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Java Continue---I don't understand how I get in an infinite loop [duplicate]

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How to handle infinite loop caused by invalid input (InputMismatchException) using Scanner
(5 answers)
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So I'm building a program which takes ints from user input. I have what seems to be a very straightforward try/catch block which, if the user doesn't enter an int, should repeat the block until they do. Here's the relevant part of the code:
import java.util.InputMismatchException;
import java.util.Scanner;
public class Except {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean bError = true;
int n1 = 0, n2 = 0, nQuotient = 0;
do {
try {
System.out.println("Enter first num: ");
n1 = input.nextInt();
System.out.println("Enter second num: ");
n2 = input.nextInt();
nQuotient = n1/n2;
bError = false;
}
catch (Exception e) {
System.out.println("Error!");
}
} while (bError);
System.out.printf("%d/%d = %d",n1,n2, nQuotient);
}
}
If I enter a 0 for the second integer, then the try/catch does exactly what it's supposed to and makes me put it in again. But, if I have an InputMismatchException like by entering 5.5 for one of the numbers, it just shows my error message in an infinite loop. Why is this happening, and what can I do about it? (By the way, I have tried explicitly typing InputMismatchException as the argument to catch, but it didn't fix the problem.

You need to call next(); when you get the error. Also it is advisable to use hasNextInt()
catch (Exception e) {
System.out.println("Error!");
input.next();// Move to next other wise exception
}
Before reading integer value you need to make sure scanner has one. And you will not need exception handling like that.
Scanner scanner = new Scanner(System.in);
int n1 = 0, n2 = 0;
boolean bError = true;
while (bError) {
if (scanner.hasNextInt())
n1 = scanner.nextInt();
else {
scanner.next();
continue;
}
if (scanner.hasNextInt())
n2 = scanner.nextInt();
else {
scanner.next();
continue;
}
bError = false;
}
System.out.println(n1);
System.out.println(n2);
Javadoc of Scanner
When a scanner throws an InputMismatchException, the scanner will not pass the token that caused the exception, so that it may be retrieved or skipped via some other method.

YOu can also try the following
do {
try {
System.out.println("Enter first num: ");
n1 = Integer.parseInt(input.next());
System.out.println("Enter second num: ");
n2 = Integer.parseInt(input.next());
nQuotient = n1/n2;
bError = false;
}
catch (Exception e) {
System.out.println("Error!");
input.reset();
}
} while (bError);

another option is to define Scanner input = new Scanner(System.in); inside the try block, this will create a new object each time you need to re-enter the values.

To follow debobroto das's answer you can also put after
input.reset();
input.next();
I had the same problem and when I tried this. It completely fixed it.

As the bError = false statement is never reached in the try block, and the statement is struck to the input taken, it keeps printing the error in infinite loop.
Try using it this way by using hasNextInt()
catch (Exception e) {
System.out.println("Error!");
input.hasNextInt();
}
Or try using nextLine() coupled with Integer.parseInt() for taking input....
Scanner scan = new Scanner(System.in);
int num1 = Integer.parseInt(scan.nextLine());
int num2 = Integer.parseInt(scan.nextLine());

To complement the AmitD answer:
Just copy/pasted your program and had this output:
Error!
Enter first num:
.... infinite times ....
As you can see, the instruction:
n1 = input.nextInt();
Is continuously throwing the Exception when your double number is entered, and that's because your stream is not cleared. To fix it, follow the AmitD answer.

#Limp, your answer is right, just use .nextLine() while reading the input. Sample code:
do {
try {
System.out.println("Enter first num: ");
n1 = Integer.parseInt(input.nextLine());
System.out.println("Enter second num: ");
n2 = Integer.parseInt(input.nextLine());
nQuotient = n1 / n2;
bError = false;
} catch (Exception e) {
System.out.println("Error!");
}
} while (bError);
System.out.printf("%d/%d = %d", n1, n2, nQuotient);
Read the description of why this problem was caused in the link below. Look for the answer I posted for the detail in this thread.
Java Homework user input issue
Ok, I will briefly describe it. When you read input using nextInt(), you just read the number part but the ENDLINE character was still on the stream. That was the main cause. Now look at the code above, all I did is read the whole line and parse it , it still throws the exception and work the way you were expecting it to work. Rest of your code works fine.

Java Error when text input instead of number

I am just trying to get code to work where the code asks again for an answer, if text or a symbol is entered, instead of a required integer:
import java.util.Scanner;
class timecalc {
int hrs = 0;
int min = 0;
static int hourflag = 0;
static int minflag = 0;
Scanner sc = new Scanner(System.in);
public int getHours() {
try {
hourflag = hourflag + 1;
if (hourflag > 1) {
System.out.println("Invalid month Please enter hours again:");
}
System.out.println("Enter month:");
return hrs = sc.nextInt();
} catch (InputMisMAtchException e) {
System.out.println("entered invalid input " + e);
}
}
Have reviewed answers already given but cant get a workable solution
Any ideas?

I won't give you the entire code, but just a hint or psuedo-code. As an exercise you can implement it as per your requirement.
System.out.println("Enter month:");
while (true) {
try {
int min = sc.nextInt();
break;
} catch (InputMismatchException ex) {
System.err.println("Invalid input, please enter again");
sc.nextLine(); // <----- advance the scanner
}
}
Here the logic is to loop until we get the right input. If it is an invalid input, the loop never breaks.
Also as a side-note, I would recommend you to create just one method to fetch correct inputs and call it respectively from other methods. Rather than duplicating this logic everywhere.

Exception handling with a do-while loop in Java

The algorithm should take in 3 integers to an ArrayList. If the input is not an integer, then there should be a prompt. When I execute my code the catch clause is executed, but the program runs into a infinite loop. Could someone guide me into the right direction, I appreciate the help. :-D
package chapter_08;
import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
public class IntegerList {
static List<Integer> numbers = new ArrayList<Integer>();
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int counter = 1;
int inputNum;
do {
System.out.print("Type " + counter + " integer: " );
try {
inputNum = input.nextInt();
numbers.add(inputNum);
counter += 1;
}
catch (Exception exc) {
System.out.println("invalid number");
}
} while (!(numbers.size() == 3));
}
}

That is because when the next int is read using nextInt() and it fails, the Scanner still contains the typed contents. Then, when re-entering the do-while loop, input.nextInt() tries to parse it again with the same contents.
You need to 'flush' the Scanner contents with nextLine():
catch (Exception exc) {
input.nextLine();
System.out.println("invalid number");
}
Notes:
You can remove the counter variable, because you're not using it. Otherwise, you could replace counter += 1 by counter++.
You can replace while (!(numbers.size() == 3)) with while (numbers.size() != 3), or even better: while (numbers.size() < 3).
When catching exceptions, you should be as specific as possible, unless you have a very good reason to do otherwise. Exception should be replaced by InputMismatchException in your case.

If inputNum = input.nextInt(); cannot be fit into an int and a InputMismatchException is raised, the input of the Scanner is not consumed.
So after the catch, it loops and it goes again here :
inputNum = input.nextInt();
with exactly the same content in the input.
So you should execute input.nextLine(); in the catch statement to discard the current input and allow a new input from the user.
Besides it makes more sense to catch InputMismatchException rather than Exception as other exception with no relation with a mismatch could occur and it would not be useful to display to the user "invalid number " if it is not the issue :
catch (InputMismatchException e){
System.out.println("invalid number ");
input.nextLine();
}

You should to use a break; in your catch(){} like so :
try {
inputNum = input.nextInt();
numbers.add(inputNum);
counter += 1;
} catch (Exception e) {
System.out.println("invalid number ");
break;
}
So if one input is not correct break your loop.

try changing
inputNum = input.nextInt();
to
String inputText=input.next();
inputNum = Integer.valueOf(inputText);
it works perfectly well.

You need to move the scanner to the next line. Add this line of code below the error message in the catch section.
input.nextLine();

How can I use hasNextInt() to catch an exception? I need Int but if input is character, that is bad

I have been trying to stop the exceptions but I cannot figure out how.
I tried parseInt, java.util.NormalExceptionMismatch etc.
Does anyone have any insight how to fix this problem? Formatting is a bit off due to copy and paste.
do
{
System.out.print(
"How many integers shall we compare? (Enter a positive integer):");
select = intFind.nextInt();
if (!intFind.hasNextInt())
intFind.next();
{
// Display the following text in the event of an invalid input
System.out.println("Invalid input!");
}
}while(select < 0)
Other methods I have tried :
do
{
System.out.print(
"How many integers shall we compare? (Enter a positive integer):");
select = intFind.nextInt();
{
try{
select = intFind.nextInt();
}catch (java.util.InputMismatchException e)
{
// Display the following text in the event of an invalid input
System.out.println("Invalid input!");
return;
}
}
}while(select < 0)

It seems to me that you want to skip everything until you get an integer. This code here skips any input except an integer.
As long as there is no integer available (while (!in.hasNextInt())) discard the available input (in.next). When integer is available - read it (int num = in.nextInt();)
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (!in.hasNextInt()) {
in.next();
}
int num = in.nextInt();
System.out.println("Thank you for choosing " + num + " today.");
}
}

Quick sample of how to catch exceptions:
int exceptionSample()
{
int num = 0;
boolean done = false;
while(!done)
{
// prompt for input
// inputStr = read input
try {
num = Integer.parseInt(inputStr);
done = true;
}
catch(NumberFormatException ex) {
// Error msg
}
}
return num;
}

IMO, the best practice is to use nextLine() to get a String input, then parseInt it to get the integer. If unparsable, just complain back to the user and request re-entry.
Remember you may have to do a second nextLine() (discard the input) to clear up the buffer.

While Not Condition

How can I say the following:
while(input is not an int){
do this
}
I tried this code but I know it's wrong:
int identificationnumber;
Scanner sc3 = new Scanner(System.in);
identificationnumber = sc3.nextInt();
while( identificationnumber != int){ // this line is wrong
Scanner sc4 = new Scanner(System.in);
identificationnumber = sc4.nextInt();
}
Any suggestions please.Thanks.

Javadocs are your friend: http://docs.oracle.com/javase/6/docs/api/java/util/Scanner.html
nextInt() throws an exception if the next token isn't an int. You're probably looking for hasNextInt()
Also, why are you creating a new Scanner every time you loop? (Or at all - you already have one before the loop)

try :
while (! scanner.hasNextInt()) { // while the next token is not an int...
scanner.next(); // just skip it
}
int i = scanner.nextInt(); // then read the int

Scanner throws an exception before getting to that line
http://docs.oracle.com/javase/1.5.0/docs/api/java/util/Scanner.html#nextInt()
The following code will work:
int i = 0;
while(true){
Scanner scan = new Scanner(System.in);
try{
i = scan.nextInt();
}catch (Exception e) {
break;
}
}

You want this?
String identificationnumber;
Scanner scanner = new Scanner(System.in);//Only one Scanner is needed
while (scanner.hasNext()) { // Is there has next input?
identificationnumber = scanner.next();//Get next input
try {
Integer.parseInt(identificationnumber);//Try to parse to integer
System.out.println(identificationnumber + " is a number!");
} catch (NumberFormatException e) {
System.out.println(identificationnumber + " is not a number!");
}
}

By writing sc3.nextInt() I assume you always get an int back, so checking for a non int seems a bit strange.
Maybe its better to return a string with the number inside. If the string is empty stop (you can just check against "") and otherwise convert it to an integer.

Use nextInt() method of scanner class.
It throws,
InputMismatchException - if the next token does not match the
Integer regular expression, or is out of range

You should be doing this:
if (sc3.hasNextInt())
Check this out: How to use Scanner to accept only valid int as input
With regards to class/type comparison, read this: What is the difference between instanceof and Class.isAssignableFrom(...)?