I want to take an integer input from the user. I am using a try-catch exception so that if the input is not an int it asks the user for another input. I have written the following code:
while (flag==true) {
try {
System.out.println("Enter the encryption key!");
k = input.nextInt();
flag = false;
}
catch (InputMismatchException ex) {
System.out.println("Please enter integer value!(");
}
}
But when an exception occurs the loop keeps on printing
"Enter the encryption key!"
"Please enter integer value!"
infinitely because the scanner input does not wait to take another input.
How can I tackle with this problem?
You could try to use
input.nextLine()
because then it waits until you press enter. But then you need to parse it to an int if i remember correctly.
You should call next inside the catch clause, this will solve the problem.
See the docs - 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.
Without having next in the catch clause, input.nextInt(); will keep reading the same token, adding next will consume that token.
You should go with suggestion of Maroun Maroun that looks more clean to me.
One other way is to re-initialize the Scanner in loop.
boolean flag = false;
while (!flag) {
try {
input = new Scanner(System.in);//Initialize it here
//....
Related
I am creating a program that takes input from the user.Now I want to re-run the input option if a user enters an invalid command. Like here only integer input are preferred but if the user inputs text, then I want to catch the exception and rerun the method.
I have tried this but it turns me into an infinite loop.What would be the best way to achieve this. It would be great if someone helps me on this.
boolean retry = true;
while(retry){
try{
//considering response from the user
System.out.print("\nEnter Your Command : ");
f_continue = processresponse(scanner.nextInt());
retry = false;
}catch(InputMismatchException e){
System.err.println("\nPlease Input a Valid Command from 1-5 or 0 for Exit");
}
}
If you enter something other than an integer, nextInt() will throw an exception, and what was entered will still be in the input buffer, so the next time you call nextInt(), you will read the same bad input. You should first call hasNextInt() to see if the input is an integer; if not, just call nextLine() to skip the bad input (and write a message telling the user to re-try).
Or you can just call scanner.nextLine() in the catch block.
I have a method that a wrote. This method just scans for a user entered integer input. If the user enters a character value it will throw an input mismatch exception, which is handled in my Try-Catch statement. The problem is that, if the user inputs anything that is not a number, and then an exception is thrown, I need the method to loop back around to ask the user for input again. To my understanding, a Try catch statement automatically loops back to the Try block if an error is caught. Is this not correct? Please advise.
Here is my method (it's pretty simple):
public static int getMask() {
//Prompt user to enter integer mask
Scanner keyboard = new Scanner(System.in);
int output = 0;
try{
System.out.print("Enter the encryption mask: ");
output = keyboard.nextInt();
}
catch(Exception e){
System.out.println("Please enter a number, mask must be numeric");
}
return output;
}//end of getMask method
Here is how the method is implemented into my program:
//get integer mask from user input
int mask = getMask();
System.out.println("TEMP mask Value is: " + mask);
/***********************************/
Here is my updated code. It creates an infinate loop that I can't escape. I don't understand why I am struggling with this so much. Please help.
public static int getMask() {
//Prompt user to enter integer mask
Scanner keyboard = new Scanner(System.in);
int output = 0;
boolean validInput = true;
do{
try {
System.out.print("Enter the encryption mask: ");
output = keyboard.nextInt();
validInput = true;
}
catch(InputMismatchException e){
System.out.println("Please enter a number, mask must be numeric");
validInput = false;
}
}while(!(validInput));
return output;
/********************/FINAL_ANSWER
I was able to get it finally. I think I just need to study boolean logic more. Sometimes it makes my head spin. Implementing the loop with an integer test worked fine. My own user error I suppose. Here is my final code working correctly with better exception handling. Let me know in the comments if you have any criticisms.
//get integer mask from user input
int repeat = 1;
int mask = 0;
do{
try{
mask = getMask();
repeat = 1;
}
catch(InputMismatchException e){
repeat = 0;
}
}while(repeat==0);
To my understanding, a Try catch statement automatically loops back to the Try block if an error is caught. Is this not correct?
No this is not correct, and I'm curious as to how you arrived at that understanding.
You have a few options. For example (this will not work as-is but let's talk about error handling first, then read below):
// Code for illustrative purposes but see comments on nextInt() below; this
// is not a working example as-is.
int output = 0;
while (true) {
try{
System.out.print("Enter the encryption mask: ");
output = keyboard.nextInt();
break;
}
catch(Exception e){
System.out.println("Please enter a number, mask must be numeric");
}
}
Among others; your choice of option usually depends on your preferred tastes (e.g. fge's answer is the same idea but slightly different), but in most cases is a direct reflection of what you are trying to do: "Keep asking until the user enters a valid number."
Note also, like fge mentioned, you should generally catch the tightest exception possible that you are prepared to handle. nextInt() throws a few different exceptions but your interest is specifically in an InputMismatchException. You are not prepared to handle, e.g., an IllegalStateException -- not to mention that it will make debugging/testing difficult if unexpected exceptions are thrown but your program pretends they are simply related to invalid input (and thus never notifies you that a different problem occurred).
Now, that said, Scanner.nextInt() has another issue here, where the token is left on the stream if it cannot be parsed as an integer. This will leave you stuck in a loop if you don't take that token off the stream. To that end you actually want to use either next() or nextLine(), so that the token is always consumed no matter what; then you can parse with Integer.parseInt(), e.g.:
int output = 0;
while (true) {
try{
System.out.print("Enter the encryption mask: ");
String response = keyboard.next(); // or nextLine(), depending on requirements
output = Integer.parseInt(response);
break;
}
catch(NumberFormatException e){ // <- note specific exception type
System.out.println("Please enter a number, mask must be numeric");
}
}
Note that this still directly reflects what you want to do: "Keep asking until the user enters a valid number, but consume the input no matter what they enter."
To my understanding, a Try catch statement automatically loops back to the Try block if an error is caught. Is this not correct?
It is indeed not correct. A try block will be executed only once.
You can use this to "work around" it (although JasonC's answer is more solid -- go with that):
boolean validInput = false;
while (!validInput) {
try {
System.out.print("Enter the encryption mask: ");
output = keyboard.nextInt();
validInput = true;
}
catch(Exception e) {
keyboard.nextLine(); // swallow token!
System.out.println("Please enter a number, mask must be numeric");
}
}
return output;
Further note: you should NOT be catching Exception but a more specific exception class.
As stated in the comments, try-catch -blocks don't loop. Use a for or while if you want looping.
This question already has answers here:
How to handle infinite loop caused by invalid input (InputMismatchException) using Scanner
(5 answers)
Closed 5 years ago.
Help, I am completely new to java and I am trying to create a loop that will ask for an input from the user which is to be a number. If the user enters anything other than a number I want to catch the exception and try again to get the correct input. I did this with a while loop however it does not give the opportunity after the error for the user to type in anything it loops everything else but that. Please help me to see understand what is wrong and the correct way to do this... Thank you. This is what I have:
import java.util.Scanner;
import java.util.InputMismatchException;
public class simpleExpressions {
public static void main (String[] args) {
Scanner keyboard = new Scanner(System.in);
while ( true ) {
double numOne;
System.out.println("Enter an Expression ");
try {
numOne = keyboard.nextInt();
break;
} catch (Exception E) {
System.out.println("Please input a number only!");
} //end catch
} //end while
} //end main
while ( true )
{
double numOne;
System.out.println("Enter an Expression ");
try {
numOne = keyboard.nextInt();
break;
}
catch (Exception E) {
System.out.println("Please input a number only!");
}
This suffers from several problems:
numOne hasn't been initialized in advance, so it will not be definitely assigned after the try-catch, so you won't be able to refer to it;
if you plan to use numOne after the loop, then you must declare it outside the loop's scope;
(your immediate problem) after an exception you don't call scanner.next() therefore you never consume the invalid token which didn't parse into an int. This makes your code enter an infinite loop upon first encountering invalid input.
Use keyboard.next(); or keyboard.nextLine() in the catch clause to consume invalid token that was left from nextInt.
When InputMismatchException is thrown Scanner is not moving to next token. Instead it gives us opportunity to handle that token using different, more appropriate method like: nextLong(), nextDouble(), nextBoolean().
But if you want to move to other token you need to let scanner read it without problems. To do so use method which can accept any data, like next() or nextLine(). Without it invalid token will not be consumed and in another iteration nextInt() will again try to handle same data throwing again InputMismatchException, causing the infinite loop.
See #MarkoTopolnik answer for details about other problems in your code.
You probably want to use a do...while loop in this case, because you always want to execute the code in the loop at least once.
int numOne;
boolean inputInvalid = true;
do {
System.out.println("Enter an expression.");
try {
numOne = keyboard.nextInt();
inputInvalid = false;
} catch (InputMismatchException ime) {
System.out.println("Please input a number only!");
keyboard.next(); // consume invalid token
}
} while(inputInvalid);
System.out.println("Number entered is " + numOne);
If an exception is thrown then the value of inputInvalid remains true and the loop keeps going around. If an exception is not thrown then inputInvalid becomes false and execution is allowed to leave the loop.
(Added a call to the Scanner next() method to consume the invalid token, based on the advice provided by other answers here.)
I am trying to run a simple code which will take an account number as the long type from the user through nextLong() method.Now if the user has given the account_number(long accNo) as string_type or character_type instead of long_type value then the InputMismatchException is thrown.That is also fine,because I am aware of the fact that nextLong() method can throw InputMismatchException along with NoSuchElementException and IllegalStateException.But after that when I am expecting that after getting InputMismatchException the loop will revisit and ask me to give the account_number(long accNo) as long value again! then the problem occurs.It is not asking me to give any value,instead an infinite loop is running along with the exception again & again.!!
package genericsandcollection;
import java.util.*;
public class ScannerTest {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc=new Scanner(System.in);
boolean b=true;
while(b){
try{
System.out.println("Enter Your Account Number..: ");
long accNo=sc.nextLong();
System.out.println(accNo);
b=false;
}catch(InputMismatchException | IllegalStateException e){
e.printStackTrace();
System.out.println(e);
System.out.println("Wrong syntax");
}
}
}
}
Why is it happening ? Is not nextLong() smart enough to handle a situation inside a loop ! If I use Long.parseLong(sc.next()) instead of sc.nextLong(),then everything is alright as parseLong() is throwing NumberFormatException i.e. until I will give the account number as long_type it keeps asking the user to give the account_number as long_type value.Really Strange !!!! If anybody has any concern,please help.Thank you.
Do not close the scanner in the catch block or you won't be able to use it again.
You shouldn't close the Scanner because as well as meaning you cannot use the Scanner anymore it will also close System.in. Your infinite loop is because using any other method than nextLine will leave a line break (from pressing 'enter' to submit) at the end of the input buffer. This will cause nextLong to always throw an exception.
Calling nextLine will advance the Scanner. The following is a short example that shows this (adapted from a very similar answer I wrote).
do {
try {
accNo = sc.nextLong();
break;
} catch (InputMismatchException e) {
} finally {
sc.nextLine(); // always advances (even after the break)
}
System.out.print("Input must be a number: ");
} while (true);
I have the following code which I'm using to get the user to input an integer, check to make sure the input is valid, and if not, ask for input again. When the code is run, everything works fine until some invalid input is given, at which point the code loops without pausing to ask for input again until a stack overflow occurs, and I have no idea why. The code:
//Create the scanner object
private static Scanner in = new Scanner(System.in);
//Function to get input in integer form, complete with exception handling
//A value of -1 means the user is done inputing
public static int getInt()
{
int num;
//Begin try block
try
{
//Get the user to input an integer
num = in.nextInt();
//Make sure the integer is positive. Throw an exception otherwise
if (num < -1)
throw new InputMismatchException();
}
//If an exception occurred during the inputting process, recursively call this function
catch (InputMismatchException e)
{
System.out.println("Error: Input must be a positive integer, or -1.");
System.out.print("Enter a score: ");
num = getInt();
}
//Return the inputed number
return num;
}
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
So sayeth the javadoc, i.e. the string that is not a number is not removed from the input automatically. Call in.next() manually to discard it.
You should call next(); to skip the invalid input before calling nextInt(); once more.
Is it necessary for the function to be recursive? Seems like this should be easy enough to do with a simple loop.
getInt() -> if (num<-1) throw Exception() -> catch (exception) -> num = getInt()
how do you expect NOT to loop ?