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.
Related
I coded a program, that calculates the gcd (greatest common divisor) and lcm (least common multiple). Everything works fine except the try {...} catch(...) {...}. Here is the part of the code that doesn't work as I want it to:
try {
num1 = Integer.parseInt(sc.nextLine());
}
catch(Exception e) {
System.out.println("Your input is not an integer (number w/o decimals)! Try again.");
System.out.print("Enter your first number: ");
num1 = Integer.parseInt(sc.nextLine());
}
When I input e.g. letters, it says:
Your input is not an integer (number w/o decimals)! Try again.
Enter your first number:
But when I type letters the second time, the program crashes:
Exception in thread "main" java.lang.NumberFormatException: For input string: "asdf"
at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:68)
at java.base/java.lang.Integer.parseInt(Integer.java:658)
at java.base/java.lang.Integer.parseInt(Integer.java:776)
at GCDLCMgetter.main(GCDLCMgetter.java:56)
It is probably a very simple mistake I made but I can't figure it out...
Thank you
Your second parseInt method call is not in try catch block. You need to use a loop for this kind of logic.
It's because your second prompt is inside the catch block. Instead of prompting again inside the catch block, you want to wrap the entire code section in a loop so it comes back around to the try block for the prompt again.
Something like:
boolean repeat = true;
while(repeat){
try{
//Prompt for input
repeat = false;
}catch(Exception e) {
//Display error message
}
}
When the first time you give letters it goes into catch block. Displays the error message. And then executes the line num1 = Integer.parseInt(sc.nextLine());
Again you have entered the letters, but this time there is no try-catch block to handle this. So it throws error.
In your code, it gets executed twice:
Reads a Line at try{...}
There is an Exception
The Exception is handled by the catch(Exception e){...}
The num1 = Integer.parseInt(sc.nextLine()); inside the catch(Exception e){...} cannot be handled. It's not placed inside try{}.
Execution finished because the last exception cannot be handled by any catch.
It seems you're using Scanner, I would recommend you using a loop whis way:
while (sc.hasNextLine()){
try {
System.out.print("Enter your first number: ");
num1 = Integer.parseInt(sc.nextLine());
}
catch(Exception e) {
System.out.println("Your input is not an integer (number w/o decimals)! Try again.");
}
}
If you're managing integers, it would be interesting using Scanner.nextInt()
while (sc.hasNextInt()){
try {
System.out.print("Enter your first number: ");
num1 = sc.nextInt());
}
catch(Exception e) {
System.out.println("Your input is not an integer (number w/o decimals)! Try again.");
}
}
I have read user input that must be only of type int, the problem comes when the user enters letter instead of a int. I know how to handle the exception, but I would like to return the scanner read where the user has made a mistake. How can I do?
I already tried with an infinite loop, but it does not work.
try{
System.out.print("enter number: ");
value = scanner.nextInt();
}catch(InputMismatchException e){
System.err.println("enter a number!");
}
While other answers give you correct idea to use loop you should avoid using exceptions as part of your basic logic. Instead you can use hasNextInt from Scanner to check if user passed integer.
System.out.print("enter number: ");
while (!scanner.hasNextInt()) {
scanner.nextLine();// consume incorrect values from entire line
//or
//tastiera.next(); //consume only one invalid token
System.out.print("enter number!: ");
}
// here we are sure that user passed integer
int value = scanner.nextInt();
A loop is the right idea. You just need to mark a success and carry on:
boolean inputOK = false;
while (!inputOK) {
try{
System.out.print("enter number: ");
numAb = tastiera.nextInt();
// we only reach this line if an exception was NOT thrown
inputOK = true;
} catch(InputMismatchException e) {
// If tastiera.nextInt() throws an exception, we need to clean the buffer
tastiera.nextLine();
}
}
I have read user input that must be only of type int, the problem comes when the user enters letter instead of a int. I know how to handle the exception, but I would like to return the scanner read where the user has made a mistake. How can I do?
I already tried with an infinite loop, but it does not work.
try{
System.out.print("enter number: ");
value = scanner.nextInt();
}catch(InputMismatchException e){
System.err.println("enter a number!");
}
While other answers give you correct idea to use loop you should avoid using exceptions as part of your basic logic. Instead you can use hasNextInt from Scanner to check if user passed integer.
System.out.print("enter number: ");
while (!scanner.hasNextInt()) {
scanner.nextLine();// consume incorrect values from entire line
//or
//tastiera.next(); //consume only one invalid token
System.out.print("enter number!: ");
}
// here we are sure that user passed integer
int value = scanner.nextInt();
A loop is the right idea. You just need to mark a success and carry on:
boolean inputOK = false;
while (!inputOK) {
try{
System.out.print("enter number: ");
numAb = tastiera.nextInt();
// we only reach this line if an exception was NOT thrown
inputOK = true;
} catch(InputMismatchException e) {
// If tastiera.nextInt() throws an exception, we need to clean the buffer
tastiera.nextLine();
}
}
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
}