Develop Reference

Stuck in Infinite loop, why it does not wait for scanner to re-enter the values [duplicate]

my question is short and sweet. I do not understand why my program infinitely loops when catching an error. I made a fresh try-catch statement but it looped and even copied, pasted and modified the appropriate variables from a previous program that worked. Below is the statement itself and below that will be the entire program. Thank you for your help!
try {
input = keyboard.nextInt();
}
catch(Exception e) {
System.out.println("Error: invalid input");
again = true;
}
if (input >0 && input <=10)
again = false;
}
Program:
public class Blanco {
public static int input;
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
nameInput();
}
/**
*
* #param name
*/
public static void nameInput() {
System.out.println("What is the name of the cartoon character : ");
Scanner keyboard = new Scanner(System.in);
CartoonStar star = new CartoonStar();
String name = keyboard.next();
star.setName(name);
typeInput(keyboard, star);
}
public static void typeInput(Scanner keyboard, CartoonStar star) {
boolean again = true;
while(again){
System.out.println("What is the cartoon character type: 1 = FOX,2 = CHICKEN,3 = RABBIT,4 = MOUSE,5 = DOG,\n"
+ "6 = CAT,7 = BIRD,8 = FISH,9 = DUCK,10 = RAT");
try {
input = keyboard.nextInt();
}
catch(Exception e) {
System.out.println("Error: invalid input");
again = true;
}
if (input >0 && input <=10)
again = false;
}
switch (input) {
case 1:
star.setType(CartoonType.FOX);
break;
case 2:
star.setType(CartoonType.CHICKEN);
break;
case 3:
star.setType(CartoonType.RABBIT);
break;
case 4:
star.setType(CartoonType.MOUSE);
break;
case 5:
star.setType(CartoonType.DOG);
break;
case 6:
star.setType(CartoonType.CAT);
break;
case 7:
star.setType(CartoonType.BIRD);
break;
case 8:
star.setType(CartoonType.FISH);
break;
case 9:
star.setType(CartoonType.DUCK);
break;
case 10:
star.setType(CartoonType.RAT);
break;
}
popularityNumber(keyboard, star);
}
public static void popularityNumber(Scanner keyboard, CartoonStar star) {
System.out.println("What is the cartoon popularity number?");
int popularity = keyboard.nextInt();
star.setPopularityIndex(popularity);
System.out.println(star.getName() + star.getType() + star.getPopularityIndex());
}
}

Your program runs forever because calling nextInt without changing the state of the scanner is going to cause an exception again and again: if the user did not enter an int, calling keyboard.nextInt() will not change what the scanner is looking at, so when you call keyboard.nextInt() in the next iteration, you'll get an exception.
You need to add some code to read the garbage the user entered after servicing an exception to fix this problem:
try {
...
} catch(Exception e) {
System.out.println("Error: invalid input:" + e.getMessage());
again = true;
keyboard.next(); // Ignore whatever is entered
}
Note: you do not need to rely on exceptions in this situation: rather than calling nextInt(), you could call hasNextInt(), and check if the scanner is looking at an integer or not.

How to switch order of output?

How do I switch the order of the outcome? I would liked to be asked to enter output after the menu.
Welcome to the Library! Please make a selection from the menu:
1. View.
2. Show.
Enter a choice: 1
However, I am made to enter input first, and I see:
Enter a choice: 1
Welcome to the Library! Please make a selection from the menu:
1. View.
2. Show.
This is my code:
import java.util.*;
public class Store {
public static void main(String[] args) {
new Store().use();
}
public void use() {
char choice;
while ((choice = readChoice()) != 'X') {
switch (choice) {
case 1: view(); break;
case 2: show(); break;
default: help(); break;
}
}
}
private char readChoice() {
return In.nextChar();
}
private String view() {
return "";
}
private String show() {
return "";
}
}
private void help() {
System.out.println("Welcome! Please make a selection from the menu:");
System.out.println("1. View.");
System.out.println("2. Show."); }

Just add help() on top of the use() method:
public void use() {
help();
char choice;
while ((choice = readChoice()) != 'X') {
switch (choice) {
case 1: view(); break;
case 2: show(); break;
default: help(); break;
}
}
}
You also need to change 1 and 2 into '1' and '2' respectively, because you are switching over a char. The fact that this compiles is because the compiler applies a narrowing primitive conversion to convert int to char.

I think there are three reasons that the code does not do what you expect.
You are made to enter input first, because before printing anything on the console, the readChoice() function is called. Which waits until it reads one character from the console and then returns. So you must call Help() function once before the while loop.
I guess the switch-case will not do what you expect. I mean the view() and show() functions are not called when you enter 1 or 2. The reason is that you read 1 and 2 as characters not integers. So the switch-case should change to this:
switch (choice) {
case '1': view(); break; //1 changed to '1'
case '2': show(); break; //2 changed to '2'
default: help(); break;
}
I think you might have forgotten to print "Enter a choice:" before reading the character. (I used System.out.print() rather than System.out.println() because it seems that "Enter a choice:" and the choice entered should be in the same line)
private char readChoice() {
System.out.print("Enter a choice:");
return In.nextChar();
}
this is the entire code, hope it works correctly (I put comments so you see what changes I made):
import java.util.*;
public class Store {
public static void main(String[] args) {
new Store().use();
}
public void use() {
char choice;
help();//called help() once before the loop
while ((choice = readChoice()) != 'X') {
switch (choice) { //cases changed
case '1': view(); break;
case '2': show(); break;
default: help(); break;
}
}
}
private char readChoice() {
//printing "Enter a choice:"
System.out.print("Enter a choice: ");
//I used Scanner class to read the next char, because I don't have 'In' class to use.
//you might write "return In.nextChar();" insead of the following lines
Scanner reader = new Scanner(System.in);
char c = reader.next().charAt(0);
return c;
}
private String view() {
System.out.println("you selected view"); //to see this function is called
return "";
}
private String show() {
System.out.println("you selected show"); //to see this function is called
return "";
}
private void help() {
System.out.println("Welcome! Please make a selection from the menu:");
System.out.println("1. View.");
System.out.println("2. Show.");
}
};

Java character option pane

I'm doing a basic calculator, I got confuse how to make this do a do while loop for it to execute again. Any suggestions? I'm just starting to learn Java so I'm a beginner. I only have trouble at the "Do you want to try again" Part. Is there a character parse like Integer.parseInt for character data types? or string data type? Thank you
public static void main(String[] args) {
Calc2 op = new Calc2();
Scanner scan = new Scanner(System.in);
char ans = 0;
do {
String c = JOptionPane.showInputDialog("Calculator\n" + "1.Addition\n" + "2.Subtraction\n" + "3.Multiplication\n" + "4.Division\n");
int n1 = Integer.parseInt(c);
switch (n1) {
case 1:
op.add();
break;
case 2:
op.diff();
break;
case 3:
op.prod();
break;
case 4:
op.quo();
break;
default:
JOptionPane.showMessageDialog(null, "Invalid Input", "Calculator", JOptionPane.PLAIN_MESSAGE);
break;
}
String s1 = JOptionPane.showInputDialog("Try again? [Y/N]");
int a = Integer.parseInt(s1);
} while (ans == 'y' || ans == 'Y');
}

You don't need to parse a character and should remove:
int a = Integer.parseInt(s1);
and change the following line:
} while (ans == 'y' || ans == 'Y');
to:
} while (JOptionPane.showInputDialog("Try again? [Y/N]").equalsIgnoreCase("y"));
and remove the unnecessary lines:
Scanner scan = new Scanner(System.in);
char ans = 0;

Loop back to program after an exception is thrown

public void runMenu() {
int x = 1;
Scanner Option = new Scanner (System.in);
int Choice = 0;
do {
try {
System.out.println("Choose Option");
System.out.println("");
System.out.println("1: Create Account");
System.out.println("2: Check Account");
System.out.println("3: Take Action");
System.out.println("4: Exit");
System.out.println("Please choose");
Choice = Option.nextInt();
switch (Choice) { //used switch statement instead of If else because more effective
case 1:
CreateAccount();
break; //breaks iteration
case 2:
selectAccount();
break;
case 3:
Menu();
int choice = UserInput();
performAction(choice);
break;
case 4:
System.out.println("Thanks for using the application");
System.exit(0);
default:
System.out.println("Invalid Entry");
throw new Exception();
}
} catch (Exception e) {
System.err.println("Enter Correct Input");
return;
}
} while (true);
}
I am trying to make it when users enter incorrect input type like a letter , the exception is caught and then returns back to the menus, right now it catches the exception but it doesnt stop running I have to force stop the program. So I added a return but that just displays the exception error and stops, how can I make it return back to the menus?

That is because you're returning from the method itself in the catch block.
And Do not throw exceptions like that. Just use some boolean to know if the choice is valid and loop until the choice is entered correctly.Prefer not to use while(true), instead rely on a boolean flag everytime like below,
public void runMenu() {
int x = 1;
Scanner Option = new Scanner (System.in);
int Choice = 0;
boolean isValidChoice = false;
do{
isValidChoice = false;
Choice = 0;
System.out.println("Choose Option");
System.out.println("");
System.out.println("1: Create Account");
System.out.println("2: Check Account");
System.out.println("3: Take Action");
System.out.println("4: Exit");
System.out.println("Please choose");
if(Option.hasNextInt()){
Choice= Option.nextInt();
isValidChoice = true;
}
switch (Choice)
{
case 1:
CreateAccount();
break;
case 2:
selectAccount();
break;
case 3:
Menu();
int choice = UserInput();
performAction(choice);
break;
case 4:
System.out.println("Thanks for using the application");
System.exit(0);
default:
isValidChoice = false; //if invalid choice, then set flag to loop
System.out.println("Invalid Entry");
}
} while (!isValidChoice);
}

Move the "try {" after the "System.out.println("Please choose");" line.

you just need to remove the return in the catch. also just as a tip, you can get rid of the do while and just have a while loop, because the loop is never ending.
} catch (Exception e) {
System.err.println("Enter Correct Input");
}

Okay so I'm pretty sure this should work:
Create a boolean value outside of while loop that is holds if there was a valid input
boolean validInput = true;
In default set this value to false (meaning there is an invalid input)
default:
System.out.println("Invalid Entry");
validInput = false;
throw new Exception();
Make sure the catch statement is still in the do loop because the throw clause will halt normal execution and transition into exception execution. Next the while tester will test if there was a valid input
while(!validInput)
Lastly go up to the top of the do loop and set validInput to true. This will make it so that each time you clear the previous incorrect input.
This should work.

.hasNextInt() in switch statement

I'm basically trying to validate so that you can only enter an Integer. This is what I have at the moment, but if I type letters it goes through the switch and just leaves the result as blank.
I want it so that if anything other than an integer is entered it will go to default in the switch.
Any help would be great. Thanks!
while(loop && kb.hasNextInt())
{
choice = kb.nextInt();
switch(choice)
{
case 1 :
language = "FRENCH";
loop = false;
break;
case 2 :
language = "GERMAN";
loop = false;
break;
case 3 :
language = "SPANISH";
loop = false;
break;
default :
System.out.println("That is not a correct choice. Please try again!");
break;
}
}

If the next input is not an integer,
then .hasNextInt() will return false,
and therefore the loop will terminate early.
If you want to allow text input and respond to it,
then you need to read line by line, text instead of numbers,
and parse the line read with Integer.parseInt.
If the line cannot be parsed, you will get a NumberFormatException.
You can catch it, and handle appropriately.
while (loop && scanner.hasNextLine()) {
String line = scanner.nextLine();
try {
choice = Integer.parseInt(line);
} catch (NumberFormatException e) {
System.out.println("That is not an integer. Please try again!");
continue;
}
switch (choice) {
case 1:
language = "FRENCH";
loop = false;
break;
case 2:
language = "GERMAN";
loop = false;
break;
case 3:
language = "SPANISH";
loop = false;
break;
default:
System.out.println("That is not a correct choice. Please try again!");
break;
}
}

This is because a letter will cause your while(loop && kb.hasNextInt()) to be false. I suggest put an if statement with the hasNextInt() within the while loop.
Example (using a while loop instead of if statement to really try getting the number):
while(loop)
{
// validate int using while loop
while(!kb.hasNextInt())
{
System.out.println("you must enter a number! ");
kb.next();
}
choice = kb.nextInt();
switch(choice)
{
case 1 :
language = "FRENCH";
loop = false;
break;
case 2 :
language = "GERMAN";
loop = false;
break;
case 3 :
language = "SPANISH";
loop = false;
break;
}
}
System.out.println("Thank You " + studentID + " you have been registered for " + language);

This code will blow before it even begins if the user did not enter a number as the while required kb.hasNextInt() to be true (have a number) to even run.
What I do is that I usually put the validation around where I receive the input:
int choice;
Boolean retry = null;
while(retry == null) {
try{
String input = scanner.nextLine();
choice = Integer.parseInt(input);
retry = false;
}catch(NumberFormatException e){
System.out.println("Please enter a number from 1 to 4.");
}
}
switch(choice){
case 1:
// Do stuff
break;
case 2:
// Do stuff
break;
case 3:
// Do stuff
break;
case 4:
// Do stuff
break;
default:
System.out.println("Something went wrong!");
}