Develop Reference

NumberFormatException in try catch recursion doesn't work after 2 or more "mistakes"

So I am trying to write a method that checks if scanner input is an int, and loops errormessage until the user inputs an int. The method below works aslong as the usser doesn't give more than 1 wrong input. If I type muliple letters and then an int, the program will crash. I think it might have something to do with my try catch only catching 1 exception but not sure, and cant seem to get it to work. Does anyone know how I can fix this?
calling on method:
System.out.println("Write the street number of the sender: ");
int senderStreetNumber = checkInt(sc.nextLine);
method:
public static int checkInt (String value){
Scanner sc = new Scanner(System.in);
try{
Integer.parseInt(value);
} catch(NumberFormatException nfe) {
System.out.println("ERROR! Please enter a number.");
value = sc.nextLine();
checkInt(value);
}
int convertedValue = Integer.parseInt(value);
return convertedValue;
}

Something like this. Did not code it in an IDE, just from brain to keyboard.
Hope it helps. Patrick
Scanner sc = new Scanner(System.in);
int senderStreetNumber;
boolean ok = false;
while(!ok) {
System.out.println("Write the street number of the sender: ");
try {
senderStreetNumber = Integer.parseInt(sc.nextLine());
ok = true;
} catch (NumberFormatException nfe) {
System.out.println("ERROR! Please enter a number.");
}
}

Your logic of recursive is not good.
Let me try to explain your error...
The first time you get in the function you "check if the value is a int)
if not you to recurcive.
Lets say the second time is good.
then you cto the converted value
then the recurvice kicks in and you come back to the first time you get in the fucntion.
Then it does again the converted value and you don't catch that Exception so your application crash

This works.., just modified your program..tested
public static int checkInt(String value) {
Scanner sc = new Scanner(System.in);
try {
return Integer.parseInt(value);
}catch (Exception e) {
System.out.println("Error please enter correct..");
value = sc.nextLine();
return checkInt(value);
}
//int convertedValue = Integer.parseInt(value);
//return convertedValue;
}

About some simple exception handling in Java

There are several questions I would like to ask, please refer the comment part I have added in the code, Thanks.
package test;
import java.util.InputMismatchException;
import java.util.Scanner;
public class Test {
/* Task:
prompt user to read two integers and display the sum. prompt user to read the number again if the input is incorrect */
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean accept_a = false;
boolean accept_b = false;
int a;
int b;
while (accept_a == false) {
try {
System.out.print("Input A: ");
a = input.nextInt(); /* 1. Let's enter "abc" to trigger the exception handling part first*/
accept_a = true;
} catch (InputMismatchException ex) {
System.out.println("Input is Wrong");
input.nextLine(); /* 2. I am still not familiar with nextLine() parameter after reading the java manual, would you mind to explain more? All I want to do is "Clear Scanner Buffer" so it wont loop for the println and ask user to input A again, is it a correct way to do it? */
}
}
while (accept_b == false) {
try {
System.out.print("Input B: ");
b = input.nextInt();
accept_b = true;
} catch (InputMismatchException ex) { /*3. Since this is similar to the above situation, is it possible to reuse the try-catch block to handling b (or even more input like c d e...) exception? */
System.out.println("Input is Wrong");
input.nextLine();
}
}
System.out.println("The sum is " + (a + b)); /* 4. Why a & b is not found?*/
}
}

I am still not familiar with nextLine() parameter after reading the java manual, would you mind to explain more? All I want to do is "Clear Scanner Buffer" so it wont loop for the println and ask user to input A again, is it a correct way to do it?
The use of input.nextLine(); after input.nextInt(); is to clear the remaining content from the input stream, as (at least) the new line character is still in the buffer, leaving the contents in the buffer will cause input.nextInt(); to continue throwing an Exception if it's no cleared first
Since this is similar to the above situation, is it possible to reuse the try-catch block to handling b (or even more input like c d e...) exception?
You could, but what happens if input b is wrong? Do you ask the user to re-enter input a? What happens if you have 100 inputs and they get the last one wrong?You'd actually be better off writing a method which did this for, that is, one which prompted the user for a value and returned that value
For example...
public int promptForIntValue(String prompt) {
int value = -1;
boolean accepted = false;
do {
try {
System.out.print(prompt);
value = input.nextInt();
accepted = true;
} catch (InputMismatchException ex) {
System.out.println("Input is Wrong");
input.nextLine();
}
} while (!accepted);
return value;
}
Why a & b is not found?
Because they've not been initialised and the compiler can not be sure that they have a valid value...
Try changing it something more like.
int a = 0;
int b = 0;

Yes, it's okay. And will consume the non-integer input.
Yes. If we extract it to a method.
Because the compiler believes they might not be initialized.
Let's simplify and extract a method,
private static int readInt(String name, Scanner input) {
while (true) {
try {
System.out.printf("Input %s: ", name);
return input.nextInt();
} catch (InputMismatchException ex) {
System.out.printf("Input %s is Wrong%n", input.nextLine());
}
}
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int a = readInt("A", input);
int b = readInt("B", input);
System.out.println("The sum is " + (a + b));
}

I have put comment to that question line.
package test;
import java.util.InputMismatchException;
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean accept_a = false;
boolean accept_b = false;
int a=0;
int b=0;
System.out.print("Input A: ");
while (accept_a == false) {
try {
a = input.nextInt(); // it looks for integer token otherwise exception
accept_a = true;
} catch (InputMismatchException ex) {
System.out.println("Input is Wrong");
input.next(); // Move to next other wise exception // you can use hasNextInt()
}
}
System.out.print("Input B: ");
while (accept_b == false) {
try {
b = input.nextInt();
accept_b = true;
} catch (InputMismatchException ex) {
System.out.println("Input is Wrong");
input.next();
}
}
System.out.println("The sum is " + (a + b)); // complier doesn't know wheather they have initialised or not because of try-catch blocks. so explicitly initialised them.
}
}

Check out this "nextLine() after nextInt()"
and initialize the variable a and b to zero
nextInt() method does not read the last newline character.

Java - try/catch infinite repeat

Requirement:
Accept 10 numbers, input them into an array and then invoke a method to calculate and return the smallest. This program is suppose to be error proof so when a user enters an invalid entry, it notifies the user and reprompts. I am trying to use try catch but when an invalid entry is entered, ie a character, the scanner won't reprompt.
Any ideas?
Tried:
//Variables
double [] doubleArray = new double[10];
Scanner input = new Scanner(System.in);
//Prompt
System.out.println("This program will prompt for 10 numbers and display the smallest of the group");
//Get values
for (int i = 0; i < doubleArray.length; i++) {
try {
System.out.println("Please enter entry "+ (i+1));
doubleArray[i] = input.nextDouble();
} catch (InputMismatchException e) {
// TODO: handle exception
System.out.println("Please enter a rational number");
i--;
}
}
//Invoke method and display result
System.out.println("The smallest value is: "+index(doubleArray));

I don't see any call to input.nextLine(), which means nothing is ever consuming the \n entered by the user. There's a good example on scanner.nextLine usage here. If you add a call to it in your catch block, you should be all set.

Try calling input.nextLine(); in your catch. Then the \n will be taken from the input which let's you enter the next new number.
for(int i = 0; i < doubleArray.length; ++i) {
try {
doubleArray[i] = input.nextDouble();
} catch(Exception e) {
input.nextLine();
--i;
}
}

Try something like (and make sure you consume the whole line unless you want to allow multiple numbers to be input on the same line
boolean validEntry = false;
System.out.println("Enter a rational number: ");
while (!validEnry) {
try {
double value = input.nextDouble();
validEntry = true;
doubleArray[i] = value;
} catch (Exception e) {
System.out.println("Entry invalid, please enter a rational number");
}
}
...

You have to discard the false inputted data, add input.nextLine() in the catch block.

Java: Beginner using Scanner

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.

Java stuck in an infinite loop when using Try/Catch

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