I recently asked if there is any possible way whatsoever to get an exception from assigning a value to a String variable with the Scanner (The thread is here:)
And one of the guys told me that CTRL+D would be a case where a NoSuchElementException could be thrown. This to me is kind of a special case because input.nextLine() returns a String, and a String can be basically anything a user could type on the keyboard, so one would assume that input.nextLine() would not be a concern to throw an exception.
So I decided to add some try catch blocks into a program I'm writing on the off chance that CTRL+D is pressed when the program is asking for a number.
The problem I've run into is that when I catch the CTRL+D exception, the Scanner needs to be flushed, but if I flush the Scanner, it will cause a NoSuchElementException to occur because no new line exists. I'm using this all in a while true loop, so I'm kind of stuck between a rock and a hard place.
I will post one version of the code, with the input.nextLine() commented out. If you run it as is, you will get the infinite loop that happens when the Scanner needs to be flushed. If you uncomment the input.nextLine(), that very line of code will itself cause a NoSuchElementException.
import java.util.NoSuchElementException;
private int getMainOptions(){
System.out.printf("\n********** Main Options **********");
System.out.printf("\n*%32s*", "");
System.out.printf("\n* %-30s*", "[1] Create Customer");
System.out.printf("\n* %-30s*", "[2] Create Reservation");
System.out.printf("\n* %-30s*", "[3] Display Customer");
System.out.printf("\n* %-30s*", "[4] Display Reservation");
System.out.printf("\n*%32s*", "");
System.out.printf("\n**********************************");
while(true){
try{
System.out.print("\nChoose Option: ");
if(input.hasNextInt()){
return input.nextInt();
}
System.out.print("\nInvalid option");
input.nextLine();
continue;
}
catch(NoSuchElementException e){
System.out.print("\nAn exception occurred.");
//input.nextLine();
}
}
}
Apart from creating the Scanner inside the while loop and destoying it in the catch to be recreated in the next iteration, what can be done to solve this problem?
The code goes into an infinite loop when the input.nextLine() is commented out or when input.hasNextLine() is there to check because it is in a while(true) loop with nothing to stop it since input.nextInt() is not called. if(input.hasNextInt()) will not wait for an int, but simply skip the code inside the if statement if an int is not present as input.
Instead try this:
while(true){
try{
System.out.print("\nChoose Option: ");
String in=input.nextLine();
try{
int i=Integer.parseInt(in);
return i;
}catch(NumberFormatException ex)
{
System.out.print("\nInvalid option");
}
}
catch(NoSuchElementException e){
System.out.print("\nAn exception occurred.");
//input.nextLine();
}
}
}
I hope this helps and that I am understanding the question correctly
I'm not sure you fully grasp what happens when the user presses Ctrl-D. When that happens the standard input stream is closed. There is no way to reopen a closed stream. Even if you create a new Scanner and pass it in System.in it will still throw a NoSuchElementException.
As a Linux user if I press Ctrl-D in an interactive program, I expect the program to terminate. That is really all you can do at that point.
Related
I am making a basic application where it trains your math skills. I have this code:
while (true)
{
try
{
int userAnswer;
System.out.println("Type quit to exit to the menu!");
int randInt = r.nextInt(num2);
System.out.println(num1 + " + " + randInt + " =");
userAnswer = in.nextInt();
if(userAnswer == num1 + randInt) System.out.println("Correct!");
else System.out.println("Wrong!");
break;
}
catch(Exception e)
{
}
}
When someone prints out a d or something in the answer, the try catch goes. But, then it goes to the while loop and repeatedly spams Type quit to exit to the menu and then something like 1 + 2 = infinitely... I think I know what's wrong, userAnswer has been assigned already as something that throws an exception that goes to the catch and it just keeps printing those and goes to the catch and goes back because userAnswer is already assigned. I think this is what is happening, I could be wrong. Please help!
EDIT: I forgot to make this clear, but I want the question to be re-printed again, exiting out of the loop goes to a menu where you can't get the question back, I want it to redo what's in the try catch...
You should never catch an Exception without handling it.
catch(Exception e)
{
System.out.println("An error has occured");
break;
}
This should stop your program from looping infinitely if an Exception occurs.
If user input comes as letter it will get an exception because you are trying to read(parse) as integer. So your catch clause is in the loop you have to write break in there to go out from loop.
Still i will suggest you to getline as string and than compare with your cli commands (quit in your case) than you can try to parse it as an integer and handle loop logic.
You're not breaking the while loop if there is a mismatch
while(true)
{
try
{
}
catch(InputMisMatchException e)//I suggest you to use the exact exception to avoid others being ignored
{
System.out.println("Thank you!");
break;//breaks the while loop
}
}
Yoy're not breaking the loop in case of Exception occurs.
Add break; statement in the catch block to run your program without going to infinite loop, in case exception occurs.
Since the given answers don't match your requirement I'll solve that "riddle" for you.
I guess what you didn't knew is that the scanner won't read the next token if it doesn't match the expectation. So, if you call in.nextInt() and the next token is not a number, then the scanner will throw an InputMismatchException and keeps the reader position where it is. So if you try it again (due to the loop), then it will throw this exception again. To avoid this you have to consume the erroneous token:
catch (Exception e) {
// exception handling
in.next();
}
This will consume the bad token, so in.nextInt() can accept a new token. Also there is no need to add break here.
Mind that in.next() reads only one token, which is delimited by a whitespace. So if the user enters a b c, then your code will throw three exception and therefore generate three different question befor the user can enter a number. You can avoid that by using in.nextLine() instead. But this can lead into another problem: Scanner issue when using nextLine after nextXXX, so pay attention to that :).
I have a question regarding data validation and scanners.The following piece of code checks userinput.Anything other than an integer is not allowed and the user is asked to re-enter a value.My question is that the code works only if the scanner is declared within the while loop.The program executes infinitely if the scanner is declared outside.Why is that?Thanks.
int UserInp;
boolean dataType=false;
while(dataType==false)
{
Scanner sc=new Scanner(System.in);
try
{
System.out.print("\nEnter a number: ");
UserInp=sc.nextInt();
dataType=true;
}
catch(Exception JavaInputMismatch)
{
System.out.println("Option not available.Try again.");
}
}
Interesting problem!
What happens is that the Scanner attempts to translate the non-integer to an integer, and realizes it can't -- so it throws an InputMismatchException. However, it only advances past the token if the translation was successful.
Meaning, the invalid string is still in the input buffer, and it will fail the translation every single time you loop and try to call nextInt(). You never set dataType to true, and so you loop infinitely.
To see this in action, you can grab the arbitrary content in your catch block and print it out:
catch(Exception JavaInputMismatch){
System.out.println( sc.next() );
System.out.println("Option not available.Try again.");
}
Indeed, after invalid input, we get the following:
Enter a number: hello
hello
Option not available.Try again.
Enter a number:
And we don't loop infinitely. This is because the call to next() grabbed the value from the input buffer and advanced the scanner's pointer into that buffer to the next slot, which is now empty. So nextInt() will wait for input in that case.
Oh, and the reason it works fine if you initialize in the loop is that the scanner will always start reading input fresh; scanners don't share state across instances, so the "hello" that was in the buffer for the previous iteration isn't in the buffer for the next one due to the reinitialization.
Technically, it's still in the standard input buffer, but the scanner's pointer into that buffer is beyond the invalid string because it will start reading any new input, not existing input.
To add to Purag's answer, you could alternatively use nextLine() to advance the Scanner past the current line.
So your catch block will look like this:
catch(Exception JavaInputMismatch)
{
System.out.println("Option not available.Try again.");
sc.nextLine();
}
Tricky question.
You may get it!
The answer is simple. The Scanner object is kept live till the end of the execution as it is declared outside the while loop. Look this problem in the memory level.
The Scanner object is kept live so while entering the loop next time still the value(String value) will be there in Scanner object and it doesn't listens keyboard as the exception is already thrown.So the loop keeps going.
Note : The next() method in Scanner class will accept all the types of keyboard input but not the rest of the methods such as nextInt(), nextFloat() etc..,
I'm trying to read a number for a switch case option but I'm stuck with an exception. I will try to explain the problem better in code:
do{
try{
loop=false;
int op=teclado.nextInt();
//I tryed a teclado.nextLine() here cause i saw in other Q but didn't work
}
catch(InputMismatchException ex){
System.out.println("Invalid character. Try again.");
loop=true;//At the catch bolck i change the loop value
}
}while(loop);//When loop is true it instantly go to the catch part over and over again and never ask for an int again
When I type an int it works perfectly, but the exception makes it start over. The second time, the program does not ask for the int (I think it could be a buffer and I need something like fflush(stdin) in C), and the buffer just starts writing like crazy.
You would be well-served creating a new instance of Scanner from within the catch to get the input should you fail. EDIT: You can use a Scanner.nextLine() to advance past the newline character when you fail. A do...while loop may be inappropriate for this, since it guarantees that it will execute at least once.
A construct that may help you out more is a simple while loop. This is actually a while-true-break type of loop, which breaks on valid input.
while(true) {
try {
op=teclado.nextInt();
break;
} catch(InputMismatchException ex){
System.out.println("Invalid character. Try again.");
teclado.nextLine();
}
}
This question already has answers here:
try/catch with InputMismatchException creates infinite loop [duplicate]
(7 answers)
Closed 7 years ago.
I am writing a simple java console game. I use the scanner to read the input from the console. I am trying to verify that it I ask for an integer, I don't get an error if a letter is entered. I tried this:
boolean validResponce = false;
int choice = 0;
while (!validResponce)
{
try
{
choice = stdin.nextInt();
validResponce = true;
}
catch (java.util.InputMismatchException ex)
{
System.out.println("I did not understand what you said. Try again: ");
}
}
but it seems to create an infinite loop, just printing out the catch block. What am I doing wrong.
And yes, I am new to Java
nextInt() won't discard the mismatched output; the program will try to read it over and over again, failing each time. Use the hasNextInt() method to determine whether there's an int available to be read before calling nextInt().
Make sure that when you find something in the InputStream other than an integer you clear it with nextLine() because hasNextInt() also doesn't discard input, it just tests the next token in the input stream.
Try using
boolean isInValidResponse = true;
//then
while(isInValidResponse){
//makes more sense and is less confusing
try{
//let user know you are now asking for a number, don't just leave empty console
System.out.println("Please enter a number: ");
String lineEntered = stdin.nextLine(); //as suggested in accepted answer, it will allow you to exit console waiting for more integers from user
//test if user entered a number in that line
int number=Integer.parseInt(lineEntered);
System.out.println("You entered a number: "+number);
isInValidResponse = false;
}
//it tries to read integer from input, the exceptions should be either NumberFormatException, IOException or just Exception
catch (Exception e){
System.out.println("I did not understand what you said. Try again: ");
}
}
Because of common topic of avoiding negative conditionals https://blog.jetbrains.com/idea/2014/09/the-inspection-connection-issue-2/
This question already has answers here:
try/catch with InputMismatchException creates infinite loop [duplicate]
(7 answers)
Closed 7 years ago.
I keep getting my code caught in an infinite while loop.
It is nothing to advanced, but i can not figure it out for the life of me!
Someone Please help
I have purplosely just re created the specific error without all of the if statements i have in my actual program.
package bs;
import java.util.InputMismatchException;
import java.util.Scanner;
public class bs {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
boolean continueVar = true;
while (continueVar) {
try {
System.out.println("Enter Something");
int input = sc.nextInt();
} catch (InputMismatchException i) {
System.out.println("What the f***?");
continueVar = true;
}
}
}
}
The infinite loop occurs when the Input mismatch exception is caught. I would think that it would atleast ask the user to re enter their input but instead of doing that it just continues in the loop as so:
run:
Enter Something
df
What the f***?
Enter Something
What the f***?
Enter Something
What the f***?
It acts like it is just ignoring the scanner object sc?!
No the scanner is not skipped, it's just starting at the beginning of the input. From the JavaDoc:
If the translation is successful, the scanner advances past the input that matched.
This means if the conversion isn't successfull the scanner won't advance. You'd thus have to manually skip the incorrect input using just next().
Edit: you might want to check for hasNextInt() before trying to read the input.
you loop while continueVar is true, but you never set to to false, so the loop never exits.
I think you want to set continueVar to false in the exception handler.
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.
The token that caused the mismatch is still in the scanner's buffer. You need to clear it before trying to scan again.
You can do that by calling next() in your catch block, like this:
catch (InputMismatchException i) {
System.out.println("What the f***?");
sc.next();
}
Also, you don't need to set continueVar to true again. You never set it to false, so it will stay true. Guessing that's an artifact of you removing this into a mini program.
Scanner does not advance when bad token is found. Look at Scanner.java, lines 2095-2096:
catch (NumberFormatException nfe) {
position = matcher.start(); // don't skip bad token