For a project in school, I am attempting to use the try/catch to prevent the program from crashing when the user enters a letter instead of the desired input type (i.e. a double).
public static double inputSide () {
Scanner in = new Scanner(System.in);
double side = -1;
do {
try {
System.out.println("Enter a side length (in units):");
side = in.nextDouble();
}
catch(InputMismatchException e){
System.out.println("Must input number");
}
} while (side < 0);
return side;
}
When I execute this code, it gets stuck in a loop where it outputs "Enter a side length (in units): " and "Must input number" infinitely. I am new to using try/catch, so I perhaps am simply unfamiliar with this behaviour. Anyway, if someone can help me figure out the problem, it would be much appreciated. Thanks in advance.
You have to free the buffer if you have a wrong input, in the catch block add this line:
in.next();
And everything should work:
public static double inputSide () {
Scanner in = new Scanner(System.in);
double side = -1;
do {
try {
System.out.println("Enter a side length (in units):");
side = in.nextDouble();
}
catch(InputMismatchException e){
System.out.println("Must input number");
//this line frees the buffer
in.next();
}
} while (side < 0);
return side;
}
In future, consider using
if(in.hasNextDouble()){
//read
}
instead of a try-catch block
Related
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 am trying to validate the input that decides whether or not the program does everything again or terminates. While it works enough that Y and N are the only accepted answers, it never does actually hit the catch statement - it only loops until Y or N are entered because of the while loop. I need to fix this so that the exception gets thrown for any other input, but I don't understand what I've done wrong.
public static String validateChoice(Scanner sc, String choice) {
choice = "";
while (!choice.equalsIgnoreCase("Y") && !choice.equalsIgnoreCase("N")) {
try {
System.out.print("Continue? Y/N: ");
choice = sc.nextLine();
} catch (InputMismatchException e) {
System.out.println("Invalid answer. Please answer only Y or N. Lowercase is accepted.");
sc.nextLine();
continue;
}
}
return choice;
}
Because there is no error here, catch statement cannot be reached. InputMismatchException is thrown only when the input type is incorrect. For example, you ask for nextInt() but the input is double. You are using nextLine(), so all input is accepted.
The problem is you are catching InputMismatchException which is never thrown by the code in try.
System.out.print("Continue? Y/N: ");
choice = sc.nextLine();
See this: https://docs.oracle.com/javase/10/docs/api/java/util/Scanner.html#nextLine()
So, to solve your problem, continue if user enters Y otherwise stop looping. It makes more sense to rely on a value to continue and consider everything else as NO.
if(Y){
continue;
} else { //any other value
break;
}
Full code:
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
do {
System.out.println("Do some stuff");
} while (continueCheck(sc));
System.out.println("Completed");
}
public static boolean continueCheck(Scanner sc) {
System.out.print("Continue? Y/N: ");
String choice = sc.nextLine();
return "Y".equalsIgnoreCase(choice.trim());//true only if Y or y is entered
}
I've copied part of the instructions below, and I can code pretty much every part on its own, but getting the control flow together is giving me massive doubts about my ability.
One of my biggest problems is the int gameChanger. Im supposed to immediately verify if it is a integer or not, and loop back if its not. But then Im also supposed to check to see if thebuser ever types "exit". But the input variable for my scanner instance is an integer... So Im stumped. I can use a try catch to check the missmatchexception once the input is being read in, but that doesnt solve the exit issue nor am I able to come up with solid logic to get the try catch to loop back if it indeed isnt an integer. Im thinking a do while loop but I havent gotten it to work.
Instructions:
You can whether the input is a number before attempting to consume it.
int num;
while (true) {
if (scanner.hasNextInt()) {
num = scanner.nextInt();
break;
} else {
// read whatever is there instead.
String line = scanner.nextLine();
if (line.equals("exit"))
System.exit(0);
System.out.println("Please enter a number");
}
}
System.out.println("Number entered " + num);
This gets the job done. Try it out.
import java.util.Scanner;
public class MyCode
{
public static void main(String[] args)
{
String gameInput = ".";
int gameNumber = 0;
boolean inputLoop = true;
Scanner input = new Scanner(System.in);
while(inputLoop == true)
{
try
{
System.out.print("Please enter a valid game number: ");
gameInput = input.next();
if(gameInput.equals("exit"))
{
System.out.println("Program will now end. Goodbye.");
inputLoop = false;
input.close();
}
gameNumber = Integer.parseInt(gameInput);
if(gameNumber >= 20001 && gameNumber <= 21230)
{
System.out.println("You have inputted a valid game number.");
inputLoop = false;
input.close();
}
}
catch(NumberFormatException e)
{
if(!gameInput.equals("exit"))
{
System.err.println("Invalid game number. Please try again.");
}
}
}
}
}
boolean loop = false;
double numberOfStudents;
System.out.print("Enter a number: ");
if ((scnr.nextLine().trim().isEmpty()) ) {
loop = true;
}
while (loop) {
System.out.println("Enter a number");
if (scnr.hasNextDouble() ){
System.out.println("Loop has stopped");
numberOfStudents = scnr.nextDouble();
loop = false;
}
}
System.out.println("You're outside the loop!");
I'm trying to get the program to say "Enter a number" until the user has entered an actual number (no white spaces or letters or signs). When the user has entered a number, it sets numberOfStudents equal to that number and breaks out of the loop.
But if you hit enter twice, it doesn't iterate. It only displays "Enter a number" once.
What is wrong with the loop logic? Why isn't it looping until valid input is taken?
For the actual answer to your question of "Why doesn't 'Enter a number' display more than once?" see Tom's comment (update: Tom's answer).
I've rewritten your loop in a way which preserves your code, but also makes it a little easier to handle format exceptions (though at the risk of silently swallowing an exception -- should be acceptable for this use case).
Can be up to you to use this design, here is an SO post on why empty catch blocks can be a bad practice.
public static void main(String args[])
{
boolean loop = true;
double numberOfStudents;
Scanner scnr = new Scanner(System.in);
while(loop){
System.out.print("Enter a number: ");
String input = scnr.nextLine();
try{
numberOfStudents = Double.parseDouble(input);
loop = false;
}catch(NumberFormatException e){
}
}
System.out.println("You're outside the loop!");
}
Output:
Enter a number:
Enter a number:
Enter a number:
Enter a number: 50
You're outside the loop!
First of all: Since you're reading from System.in a call to the input stream will block until the user entered a valid token.
So let's check first scan using your scnr variable:
scnr.nextLine()
nextLine() reads everything til the next line delimiter. So if you just press return, then it will successfully read it and will perform the next stuff.
The next call is:
scnr.hasNextDouble()
This call expects a "real" token and ignores white spaces, except as a delimiter between tokens. So if you just press return again it doesn't actually read that input. So it still waits for more (for the first token). That is why it stucks in your loop and you won't get another "Enter a number" output.
You can fix that by either enter a real token, like a number, or by changing the loop like trobbins said.
I hope you now understand your program flow a bit more :).
While trobbins code basically solves your problem, it's bad practice to use exceptions for flow control.
I used a small regexp to check if the value is a number. But this example is not complete, it will still crash it the user enters for example two decimal points. So you would need to create a proper number check or just use integers where the check is much easier.
Someone in the comments pointed out that people may want to enter scientific notation like 5e10, so this would also be another case to check for. If this is just some code you need as a proof of concept or something quick and dirty, you can go with the exception handling method but in production code you should avoid using exceptions this way.
double numberOfStudents;
Scanner scnr = new Scanner(System.in);
while(true) {
System.out.print("Enter a number: ");
String input = scnr.nextLine().trim();
if(input.matches("^[0-9\\.]{1,}$")) {
System.out.println("Loop has stopped");
numberOfStudents = Double.parseDouble(input);
break;
}
}
System.out.println("You're outside the loop!");
The following code should help you:
double numberOfStudents = 0;
Scanner scnr = new Scanner(System.in);
boolean readValue = false; //Check if the valid input is received
boolean shouldAskForNumber = true; //Need to ask for number again? Case for Enter
do {
if (shouldAskForNumber) {
System.out.print("Enter a number:");
shouldAskForNumber = false;
}
if (scnr.hasNextDouble()) {
numberOfStudents = scnr.nextDouble();
readValue = true;
} else {
String token = scnr.next();
if (!"".equals(token.trim())) { //Check for Enter or space
shouldAskForNumber = true;
}
}
} while (!readValue);
System.out.printf("Value read is %.0f\n", numberOfStudents);
System.out.println("You're outside the loop!");
Update
Understood the following statement in question different way:
But if you hit enter twice, it doesn't loop back. It only displays
"Enter a number" once.
The code is set to print "Enter a number" only once if the user hits RETURN/ENTER or enters space character. You may remove the special check and use the code if needed.
import java.util.Scanner;
public class Testing {
public static boolean checkInt(String s)
{
try
{
Integer.parseInt(s);
return true;
} catch (NumberFormatException ex)
{
return false;
}
}
public static void main(String[] args) {
boolean loop = false;
double numberOfStudents;
Scanner scnr = new Scanner(System.in);
String input = "";
while (!(checkInt(input))) {
System.out.println("Enter a number");
input = scnr.nextLine();
}
numberOfStudents = Integer.parseInt(input);
System.out.println("Number of students: " + numberOfStudents );
}
}
//this code is working fine, if you want you check it out.
//In your code your taking another input if the first is an int/double; if the first input is not a number then you have mentioned to take input again..
Use a debugger to see what the code is actually doing. Here's a guide on debugging in Eclipse. After you have finished debugging your code, you will probably know what the problem is.
Below code will help you
boolean loop = true;
double numberOfStudents;
Scanner scnr = new Scanner(System.in);
System.out.print("Enter a number: ");
String input = scnr.nextLine();
while(!scnr.hasNextDouble()){
System.out.print("Enter a number: ");
try{
numberOfStudents = Double.parseDouble(input);
break;
}catch(NumberFormatException e){
}
input = scnr.nextLine();
}
System.out.println("You're outside the loop!");
The following code is working,
boolean loop = true;
double numberOfStudents;
Scanner scnr=new Scanner(System.in);
while(loop) {
System.out.println("Enter a number");
if ((scnr.nextLine().trim().isEmpty()) ) {
loop = true;
}
if (scnr.hasNextDouble() ){
System.out.println("Loop has stopped");
numberOfStudents = scnr.nextDouble();
loop = false;
}
}
System.out.println("You're outside the loop!");
The output is,
run:
Enter a number
hj
po
Enter a number
lhf
Enter a number
o
Enter a number
p
Enter a number
a
Enter a number
34
Loop has stopped
You're outside the loop!
You have to scan the next line if you want to get more values form the scanner again. The code should be like:
while (loop) {
System.out.println("Enter a number");
if(!(scnr.nextLine().trim().isEmpty())){
if (scnr.hasNextDouble() ){
System.out.println("Loop has stopped");
numberOfStudents = scnr.nextDouble();
loop = false;
}
}
}
I am learning Java and am learning I/O w/ java.util.Scanner. Specifically I am learning Scanner methods.
import java.util.Scanner;
public class ScannerTest {
public static void main(String args[]) {
Scanner s = new Scanner(System.in);
int result;
while (s.hasNextInt()) {
result += s.nextInt();
}
System.out.println("The total is " + result);
}
}
Because you're checking only
while (s.hasNextInt())
You could use try catch to catch the exception (see documentation here) you get when the program quits, so you can show your error message in the catch block without making the program close.
Perhaps you should try parsing each line:
public static void main(String args[]){
int sum = 0;
final Scanner scanner = new Scanner(System.in);
System.out.println("Enter a series of integers. Press 'q' to quit.");
while(true){
final String line = scanner.nextLine();
if(line.equals("q"))
break;
try{
final int number = Integer.parseInt(line);
sum += number;
}catch(Exception ex){
System.err.printf("Invalid: %s | Try again\n", ex.getMessage());
}
}
System.out.printf("The sum is %,d" , sum);
}
The idea is to read input line by line and attempt parsing their input as an integer. If an exception is thrown (meaning they entered an invalid integer) it would throw an exception in which you could handle in what ever way you want to. In the sample above, you are simply printing the error message and prompting the user to type in another number.
You can do this for the while loop (not tested though):
while (s.hasNextLine()) {
String line = s.nextLine();
int parsedInteger;
try {
parsedInteger = Integer.parseInt(line);
} catch(NumberFormatException numEx) {
if(line.startsWith("q")) break;
else {
System.out.println("please enter valid integer or the character 'q'.");
continue;
}
}
result += parsedInteger;
}
s.close();
Instead of scanning for int's you can scan for lines and then then parse each line as an int. I feel the advantage of this approach is that if any of your int's are malformed you can then handle them appropriately by say displaying an error message to the user.
OR, based on the answer by pinckerman, you can also do this.
while (s.hasNextInt()) {
try {
result += s.nextInt();
} catch(InputMismatchException numEx) {
break;
}
}
s.close();
A smart way you can do it and I've tried before is you use Integer.parseInt(String toParse); This returns an int and will reject all non numerical chars.
while (scanner.hasNextInt()) {
int i = Integer.parseInt(scanner.nextInt());
result += i;
if (result < 2147483648 && result > -2147483648) {
try{
throw new IndexOutOfBoundsException();
catch (Exception e) {e.printStackTrace();}
}
Try the following way:
int result = input.nextInt;
this will define your variable for result.
The only problem in your code is "not initializing" result. Once you initialized code will work properly. However, please do not forget you need to tell compiler EOF. Compiler can only understand the stop of the input on the console by EOF. CTRL Z is EOF for windows Eclipse IDE.