I found this code online as a password loop game, it's working fine, but my question is: how?
What does happen in this catch block exactly?
I'm curious about this line specifically:
reader.next();
boolean loop = true;
Scanner reader = new Scanner(System.in);
System.out.println("PIN: ");
while (loop) {
try {
Integer Code = reader.nextInt();
if (Code == 8273) {
System.out.println("Access granted");
loop = false;
} else {
System.out.println("Access denied");
}
} catch (Exception e) {
System.out.println("Please enter a valid PIN!");
reader.next();
}
}
Edit : of course I did deliberately input a Non-integer input to cause the exception.
Edit2 :
When I removed that line, the program kept printing Please enter a valid PIN! For ever.
In fact, what the programmer really wanted here is to capture the next line of input, and verify whether that was a valid integer.
But the code is, admittedly, very confusing. And it relies on the fact that by default, when you "swallow" the next token with anything but .nextLine() with a Scanner, it relies on the current delimiter, which by default matches a newline.
Not good.
Here is a version which is more explicit:
String input;
int code;
while (true) {
System.out.print("PIN: ");
input = reader.nextLine();
try {
code = Integer.parseInt(input);
} catch (NumberFormatException ignored) {
// not an integer!
System.out.println("Enter a valid PIN!");
continue;
}
if (code == 8273)
break;
System.out.println("Access denied");
}
System.out.println("Access granted");
If nextInt throws an exception (because the value entered isn't an int), then the catch block is entered. The last line of which,
reader.next(); // <-- discards invalid token.
Removes the invalid token and then the loop iterates.
Also, don't box the Code1
int code = reader.nextInt();
1Using an Object type and then testing equality with == is a bad idea™. Also, by convention Java variable names start with a lower case letter.
The catch block simply catches the exception when anything other than an integer is entered. Since Code is an Integer, the input would have to be an integer. After catching the exception and printing the error, the reader moves to the next input until a proper value is entered, and the boolean loop becomes false, which ends the while loop at the end of the if statement once the correct value is entered.
Related
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.
I have a loop which breaks upon receiving the correct input from the console. I am using Scanner to read in a String from System.in, which seems to be what's giving me trouble. Here is my code:
boolean loop = true;
while(loop) {
try {
System.out.println("Enter an input (\"input a\" or \"input b\"): ");
String input = scanner.nextLine();
System.out.println("");
if (input.equals("input a")) {
System.out.println("Answer to input a.");
} else if (input.equals("input b")) {
System.out.println("Answer to input b.");
} else {
throw new IllegalArgumentException();
}
loop = false;
} catch (InputMismatchException e) {
System.out.println(e.getMessage());
System.out.println("");
} catch (IllegalArgumentException e) {
System.out.println("Input not recognized. Please enter a valid input.");
System.out.println("");
}
}
When this is called, it loops once without even waiting for input from the user, then actually stops and does what it is supposed to the second time around. IE, the output for this, without the user giving any input at all, is:
Enter an input ("input a" or "input b"):
Input not recognized. Please enter a valid input.
Enter an input ("input a" or "input b"):
If I give it a bad input (so that it loops and asks again), it does the same thing where it loops twice before waiting. I have no idea why.
Why is this happening, and what should I do to avoid it?
EDIT: Test scenarios after hasNext check:
Scenario A:
Enter an input ("input a" or "input b"): input a //my input
Input not recognized. Please enter a valid input.
Enter an input ("input a" or "input b"): //no input given here
Answer to input a.
Scenario B:
Enter an input ("input a" or "input b"): ddd //my input
Input not recognized. Please enter a valid input.
Enter an input ("input a" or "input b"): //no input given here
Input not recognized. Please enter a valid input.
Enter an input ("input a" or "input b"): input b //my input
Answer to input b.
The code which produces this:
boolean loop = true;
while(loop) {
if (scanner.hasNext()) {
try {
System.out.println("Enter an input (\"input a\" or \"input b\"): ");
String input = scanner.nextLine();
System.out.println("");
if (input.equals("input a")) {
System.out.println("Answer to input a.");
} else if (input.equals("input b")) {
System.out.println("Answer to input b.");
} else {
throw new IllegalArgumentException();
}
loop = false;
} catch (InputMismatchException e) {
System.out.println(e.getMessage());
System.out.println("");
} catch (IllegalArgumentException e) {
System.out.println("Input not recognized. Please enter a valid input.");
System.out.println("");
}
}
}
I tried out the code you've wrote down in your question, as a single question, but I can't find any problem with it at all. However, I think I know just the answer.
If you did any other input before this with primitive types or just next(), you need to flush the newline character that unfortunately those next methods leave behind. To do this, just call "scanner.nextLine()" before the statement where the method returns the inputted string and assigns it to input.
You want to make sure that you call in the method as separate; you know, on its own. It's best if you put it in your while loop at the top before you enter so that way for every iteration, the newline character is cleared. The statement will then get rid of the newline character and thus the input buffer is empty. Once that's cleared out, you can finally input your string at your first loop iteration.
How's that for an answer? Try it out and let me know if it works!
You should first check if the user has enetered any data :
if(scanner.hasNext())
{
// code logic
}
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.)
This piece of code is supposed to get an integer number from user and then finish the program. If the user inputs an invalid number, it asks user again.
After catching exception, it uses Scanner.reset() to reset the scanner, but it doesn't work. and it re-throws previous exception.
Scanner in = new Scanner(System.in);
while (true) {
try {
System.out.print("Enter an integer number: ");
long i = in.nextLong();
System.out.print("Thanks, you entered: ");
System.out.println(i);
break;
} catch (InputMismatchException ex) {
System.out.println("Error in your input");
in.reset(); // <----------------------------- [The reset is here]
}
}
I thought Scanner.reset() will reset everything and forget the exception. I put it before asking the user for a new input.
If I get the point wrong, what is the right way?
You misunderstood the purpose of the reset method: it is there to reset the "metadata" associated with the scanner - its whitespace, delimiter characters, and so on. It does not change the state of its input, so it would not achieve what you are looking for.
What you need is a call of next(), which reads and discards any String from the Scanner:
try {
System.out.print("Enter an integer number: ");
long i = in.nextLong();
System.out.print("Thanks, you entered: ");
System.out.println(i);
break;
} catch (InputMismatchException ex) {
System.out.println("Error in your input");
in.next(); // Read and discard whatever string the user has entered
}
Relying upon exceptions to catch exceptional situations is OK, but an even better approach to the same issue would be using has... methods before calling the next... methods, like this:
System.out.print("Enter an integer number: ");
if (!in.hasNextLong()) {
in.next();
continue;
}
long i = in.nextLong();
System.out.print("Thanks, you entered: ");
System.out.println(i);
break;
Per Scanner.reset() javadoc, the method only "resets" locale, radix and delimiter settings. It does not do anything to the data it already read.
I am using the following code:
while (invalidInput)
{
// ask the user to specify a number to update the times by
System.out.print("Specify an integer between 0 and 5: ");
if (in.hasNextInt())
{
// get the update value
updateValue = in.nextInt();
// check to see if it was within range
if (updateValue >= 0 && updateValue <= 5)
{
invalidInput = false;
}
else
{
System.out.println("You have not entered a number between 0 and 5. Try again.");
}
} else
{
System.out.println("You have entered an invalid input. Try again.");
}
}
However, if I enter a 'w' it will tell me "You have entered invalid input. Try Again." and then it will go into an infinite loop showing the text "Specify an integer between 0 and 5: You have entered an invalid input. Try again."
Why is this happening? Isn't the program supposed to wait for the user to input and press enter each time it reaches the statement:
if (in.hasNextInt())
In your last else block, you need to clear the 'w' or other invalid input from the Scanner. You can do this by calling next() on the Scanner and ignoring its return value to throw away that invalid input, as follows:
else
{
System.out.println("You have entered an invalid input. Try again.");
in.next();
}
The problem was that you did not advance the Scanner past the problematic input. From hasNextInt() documentation:
Returns true if the next token in this scanner's input can be interpreted as an int value in the default radix using the nextInt() method. The scanner does not advance past any input.
This is true of all hasNextXXX() methods: they return true or false, without advancing the Scanner.
Here's a snippet to illustrate the problem:
String input = "1 2 3 oops 4 5 6";
Scanner sc = new Scanner(input);
while (sc.hasNext()) {
if (sc.hasNextInt()) {
int num = sc.nextInt();
System.out.println("Got " + num);
} else {
System.out.println("int, please!");
//sc.next(); // uncomment to fix!
}
}
You will find that this program will go into an infinite loop, asking int, please! repeatedly.
If you uncomment the sc.next() statement, then it will make the Scanner go past the token that fails hasNextInt(). The program would then print:
Got 1
Got 2
Got 3
int, please!
Got 4
Got 5
Got 6
The fact that a failed hasNextXXX() check doesn't skip the input is intentional: it allows you to perform additional checks on that token if necessary. Here's an example to illustrate:
String input = " 1 true foo 2 false bar 3 ";
Scanner sc = new Scanner(input);
while (sc.hasNext()) {
if (sc.hasNextInt()) {
System.out.println("(int) " + sc.nextInt());
} else if (sc.hasNextBoolean()) {
System.out.println("(boolean) " + sc.nextBoolean());
} else {
System.out.println(sc.next());
}
}
If you run this program, it will output the following:
(int) 1
(boolean) true
foo
(int) 2
(boolean) false
bar
(int) 3
This statement by Ben S. about the non-blocking call is false:
Also, hasNextInt() does not block. It's the non-blocking check to see if a future next call could get input without blocking.
...although I do recognize that the documentation can easily be misread to give this opinion, and the name itself implies it is to be used for this purpose. The relevant quote, with emphasis added:
The next() and hasNext() methods and their primitive-type companion methods (such as nextInt() and hasNextInt()) first skip any input that matches the delimiter pattern, and then attempt to return the next token. Both hasNext and next methods may block waiting for further input. Whether a hasNext method blocks has no connection to whether or not its associated next method will block.
It is a subtle point, to be sure. Either saying "Both the hasNext and next methods", or "Both hasnext() and next()" would have implied that the companion methods would act differently. But seeing as they conform to the same naming convention (and the documentation, of course), it's reasonable to expect they act the same, and hasNext()
clearly says that it can block.
Meta note: this should probably be a comment to the incorrect post, but it seems that as a new user I can only post this answer (or edit the wiki which seems to be preferred for sytlistic changes, not those of substance).
Flag variables are too error prone to use. Use explicit loop control with comments instead. Also, hasNextInt() does not block. It's the non-blocking check to see if a future next call could get input without blocking. If you want to block, use the nextInt() method.
// Scanner that will read the integer
final Scanner in = new Scanner(System.in);
int inputInt;
do { // Loop until we have correct input
System.out.print("Specify an integer between 0 and 5: ");
try {
inputInt = in.nextInt(); // Blocks for user input
if (inputInt >= 0 && inputInt <= 5) {
break; // Got valid input, stop looping
} else {
System.out.println("You have not entered a number between 0 and 5. Try again.");
continue; // restart loop, wrong number
}
} catch (final InputMismatchException e) {
System.out.println("You have entered an invalid input. Try again.");
in.next(); // discard non-int input
continue; // restart loop, didn't get an integer input
}
} while (true);