
keepGoing = scan.next();
if(keepGoing.equalsIgnoreCase("Y")){
populate(keepGoing);
}
}
}
}
}
Hope this helps.
Happy Learning :)

I don't think you want your main() method to be throwing an exception. Typically, this is the kind of thing that you'd put in try and catch blocks.
Honestly though, for this sort of thing an if/else would work better. (Unless you're just doing this as a toy example, to learn exceptions.)
Make another method called getNumber() that throws the IllegalArgumentException, that returns an int. Then put it inside the try/catch in the main().

public static void main(String[] args)
{
String keepGoing = "y";
Scanner scan = new Scanner(System.in);
while(keepGoing.equals("y") || keepGoing.equals("Y"))
{
int val = 0;
boolean flag=true;
while(flag){
try{
System.out.println("Enter an integer: ");
val = scan.nextInt();
if (val < 0)
{
throw new IllegalArgumentException
("value must be non-negative");
}
flag = false;
} catch(IllegalArgumentException e){
System.out.println("value must be non-negative");
}
}
System.out.println("Factorial (" + val + ") = "+ MathUtils.factorial(val));
System.out.println("Another factorial? (y/n)");
keepGoing = scan.next();
}
}
}

I would suggest you add a test on the negative value and display your message on the spot, then use an else block. Also, you could use String.equalsIgnoreCase() in your loop test like
String keepGoing = "y";
Scanner scan = new Scanner(System.in);
while (keepGoing.equalsIgnoreCase("y")) {
System.out.println("Enter an integer: ");
int val = scan.nextInt();
if (val < 0) {
System.out.println("value must be non-negative");
} else { // <-- skip negative value
System.out.println("Factorial (" + val + ") = "
+ MathUtils.factorial(val));
}
System.out.println("Another factorial? (y/n)");
keepGoing = scan.next();
}
Also, an int factorial(int) method can only the first 12 correct values. You could use a long or a BigInteger like
public static BigInteger factorial(int n) {
BigInteger fac = BigInteger.ONE;
for (int i = n; i > 1; i--) {
fac = fac.multiply(BigInteger.valueOf(i));
}
return fac;
}

Similar to some other answers, I would say that your main() method should not throw an exception to display an error message to the user, because that is not the purpose of exception handling mechanisms in Java. Exception handling is designed to enable methods to signal that something happened that should not have happened, so the methods that call those methods will know that they need to deal with them. In general, exceptions are caught, not thrown, by main() methods and other user interface methods.
I would make the method factorial() throw the IllegalArgumentException, rather than your main() method in your program class. Your main() method should use try and catch to handle this exception. With this design, if someone else wanted to use your MathUtils class, they would know that your factorial() method throws an IllegalArgumentException (especially if you document your code with javadoc), and would write their code to handle the exception. In the current situation, if someone tries to call MathUtils.factorial(-1), the return value would be 1 because the for loop inside factorial() would not execute at all (because i is initially set to -1, which is not greater than 0).
This is how I would revise your code:
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String keepGoing = "y";
while(keepGoing.equalsIgnoreCase("y")) {
try { // This code might throw an exception
System.out.println("Enter an integer: ");
int val = scan.nextInt();
System.out.println("Factorial (" + val + ") = "+ MathUtils.factorial(val));
System.out.println("Another factorial? (y/n)");
keepGoing = scan.next();
} catch (IllegalArgumentException | InputMismatchException e) {
/* An InputMismatchException is thrown if the input is not an integer.
See the documentation for Scanner method nextInt() for more details.
*/
System.out.println("You must enter a non-negative integer.");
System.out.println("Try again? (y/n)");
keepGoing = scan.next();
}
}
}
}
and
public class MathUtils throws IllegalArgumentException {
public static int factorial(int n) {
if (fac < 0) {
throw new IllegalArgumentException("value must be non-negative");
}
int fac = 1;
for(int i = n; i > 0; i--) {
fac *= i;
}
return fac;
}
}

How can I check if an input is a integer or String, etc.. in JAVA?

I am wondering how I can check if a user's input is a certain primitive type (I mean integer, String, etc... I think it's called primitive type?). I want a user to input something, then I check if it's a String or not, and do a certain action accordingly. (JAVA)
I have seen some codes like this:
if (input == (String)input) { (RANDOM STUFF HERE) }
or something like
if input.equals((String) input)
And they don't work. I want to know how I can Check for only a String? (EDITED OUT)
I was wondering if there was a function that can do that? Thanks for the answers
EDIT: With the help of everyone I have created my fixed code that does what I want it to:
package files;
import java.util.*;
public class CheckforSomething {
static Scanner inputofUser = new Scanner(System.in);
static Object userInput;
static Object classofInput;
public static void main(String[] args){
try{
System.out.print("Enter an integer, only an integer: ");
userInput = inputofUser.nextInt();
classofInput = userInput.getClass();
System.out.println(classofInput);
} catch(InputMismatchException e) {
System.out.println("Not an integer, crashing down");
}
}
}
No need for answers anymore, thanks!

Use instanceof to check type and typecast according to your type:
public class A {
public static void main(String[]s){
show(5);
show("hello");
}
public static void show(Object obj){
if(obj instanceof Integer){
System.out.println((Integer)obj);
}else if(obj instanceof String){
System.out.println((String)obj);
}
}
}

You may try this with Regex:
String input = "34";
if(input.matches("^\\d+(\\.\\d+)?")) {
//okay
} else {
// not okay !
}
Here,
^\\d+ says that input starts with a digit 0-9,
()? may/or may not occur
\\. allows one period in input

Scanner input = new Scanner (System.in);
if (input.hasNextInt()) System.out.println("This input is of type Integer.");
else if (input.hasNextFloat()) System.out.println("This input is of type Float.");
else if (input.hasNextLine()) System.out.println("This input is of type string.");
else if (input.hasNextDouble()) System.out.println("This input is of type Double.");
else if (input.hasNextBoolean()) System.out.println("This input is of type Boolean.");
else if (input.hasNextLong())
System.out.println("This input is of type Long.");

Hate to bring this up after 6 years but I found another possible solution.
Currently attending a coding bootcamp and had to solve a similar problem. We introduce booleans and change their values depending on the result of the try/catch blocks. We then check the booleans using simple if statements. You can omit the prints and input your code instead. Here's what it looks like:
import java.io.IOException;
import java.util.Scanner;
public class DataTypeFinder {
public static void main(String[] args) throws IOException {
Scanner scan = new Scanner(System.in);
String input = "";
while (true) { //so we can check multiple inputs (optional)
input = scan.nextLine();
if ("END".equals(input)) { //a way to exit the loop
break;
}
boolean isInt = true; //introduce boolean to check if input is of type Integer
try { // surround with try/catch
int integer = Integer.parseInt(input); //boolean is still true if it works
} catch (NumberFormatException e) {
isInt = false; //changed to false if it doesn't
}
boolean isDouble = true; //same story
try {
double dbl = Double.parseDouble(input);
} catch (NumberFormatException e) {
isDouble = false;
}
if (isInt) {
System.out.printf("%s is integer type%n", input);
} else if (isDouble) {
System.out.printf("%s is floating point type%n", input);
} else if (input.length() == 1) { //this could be useless depending on your case
System.out.printf("%s is character type%n", input);
} else if ("true".equals(input.toLowerCase()) || "false".equals(input.toLowerCase())) {
System.out.printf("%s is boolean type%n", input);
} else {
System.out.printf("%s is string type%n", input);
}
}
}
}

class Main{
public static void main(String args[]){
String str;
Scanner sc=new Scanner(System.in);
int n,
boolean flag=false;
while(!flag){
try{
str=sc.nextLine();
n=Integer.parseInt(str);
flag=true;
}
catch(NumberFormatException e){
System.out.println("enter an no");
str=sc.nextLine();
}
}
}
}

Is this ok?
class Test
{
public static void main(String args[])
{
java.util.Scanner in = new java.util.Scanner(System.in);
String x = in.nextLine();
System.out.println("\n The type of the variable is : "+x.getClass());
}
}
Output:
subham#subham-SVE15125CNB:~/Desktop$ javac Test.java
subham#subham-SVE15125CNB:~/Desktop$ java Test
hello
The type of the variable is : java.lang.String

But Zechariax wanted an answer with out using try catch
You can achieve this using NumberForamtter and ParsePosition.
Check out this solution
import java.text.NumberFormat;
import java.text.ParsePosition;
public class TypeChecker {
public static void main(String[] args) {
String temp = "a"; // "1"
NumberFormat numberFormatter = NumberFormat.getInstance();
ParsePosition parsePosition = new ParsePosition(0);
numberFormatter.parse(temp, parsePosition);
if(temp.length() == parsePosition.getIndex()) {
System.out.println("It is a number");
} else {
System.out.println("It is a not number");
}
}
}

Try instanceof function with Integer instead of int.. each primitive also have a class

While loop wont stop looping with exception

probably missing something really silly, but my while loop will not stop printing the error message.
Could someone have a quick look and point me in the right direction?
package week5;
import java.util.*;
public class Week5 {
public static void main(String[] args) {
Scanner myKeyboard = new Scanner(System.in);
inputInt();
}
public static int inputInt(){
Scanner myKeyboard = new Scanner(System.in);
System.out.println("Enter number:");
int num;
boolean carryOn = true;
while (carryOn = true) {
{
try {
num = myKeyboard.nextInt();
carryOn = false;
}
catch (Exception e) {
System.out.println ("Integers only");
}
}
}
return 0;
}

This line is the problem
while (carryOn = true) {
Instead of using the comparison operator ==, you are using the assignment operator =. So essentially, you're setting carryOn to true every iteration, then automatically running the body of the loop since the condition essentially becomes
while (true) {
Just change that problem line to
while (carryOn == true) {

Apart from the infinite loop and the fact you always return 0; no matter what the user types, the code is far more complex than it needs to be.
// only create one scanner
private static final Scanner myKeyboard = new Scanner(System.in);
public static void main(String[] args) {
int num = inputInt();
// do something with num
}
public static int inputInt() {
System.out.println("Enter number:");
while (true) {
if (myKeyboard.hasNextInt()) {
int num = myKeyboard.nextInt();
myKeyboard.nextLine(); // discard the rest of the line.
return num;
}
System.out.println ("Integers only, please try again");
}
}

In your case:
while(carryOn==true)
or
while(carryOn)
will solve your problem