I'm trying to write a program to calculate factorial but I can't figure out why the Error message displays twice if I enter a letter instead of an integer.
I feel like the issue has to do with Line 29 c = sc.next().charAt(0);, but am not sure how to fix it. Any help is appreciated.
My program:
public class Factorials {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
char c = 'Y';
int num = 0;
do
{
System.out.print("Enter a number to calculate its factorial: ");
while (!sc.hasNextInt()) {
System.out.println("Invalid Entry - Enter only Integers! Try Again: ");
sc.nextLine();
}
int result = 1;
num = sc.nextInt();
for(int i = 1; i <= num; i++) {
result = result * i;
}
System.out.println("The factorial of " + num + " is: " + result);
System.out.println("Do you wish to continue? Y/N: ");
c = sc.next().charAt(0);
}while(c == 'y' || c == 'Y');
sc.close();
}
}
Simple fix: Change the sc.nextLine(); in your code to a sc.next() and you should be good to go. This error occurs because .nextLine() considers the enter/return key as a separate character, while .next() doesn't. (The enter key when you press it after entering either 'y' or 'n': if you try it, the error message doesn't print twice if you enter a letter the first time you run the program).
Side note: You probably want it to be a .print(/*invalid input sentence*/) instead of a .println() to go along with how you take in your other number values.
Otherwise, you're good!
Finds and returns the next complete token from this scanner.
A complete token is preceded and followed by input that matches
the delimiter pattern
As jdk doc shows, the 'sc.next' method will return when it reaches space, enter or return. So when you enter 'y' with enter, the enter character is still in buffer. You can assign sc.nextLine to a variable, like
String str = sc.nextLine();
System.out.println(str);
You can see the enter character and your input character.
Both #TheSj and #Lahiru Danushka answer could solve this problem.
add sc.nextLine(); after c = sc.next().charAt(0);
Related
i've just started java programming and was wondering on how to approach or solve this problem i'm faced with.
I have to write a program that asks a user for a number and continually sums the numbers inputted and print the result.
This program stops when the user enters "END"
I just can't seem to think of a solution to this problem, any help or guidance throughout this problem would be much appreciated and would really help me understand problems like this. This is the best i could do
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
while (true) {
System.out.print("Enter a number: ");
int x = scan.nextInt();
System.out.print("Enter a number: ");
int y = scan.nextInt();
int sum = x + y;
System.out.println("Sum is now: " + sum);
}
}
}
The output is supposed to look like this:
Enter a number: 5
Sum is now: 5
Enter a number: 10
Sum is now: 15
Enter a number: END
One solution would be to not use the Scanner#nextInt() method at all but instead utilize the Scanner#nextLine() method and confirm the entry of the numerical entry with the String#matches() method along with a small Regular Expression (RegEx) of "\d+". This expression checks to see if the entire string contains nothing but numerical digits. If it does then the matches() method returns true otherwise it returns false.
Scanner scan = new Scanner(System.in);
int sum = 0;
String val = "";
while (val.equals("")) {
System.out.print("Enter a number (END to quit): ");
val = scan.nextLine();
// Was the word 'end' in any letter case supplied?
if (val.equalsIgnoreCase("end")) {
// Yes, so break out of loop.
break;
}
// Was a string representation of a
// integer numerical value supplied?
else if (val.matches("\\-?\\+?\\d+")) {
// Yes, convert the string to integer and sum it.
sum += Integer.parseInt(val);
System.out.println("Sum is now: " + sum); // Display Sum
}
// No, inform User of Invalid entry
else {
System.err.println("Invalid number supplied! Try again...");
}
val = ""; // Clear val to continue looping
}
// Broken out of loop with the entry of 'End"
System.out.println("Application ENDED");
EDIT: Based on Comment:
Since since an integer can be signed (ie: -20) or unsigned (ie: 20) and the fact that an Integer can be prefixed with a + (ie: +20) which is the same as unsigned 20, the code snippet above takes this into consideration.
Do it like this:
public static void main(String[] args) throws Exception {
int sum = 0;
Scanner scan = new Scanner(System.in);
while (scan.hasNext()) {
System.out.print("Enter a number: ");
if (scan.hasNextInt())
sum += scan.nextInt();
else
break;
System.out.println("Sum is now: " + sum);
}
System.out.print("END");
}
This will end if the input is not a number (int).
As pointed out in the comments, if you want the program to stop when the user specifically enters "END", change the else-statement to:
else if (scanner.next().equals("END"))
break;
So I'm working on a project to count the number of vowels and consonants of a desired string. I've got almost everything working, but I can't seem to get the option to replace the string with a new one to work properly. I think it's a problem with having the new choice be applied to the Driver construct but I can't seem to figure out a way to do that without breaking the program.
import java.util.*;
public class Driver{
private String entry;
int vowels = 0;
int cons = 0;
public Driver(String input){
entry = input;
this.count();
}
public boolean isVowel(char x){
return (x == 'A' || x == 'E' || x == 'I' || x == 'O' || x == 'U' ||
x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u');
}
public boolean isConsonant(char x){
return (((x >= 'a' && x <= 'z') || (x >= 'A' && x <= 'Z')) && !isVowel(x));}
public int getVowels(){
return vowels;
}
public int getConsonants(){
return cons;
}
public void count(){
int l = entry.length();
for(int i = 0; i < l; i++){
if(this.isVowel(entry.charAt(i))){
vowels++;
}else{
if(this.isConsonant(entry.charAt(i))){
cons++;
}
}
}
}
static Scanner kb = new Scanner(System.in);
public static void main(String[] args){
int choice; //Choice from menu
int vowels = 0; //# of vowels in entry
int cons = 0; //# of consonants in entry
String entry; //User's input
System.out.print("Enter a string:");
entry = kb.nextLine();
do{
Driver sentence = new Driver(entry);
System.out.println("Enter a number");
System.out.println("1. Count the number of vowels in the string");
System.out.println("2. Count the number of consonants in the string");
System.out.println("3. Count both the vowels and consonants in the string");
System.out.println("4. Enter another string");
System.out.println("5. Exit the program:");
choice = kb.nextInt();
switch(choice){
case 1:
System.out.println("Vowels: " + sentence.getVowels());
break;
case 2:
System.out.println("Consonants: " + sentence.getConsonants());
break;
case 3:;
System.out.println("Vowels: " + sentence.getVowels() + "\nConsonants: " + sentence.getConsonants());
break;
case 4:
System.out.print("Enter a string:");
entry = kb.nextLine();
//Driver sentence = new Driver(entry);
break;
default:
System.out.println("Please enter a value input: ");
}
}while(choice != 5);
}
}
You should have the Driver Class on its own (without the main() method).
Then create another Class that will be your startup Class which will contain the main() method). For the sake of argument let's call it VowelsAndConsonants.
The problem that's causing all the grief is your keyboard entry of 1 to 5 for the Console menu. You are using the Scanner.nextInt() method which is okay as long as you know that the whatever is entered is not all consumed and remains within the Scanner buffer and this is because the Scanner.nextInt() method doesn’t read and consume the line separator that is applied when the Enter Key is hit. So, when Scanner.nextLine() is used as the next Scanner method the first thing it gets hold of is that line separator that's held within the Scanner buffer which in turn is just like hitting the Enter key and therefore it skips what was typed. On the other hand the Scanner.nextLine() method does read in and consume the line separator. One way to get around this problem is to add (in your case) kb.nextLine(); directly after the choice = kb.nextInt(); code line. This forces the line separator to be consumed and empties the buffer. In essence the menu would then look something like this:
Driver sentence = new Driver(entry);
do {
System.out.println("Enter a Menu choice (1 to 5): ");
System.out.println("1. Count the number of vowels in the string");
System.out.println("2. Count the number of consonants in the string");
System.out.println("3. Count both the vowels and consonants in the string");
System.out.println("4. Enter another string");
System.out.println("5. Exit the program:");
choice = kb.nextInt();
kb.nextLine(); // Consume
..... the rest of your do/while code .....
}
EDIT: Based on your Question in Comment: "I'm curious as to why
there is a difference in how the line separator is dealt with tho. Is
this just something that is in Java, or is there a reason that it
isn't consumed from nextInt()?"
The Scanner Class is quite involved. It is by design that the nextInt() does this since Scanner can work based on tokens. You can use nextInt() to accept multiple numbers on a single input using (for example) a whitespace delimiter and then with the Scanner.hasNextInt() or Scanner.hasNext() methods in conjuction with the Scanner.nextInt() you can individually acquire each value, for example:
Scanner kb = new Scanner(System.in).useDelimiter(" *");
System.out.println("Enter numbers delimited with a white-space: ");
while (kb.hasNextInt()) {
int x = kb.nextInt();
System.out.println(x);
}
If you were to run this code and at the prompt enter on one line:
2 4 6 8 10
(note the whitespace between numbers) you will see a display within the Console window of:
2
4
6
8
10
Through each iteration of the while loop x would equal the next integer token from the line of numbers entered. This sort of thing however is best suited for reading numbers from a text file which Scanner can do as well. Read up on the Scanner Class.
I was able to get the program to run and work with error checking to make sure that the user input is in fact an int. The issue I ran into is that I only want it to be a 3-digit int. I'm having trouble getting that into the right place:
import java.util.*;
public class listMnemonics
{
public static void main(String[] args)
{
//Defines the "keypad" similar to that of a phone
char[][] letters =
{{'0'},{'1'},{'A','B','C'},{'D','E','F'},{'G','H','I'},{'J','K','L'},
{'M','N','O'},{'P','Q','R','S'},{'T','U','V'},{'W','X','Y','Z'}};
//Creates the Scanner
Scanner scan = new Scanner(System.in);
Right here is where I need to implement that and I am running into the issue. I'm sure it's maybe only a line out of place or missing that I need, I just don't know what or where. As it sits, it will constantly ask me to enter a 3-digit number, no matter the length. Error checking for a string entered does currently work:
//Gives instructions to the user to enter 3-digit number
//Any amount of numbers will work, but instructions help
//System.out.println("Please enter a 3-digit number: ");
int j;
do
{
System.out.println("Please enter a 3-digit number: ");
while (!scan.hasNextInt()) {
System.out.println("That's not a 3-digit number! Try again!");
scan.next(); // this is important!
}
j = scan.nextInt();
}
//while (j <= 0); This works while not checking digit length
while (j != 3);
int w = (int) Math.log10(j) +1; //Found this, but not sure if it helps or not
String n = Integer.toString(w);
And here is the rest that get's it to do what I need it to:
//Determines char length based on user input
char[][] sel = new char[n.length()][];
for (int i = 0; i < n.length(); i++)
{
//Grabs the characters at their given position
int digit = Integer.parseInt("" +n.charAt(i));
sel[i] = letters[digit];
}
mnemonics(sel, 0, "");
}
public static void mnemonics(char[][] symbols, int n, String s)
{
if (n == symbols.length)
{
System.out.println(s);
return;
}
for (int i = 0; i < symbols[n].length; i ++)
{
mnemonics(symbols, n+1, s + symbols[n][i]);
}
}
}
Here's the output:
----jGRASP exec: java listMnemonics
Please enter a 3-digit number:
2345
Please enter a 3-digit number:
12
Please enter a 3-digit number:
123
Please enter a 3-digit number:
motu
That's not a 3-digit number! Try again!
With the help of MvG and pingul, this is what is currently working the way I was hoping:
import java.util.*;
import java.util.regex.Pattern;
public class listMnemonics
{
public static void main(String[] args)
{
// Defines the "keypad" similar to that of a phone
char[][] letters =
{{'0'},{'1'},{'A','B','C'},{'D','E','F'},{'G','H','I'},{'J','K','L'},
{'M','N','O'},{'P','Q','R','S'},{'T','U','V'},{'W','X','Y','Z'}};
// Creates the Scanner
Scanner scan = new Scanner(System.in);
// Gives instructions to the user to enter 3-digit number
// This 'Pattern' also guarantees that only 3 digits works.
Pattern threeDigitNumber = Pattern.compile("[0-9]{3}");
int j;
do
{
System.out.println("Please enter a 3-digit phone number: ");
// If it's not a 3-digit int, try again
while (!scan.hasNext(threeDigitNumber))
{
System.out.println("That's not a 3-digit number! Try again!");
// This is important!
scan.next();
}
j = scan.nextInt();
}
while (j <= 0);
String n = Integer.toString(j);
// Determines char length based on user input
char[][] sel = new char[n.length()][];
for (int i = 0; i < n.length(); i++)
{
// Grabs the characters at their given position
int digit = Integer.parseInt("" +n.charAt(i));
sel[i] = letters[digit];
}
mnemonics(sel, 0, "");
}
// Here is where the magic happens and creates the possible
// letter combinations based on the user input and characters
// selected in previous steps.
public static void mnemonics(char[][] symbols, int n, String s)
{
if (n == symbols.length)
{
System.out.println(s);
return;
}
for (int i = 0; i < symbols[n].length; i ++)
{
mnemonics(symbols, n+1, s + symbols[n][i]);
}
}
}
Condensed and reformatted your code reads
Scanner scan = new Scanner(System.in);
int j;
do {
System.out.println("Please enter a 3-digit number: ");
while (!scan.hasNextInt()) {
System.out.println("That's not a 3-digit number! Try again!");
scan.next(); // this is important!
}
j = scan.nextInt();
} while (j != 3);
Comparing that to the Scanner documentation we can see that the scan.next() call will read (and discard) the non-int token. Otherwise j will be the integer you read. And you continue doing so while the number you read is different from 3. Not the length of the number, but the number itself. So if you want to end the loop, enter 3. If you want to do so while following the prompt, enter 003.
If that's not what you want to check, then change the end of loop condition. Or perhaps change the way you test for three-digit numbers, by using regular expressions to match these.
Scanner scan = new Scanner(System.in);
Pattern threeDigitNumber = Pattern.compile("\\d\\d\\d");
int j;
System.out.println("Please enter a 3-digit number: ");
while (!scan.hasNext(threeDigitNumber)) {
if (scan.hasNext()) {
System.out.println(scan.next() + " is not a 3-digit number! Try again!");
} else {
System.out.println("Input terminated unepxectedly");
System.exit(1);
}
}
j = scan.nextInt();
As comments correctly indicate, the pattern "\\d\\d\\d" could just as well be written as "[0-9]{3}", or as "\\d{3}" or "[0-9][0-9][0-9]". Using {…} might be useful in situations where the number of digits is a variable.
The documentation for Scanner.hasNext(Pattern) requires the pattern to match the input. This apparently follows the Matcher.matches() semantics of matching the whole string against the pattern, as opposed to Matcher.find() which checks whether the string contains any part matching the pattern. So the input does not have to be enclosed in ^ and $, as I assumed at first, and in fact should not be using these unless the pattern is compiled with the Pattern.MULTILINE flag.
You may want to call Scanner.useDelimiter to delimit using line breaks only.
Scanner.useDelimiter("[\\r\\n]+")
so my problem is that I need to get the user to enter a string. then they will enter a character that they want counted. So the program is supposed to count how many times the character they entered will appear in the string, this is my issue. If someone can give me some information as to how to do this, it'll be greatly appreciated.
import java.util.Scanner;
public class LetterCounter {
public static void main(String[] args) {
Scanner keyboard= new Scanner(System.in);
System.out.println("please enter a word");//get the word from the user
String word= keyboard.nextLine();
System.out.println("Enter a character");//Ask the user to enter the character they wan counted in the string
String character= keyboard.nextLine();
}
}
Here is a solution taken from this previously asked question and edited to better fit your situation.
Either have the user enter a char, or take the first character from
the string they entered using character.chatAt(0).
Use word.length to figure out how long the string is
Create a for loop and use word.charAt() to count how many times your character appears.
System.out.println("please enter a word");//get the word from the user
String word= keyboard.nextLine();
System.out.println("Enter a character");//Ask the user to enter the character they want counted in the string
String character = keyboard.nextLine();
char myChar = character.charAt(0);
int charCount = 0;
for (int i = 1; i < word.length();i++)
{
if (word.charAt(i) == myChar)
{
charCount++;
}
}
System.out.printf("It appears %d times",charCount);
This should do it. What it does is that it gets a string to look at, gets a character to look at, iterates through the string looking for matches, counts the number of matches, and then returns the information. There are more elegant ways to do this (for example, using a regex matcher would also work).
#SuppressWarnings("resource") Scanner scanner = new Scanner(System.in);
System.out.print("Enter a string:\t");
String word = scanner.nextLine();
System.out.print("Enter a character:\t");
String character = scanner.nextLine();
char charVar = 0;
if (character.length() > 1) {
System.err.println("Please input only one character.");
} else {
charVar = character.charAt(0);
}
int count = 0;
for (char x : word.toCharArray()) {
if (x == charVar) {
count++;
}
}
System.out.println("Character " + charVar + " appears " + count + (count == 1 ? " time" : " times"));
I'm trying to parse a char from console input using in.nextLine() and from there take charAt(0). My problem is after asking the user to enter a string to perform the in.nextLine() on, it skips the input and yields an error due to trying to get the first character of a null string.
System.out.print("Select an operator (+, -, *, /), 'c' or 'C' to clear, or 'q' to quit: ");
String temp = in.nextLine();
char tempOperator = temp.charAt(0);
the error is
java.lang.StringIndexOutOfBoundsException: String index out of range: 0
at java.lang.String.charAt(Unknown Source)
full program is available here
General comments and suggestions always welcome.
Thanks in advance.
When you do cValue = in.nextDouble();, it reads the next token(full value) and parses it to double. If the return key was pressed at this time, \n is the next token in the buffer to read.
When you do: String temp = in.nextLine();, it reads the \n from the buffer of the previous enter and charAt(0) fails as it has read only empty("") string.
To overcome the issue, either skip the previous buffer by adding an addition in.nextLine(), add \n \r as skip pattern as below (this is the pattern defined in Scanner.class as LINE_SEPARATOR_PATTERN):
in.skip("\r\n|[\n\r\u2028\u2029\u0085]");
or put a small while loop as below:
String temp = "";
while((temp.length() < 0){
temp = in.nextLine();
}
One thing you might consider is instead of doing this:
String temp = in.nextLine();
Do this:
String temp = in.next();
That of course is assuming that you only want a single character from the user input, which it looks like that is all you are asking the user for. This will get rid of the exception you are receiving. However, you may want to read the whole line first to see what the length is, if you would like to throw an error when the user types more than one character.
Well, the problem is in the line in.nextDouble().
Scanner#nextDouble reads the next token from the user as double. So, when you pass a value 4.5 as input, then the nextDouble just reads the token 4.5 and skips the linefeed after that input that user have entered. Now this linefeed(Since this is a new line), will be regarded as input for the next Scanner.nextLine. And hence your in.nextLine after in.nextDouble is reading the newline left over by the previous in.nextDouble.
So, the workaround is the same that #tjg184 has pointed out in his answer. Add an empty in.nextLine after in.nextDouble that will read that newline left over. So, that your coming in.nextLine reads the input actually passed.
cValue = in.nextDouble();
in.nextLine(); // Add this before your `while` which will read your `newline`
while (true) {
//continue;
}
Whereas, Scanner.nextLine reads the complete input till the newline. So, it will not leave any newline to be read by the next read by user.
Although this answer was not needed, as the #tjg184's answer does the same. I just added it to give a better explanation of what exactly is happening.
Looking at your program, it's because of String temp = in.nextLine();. When the user hits the "Enter" key, it essentially is skipping this statement since the user typed hit "Enter". Try the following. Note, I only added the following line String enter = in.nextLine();
import java.util.Scanner;
public class Calculator
{
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
char operator = 'z'; //initialize operator
double cValue; //current value
double rhValue; //right hand value
boolean cont = true;
System.out.print("Enter starting value: ");
cValue = in.nextDouble();
// disregard the "Enter" key
String enter = in.nextLine();
while(true)
{
System.out.print("Select an operator (+, -, *, /), 'c' or 'C' to clear, or 'q' to quit: ");
String temp = in.nextLine();
char tempOperator = temp.charAt(0);
if (tempOperator == 'c' || tempOperator == 'C')
{
cValue = 0.0;
System.out.println("Current value is: " + cValue);
System.out.println();
}
else if(tempOperator == 'q')
{
System.out.println("Final result: " + cValue);
System.exit(1);
}
else if(tempOperator == '+' || tempOperator == '-' || tempOperator == '*' || tempOperator == '/')
{
operator = tempOperator;
break;
}
else
throw new IllegalArgumentException("operator not valid");
}
System.out.println("Enter a right hand value (type double): ");
rhValue = in.nextDouble();
System.out.println("Math expression: answer " + operator + "= " + rhValue);
switch(operator)
{
case '+': cValue =+ rhValue;
break;
case '-': cValue =- rhValue;
break;
case '*': cValue = cValue * rhValue;
break;
case '/': cValue = cValue / rhValue;
break;
}
System.out.println("Current value is: " + cValue);
}
}