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Java Method Example clarification

I apologize if this question is uber-simplistic, but I'm still in the early stages of learning Java. I have an example program that calls other methods within the class, and I'm not totally following a few of the elements - hoping someone can clarify. It's a simply random number guessing game and it works fine, but I want to better understand some components. Specifically:
There is a boolean variable (validInput) that is declared but never appears to be used anywhere in the methods
There are 2 methods (askForAnotherRound and getGuess) with a 'while' loop that just has 'true' as the variable(?) - i.e. "while (true)."
This code is directly from the example in the book and, again, it works. I just want to better understand those 2 elements above. (I think the validInput variable is not useful as when I 'comment out' that line the code still executes). I'm curious, though, about the "while (true)" element. There is an option to set, in the askForAnotherRound, to set the return value to false (ending the program). Are boolean methods defaulted to 'true' when they are first executed/called?
Again...understand this is probably a super-simple question for most folks on here, but as a newb I just want to understand this as best I can...
Thanks!!!
// Replicates the number guessing game using 4 separate methods.
import java.util.Scanner;
public class GuessingGameMethod2
{
static Scanner sc = new Scanner(System.in);
public static void main(String[] args)
{
System.out.println("Let's play a guessing game!");
do
{
playARound();
}while (askForAnotherRound());
System.out.println("Thanks for playing!");
}
public static void playARound()
{
boolean validInput;
int number, guess;
String answer;
//Pick a Random Number
number = getRandomNumber();
//Get a guess
System.out.println("\nI'm thinking of a number between 1 and 10.");
System.out.print("What do you think it is? ");
guess = getGuess();
//Check the guess
if (guess == number)
System.out.println("You're right!");
else
System.out.println("You're wrong! The number was " + number + ".");
}
public static int getRandomNumber()
{
return (int)(Math.random() * 10) + 1;
}
public static int getGuess()
{
while (true)
{
int guess = sc.nextInt();
if (guess < 1 || guess > 10)
{
System.out.print("I said, between 1 and 10. Try again");
}
else
return guess;
}
}
public static boolean askForAnotherRound()
{
while (true)
{
String answer;
System.out.print("\nPlay again? Y or N: ");
answer = sc.next();
if (answer.equalsIgnoreCase("Y"))
return true;
else if (answer.equalsIgnoreCase("N"))
return false;
}
}
}

I don't see boolean validInput being used either. But if it were to be used somewhere it would probably be to check that you guess satisfies 1 <= guess <= 10.
When it comes to askForAnotherRound and getGuess here's what you should know:
while(true) is always executed. One way you can get out of the while loop is if you use the break statement or if the loop is in a function you can return something. The method askForAnotherRound() will always return either true or false. Depending on the returned value of askForAnotherRound() you will either play another round or not. Note that when you have
`do{
...
someActions()
...
}while(booleanValue)`
someActions() will be executed at least once before it checks for the value of booleanValue which, if it turns out false you'll exit out of the do/while loop. Boolean methods do not default to anything, you have to give them a value.
Hope this helps! I'm also in the process of learning Java right now, so good luck!

As I see you're absolutely true about validInput - it isn't used. May be it will be used in the following chapters.
As for askForAnotherRound() - no, boolean methods don't evalute to true, by default. Even more, Java compiler throw an error if it find a method which does not return value and finish it execution in some cases.
while(true) - it's infinite loop, so it will be executed untill some instruction which interrupts loop, in general it's return statement.
askForAnotherRound() do the following:
- asks user if he/she want to play again
- returns true if user input "Y"
- returns false if user input "Y"
- asks again in all other cases(so it doesn't finish execution) and etc.
Hope it'll help

The validInput is indeed worthless.
The infinite loops are required to read from the console to get a valid input. e.g
while (true)
//start infinite loop
{
int guess = sc.nextInt();
if (guess < 1 || guess > 10)
{
//continue the loop the input is not between 1-10
System.out.print("I said, between 1 and 10. Try again");
}
else
//break out of infinite loop, valid int
return guess;
}
If we take this method without the infinite loop, and i recommend trying this, it will simply return the value read even if it was not valid.
For example.
return sc.nextInt();
will allow any int returned, if we returned anything outside of the bounds in the current impl it would loop again until you enter a value between 1-10
The same is also true for ask for next round, its looping until a valid input is given.
I would bet the next exercises will use the validInput var as both these methods loop until a valid input is given.

You are right about validInput. It is not used. And probably missed after some code change. Should be removed.
while(true) - true is not variable but a boolean constant. It will basically make program run for ever in this case unless somebody kills program. Another alternative would have been to use break to exit out of loop on some condition.

Validating user input in java?

My guessing game takes either 5, 10, or 20 guesses from a user and they are supposed to try to guess a random number chosen by the computer. Everything in my code is working except this: when the code asks the user whether they want 5, 10, or 20 guesses, if the user were to enter 15, for example, which is not one of the options, it goes on and starts asking for their guesses. I need some type of validation that will make sure they enter one of the options. I'm not sure where or how to include this in the correct way since I am new to programming. I've tried several different ways but get errors for all. What I need is if the user puts a number that is not one of the options, it should just ask them again until they input one of the options. Can someone show me how I should do this?

First of all if (answer.length() ==3) makes no sense.
Maybe you meant:
if(answer.equals("yes"))
Besides, to accomplish what you want I would use a Set containing the valid guesses numbers. It is scalable and makes much more sense than checking against multiple values in an if clause. It will look like this:
Set<Integer> validNumberOfGuesses = new HashSet<Integer>(Arrays.asList(5, 10, 20));
int numberOfGuesses = scan.nextInt();
while (!validNumberOfGuesses.contains(numberOfGuesses)) {
/* ask again */
System.out.println(numberOfGuesses + " is not a valid number of guesses, please try again");
numberOfGuesses = scan.nextInt();
}

Take input from the user inside a loop. For example:
System.out.print("How many guesses would you like? (5, 10, 20)");
do {
int numberOfGuesses = scan.nextInt();
//on correct guess, break out of the loop
if(numberOfGuesses == 5 || numberOfGuesses == 10 || numberOfGuesses == 20)
break;
System.out.print("Please enter a guess having one of these values (5, 10, 20)");
} while (true);
Unless the user, enters one of the three values, he/she will kept being prompted to enter a correct guess value.

Java has the continue keyword that jumps to start of the current loop when run. See The Continue Statement documentation.
Once you have your user input you can do something like
if (numberOfGuesses != 5 && numberOfGuesses != 10 && numberOfGuesses != 20) {
continue; // jumps to start of while loop block, without running conditional
}

When you receive the "numberOfGuesses" you should check the value of that number before moving on. Otherwise you just move on in your code because you don't actually validate the number.
It may be a good idea to creat a function that returns a boolean value and then you can check the number there.
boolean isValidOption(int number)
In the function you want to perform some comparison and validate. Since you have three options you can opt for something like
if (number == 5 || ... )
You can consider how you'll verify the value as there are many ways. Just compare with valid numbers you know you want, you can do some if statements, or place the numbers in an array and compare the value while iterating through the array, and so on. Hope that helps you get started and happy coding!
Edit: Lastly I should have mentioned, but you need to consider the flow of your code. A loop of somesort like while(!isValidOption()) for your check should be use. Loop around the instructions until the user enters a valid option. You need to consider order of operations here in your code and understand the computer doesn't think for you. It does what you tell it, so understand what you are trying to tell it here. I want to step into my game, if and only if, the condition of isValidOption is met for example.

What you need to do is to stay in the loop until you get input that satisfy your demands for example you can use the following function
private int getNumberOfGuesses(Scanner scan) {
int numberOfGuesses;
boolean numberOfGuesesIsValid;
do {
System.out.println("How many guesses would you like? (5, 10, 20)");
numberOfGuesses = scan.nextInt();
numberOfGuesesIsValid = numberOfGuesses == 5 || numberOfGuesses == 10 || numberOfGuesses == 20;
if (!numberOfGuesesIsValid) {
System.out.print("Wrong option !!!");
}
} while (!numberOfGuesesIsValid);
return numberOfGuesses;
}

you can write your code inside a loop to make sure the value is either 5,10 or 20
while(numberOfGuesses!=5||numberOfGuesses!=10||numberOfGuesses=!20);
and the condition if(answer.length()==3 can cause errors. it means it will work every time the input is of length 3,even "noo"

Input Validation regarding integers corresponding to days of the week

I have been reading through Java documentation, but I cannot really seem to find a solution. My program asks for a number of 1-7 corresponding to a day of the week. I want the program to return an error "Sorry, not a valid number" when they enter a wrong number, such as 8,9,20 etc (since they do not correspond to a specific day of the week) My method is not working, any input?
while (!stdln.hasNextInt(1,2,3,4,5,6,7)) {
System.out.println("Sorry, not a valid number.");
stdln.next();
}
I tried
while (dayOfTheWeek > 7) {
System.out.println("Sorry, try again");
stdln.next();
}
But this gets me stuck in a loop

This should do what you want. It will get an int from the user and report when the int out of range. You need to add a terminating condition that will break the loop or it will run until you terminat the application:
while (stdIn.hasNextInt()) {
dayOfTheWeek = stdIn.nextInt();
if (dayOfTheWeek == 0) {
System.out.println("Zero was entered. Exiting");
break;
}
if (dayOfTheWeek < 1 || dayOfTheWeek > 7) {
System.out.println("Sorry, not a valid number.");
}
// do something else with dayOfTheWeek
}
This method call in your code is invalid. This method doesn't take 7 integers as arguments :
stdln.hasNextInt(1,2,3,4,5,6,7)
Also keep in mind that next() method will get every char you input including new lines. You will also have to assign the result of next to a variable for it to be any use.

while loop does not preform else statement and repeat loop

When compiled upon not entering the right selection it doesnt display to re-enter or repeat the loop
int counterA = 0;
while (counterA < 0) {
if (conversionSelection.equalsIgnoreCase("binary"))
counterA++;
if (conversionSelection.equalsIgnoreCase("octal"))
counterA++;
else System.out.println("Error. Please enter weither to convert the Hex to Octal or Binary:");
conversionSelection = keyboard.nextLine();
}

The loop should not execute at all since the condition fails right from the get-go.
There is also a dangling else statement. So you need to have an else-if for your second if statement if you want the else to properly execute. While it may not currently produce errors, it is an important skill to prevent dangling 'else's'.

Your code will never enter the while loop since counterA = 0 and the while loop condition is for it to be < 0. You want counterA to be less than zero, or the condition to include counterA's value in its set (e.g. while (counterA <= 0)).

This loop will never start because 0 is not smaller than 0.

WHILE repeats before ELSE is evaluated

I've nested an IF ELSE statement inside a WHILE statement, but am confused as to why the WHILE is interpreted before the ELSE (when the IF fails). A user is asked to enter a number from 1-10 (inclusive). If the number is inside that range, the program ends. If it's outside of that range, I want to display an error and then prompt them to again enter a number.
It works well if I put the "prompt" before the WHILE, but then I have to put it again inside the ELSE statement for it to show up again. I found this question, but it didn't appear to answer the issue I'm running into. I'm admittedly a Java novice, so I apologize if I'm missing some fundamental aspect of Java.
import java.util.Scanner;
public class Rangecheck {
private static int userNumber; //Number input by user
private static boolean numberOK = false; //Final check if number is valid
//String that will be reused in the DO statement
private static String enterNumber = "Please enter a number from 1 to 10: ";
public static void main(String[] args) {
//Print string
while(!numberOK) // Repeat until the number is OK
{ System.out.println(enterNumber);
Scanner input_UserNumber = new Scanner(System.in); //input integer
userNumber = input_UserNumber.nextInt();
if (10>= userNumber && userNumber >= 1) //Check if 10>=input>=1
{
/*
** If number was valid, congratulate the user and mark numberOK true
*/
System.out.println("Good job! The number you entered is "+userNumber+".");
numberOK = true; // Congratulate user / exit loop if successful
}
else ; //if (10 < userNumber && userNumber < 1)
{
System.err.println("The number entered was not between 1 and 10!");
System.err.print(enterNumber); // Error; user retries until successful
}
}
}
}
I'd expect the System.err.println() to be evaluated in the else statement and then the whileto be evaluated, so that this gets returned:
The number entered was not between 1 and 10!
Please enter a number between 1 and 10:
I've sort of worked around this by putting enterNumber just before while, then putting a second
println in the else statement immediately following the error. It returns what I expect, but I believe I'm fundamentally misunderstanding something.

Let's suppose you have the following code:
while (whileCondition) {
//Inside while, before if
if (ifCondition) {
//Inside if
} else {
//Inside else
}
}
This cycle will execute repeatedly, until the whileCondition becomes false. Whenever the ifCondition is true, the operations inside if will be executed, otherwise the operations inside else will be executed.
Back to your problem:
Your line of
System.out.println(enterNumber);
is at the start of the while. So before the code even gets to your if, the content of enterNumber is displayed on the console. Later, your if is evaluated and if you enter, let's say 22, the condition given to the if will be false and the content of the else block will be evaluated.

The else statement is repeated before the while statement. There can however sometimes be a problem with Streams. A Stream does not necessarily print the data immediately. Especially in the case one uses two different streams, the order of printing the output data can be interleaved.
You can use:
System.err.flush();
to ensure the data is written to the console first.

Add this statement in else body:
numberOk=false;