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();
Related
I'm trying to ask the user for a number and if they enter anything wrong (not an int between 1 and 99) then catch (to prevent crash if string) and loop until enter a right number. My loop is stuck in an endless loop somehow. Note: I do have the Scanner imported and the exception imported too.
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String result;
int number;
boolean done = false;
while (true) {
try {
System.out.println("Please select a number from 1 to 99.");
number = input.nextInt();
input.nextLine();
if (number >= 1 || number <= 99) {
result = checkNumber(number);
System.out.println(result);
break;
}
} catch (InputMismatchException exception) {
}
}
}
input.nextInt() won't consume anything not an int. It will throw an exception. You ignore the exception and try to consume an int. It's still not an int. Same exception. Infinite loop. Add another input.nextLine() in your catch block.
This question already has answers here:
How to handle infinite loop caused by invalid input (InputMismatchException) using Scanner
(5 answers)
Closed last month.
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.
I have a WHILEloop which checks for marks for a particular student. However, it does not loop if the value is invalid (input less than 0 and more than 100):
int marks= -1;
System.out.print("Student Marks (/100): ");
while (((marks< 0) || (marks> 100))) {
try {
marks = Integer.parseInt(sc.nextLine());
break;
} catch (NumberFormatException nfe) {
System.err.println("Error: Invalid Mark(s)");
System.out.print("Student Marks (/100): ");
}
}
It does catches exception if characters other than numbers are entered.
But it does not loop again if value if less than 0 or more than 100.
I have tried making many changes to it but to no result.
Any help given is appreciated!
You should remove the break statement, since it breaks you out of the loop regardless of what value of marks was input.
Always use continue instead of break if you want to keep the loop running.
You may check the marks inside the while loop with an if condition and here you may use break -
import java.util.Scanner;
public class TakeInput{
public static void main(String args[]){
int marks= -1;
Scanner sc = new Scanner(System.in);
System.out.print("Student Marks (/100): ");
while (sc.hasNext()) {
try {
marks = Integer.parseInt(sc.nextLine());
if(marks<0 || marks>100){
break;
}
//do something
// with the marks
//take new marks
System.out.print("Student Marks (/100): ");
} catch (NumberFormatException nfe) {
System.err.println("Error: Invalid Mark(s)");
System.out.print("Student Marks (/100): ");
}
}
}
}
Now as long as you enter anything except a number n when n<0 || n>100 will continue the loop. Any NumberFormatExeption take you to the catch block.
If you enter 34 then it goes to the while block and prompt for a next number.
Then if you enter 56 then it dose the same thing.
When you enter any String rather than a number than it goes to the catch block
This process continues until you enter a invalid number (n>100 || n<100). Pressing Ctrl+C also exit you from the loop.
Hope it will help you.
Thanks a lot.
The code below asks the user how many racers he/she would like.
while (true) { // loops forever until break
try { // checks code for exceptions
System.out.println("How many racers should" + " participate in the race?");
amountRacers = in.nextInt();
break; // if no exceptions breaks out of loop
}
catch (InputMismatchException e) { // if an exception appears prints message below
System.err.println("Please enter a number! " + e.getMessage());
continue; // continues to loop if exception is found
}
}
If a number is entered at amoutnRacers = in.nextInt(); the code breaks out of the loop and the rest of the program runs fine; however, when I enter something such as "awredsf" it should catch that exception, which it does. Instead of prompting the user again it loops continuously, which to me does not make sense.
The program prints like this when looping continuously:
How many racers should participate in the race?
How many racers should participate in the race?
How many racers should participate in the race?
How many racers should participate in the race?
How many racers should participate in the race?
How many racers should participate in the race?
How many racers should participate in the race?Please enter a number! null
Please enter a number! null
Please enter a number! null
Please enter a number! null
Please enter a number! null
Please enter a number! null
Please enter a number! null
...
I do not understand what is going on amountRacers = in.nextInt(); so why is the user not able to enter a number?
Just add input.next() once you catch InputMismatchException.
catch (InputMismatchException e) { //if an exception appears prints message below
System.err.println("Please enter a number! " + e.getMessage());
input.next(); // clear scanner wrong input
continue; // continues to loop if exception is found
}
You need to clear the wrong input, which scanner automatically does not.
Today i solved this problem :-) This is my code. I think that i help
public int choice () throws Exception{
Scanner read = new Scanner(System.in));
System.out.println("Choose the option from the upper list");
int auxiliaryChoiceMenu = 5;
int auxiliaryVariable = -1;
boolean auxiliaryBoolean = false;
while (!auxiliaryBoolean) {
try {
auxiliaryVariable = read.nextInt();
read.nextLine();
} catch (Exception e) {
System.out.println("incorrect data, try again"+e);
read.nextLine();
continue;
}
if (auxiliaryVariable<0 || auxiliaryVariable>auxiliaryChoiceMenu){
System.out.println("incorrect data, try again");
} else {
auxiliaryBoolean = true;
}
choiceMenu = auxiliaryVariable;
}
return choiceMenu;
//choicemenu is a external variable
}
You may need to create a Scanner class for getting standard input streamed from the keyboard. You should have a statement somewhere in your code that creates an instance of a Scanner class like: Scanner in = new Scanner(System.in);
so the " in " variable in your statement: amountRacers = in.nextInt(); waits and scans for entered input from the keyboard and stores it.
Why use a loop with a try and catch?
My advice would be to always use a try and catch with either a while or do while loop, so you can ask the user to repeat his/her input. It also depends which loop you already use and/or on how your code is structured.
For example if you already have a do while loop then I would advice you to simply adjust/modify your existing loop.
I will post some examples on how you can use a try and catch with a loop to repeat the input after a user has provided a wrong one.
See examples below:
Example 1
Scanner input = new Scanner(System.in);
int exampleInput = 0;
do {
try {
System.out.print("\nEnter an integer from 1 to 25: ");
exampleInput = input.nextInt();
}
catch (InputMismatchException e) { //if an exception appears prints message below
System.err.println("Wrong input! Enter an integer from 1 to 25");
input.next(); // Clear scanner buffer of wrong input
}
} while (exampleInput < 1 || exampleInput > 25);
System.out.println("Print exampleInput: " + exampleInput);
Example 2
Scanner input = new Scanner(System.in);
int exampleInput; // Here you don't need to initialize this variable because you don't need it as a condition for the loop.
boolean isDone = false;
do {
try {
System.out.print("\nEnter an integer: ");
exampleInput = input.nextInt();
isDone = true;
}
catch (InputMismatchException e) { //if an exception appears prints message below
System.err.println("Wrong input! Enter an integer");
input.next(); // Clear scanner buffer of wrong input
}
} while (!isDone);
System.out.println("Print exampleInput: " + exampleInput);
Example 3
Scanner input = new Scanner(System.in);
int exampleInput; // Here you don't need to initialize this variable because you don't need it as a condition for the loop.
boolean isDoneLoop2 = false;
while (!isDoneLoop2) {
try {
System.out.print("\nEnter an integer: ");
exampleInput = input.nextInt();
isDoneLoop2 = true;
}
catch (InputMismatchException e) { //if an exception appears prints message below
System.err.println("Wrong input! Enter an integer");
input.next(); // Clear scanner buffer of wrong input
}
}
System.out.println("Print exampleInput: " + exampleInput);
This works for me.
while (true) {
try {
System.out.print("Ingrese la cantidad de puestos de atención: ");
int puestos = Integer.parseInt(scn.nextLine());
break;
}
catch (NumberFormatException e) {
System.out.println("Ingrese un valor correcto");
scn.reset();
continue;
}
}
This question already has answers here:
How to handle infinite loop caused by invalid input (InputMismatchException) using Scanner
(5 answers)
Closed last month.
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.