What did I miss? No output is shown - javascript

//Q1.7
class Toing {
public static void main(String [] args){
}
public int distance(int sq1,int sq2){
int x1 = sq1%8;
int y1 = sq1/8;
int x2 = sq2%8;
int y2 = sq2/8;
double a = Math.sqrt((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1));
if(a % Math.sqrt(2) >= 1.41 && a % Math.sqrt(2) <= 1.42){
return (int) (Math.round(a/Math.sqrt(2)));
}
else if(sq2 == sq1){
return 0;
}
else {
return (int)a;
}
}
//Q1.8
public boolean sameColumn(int sq1,int sq2) {
return ((sq2 - sq1) % 8 == 0);
}
}
Hello, can someone help me if there's anything that I miss while doing these questions as when I run the command, it only shows "Run Toing". Is it suppose to show the answer or we just need to have the code only? As this question is about moving etc for example in
Q.1.7) the distance(0,63)=>6 . Should my output when I run the command show 6?
Q.1.8) sameColumn(10,12)=>true. Should my output when I run the command show true?
If yes, what do I miss writing it down on my commands as I cant think anymore as I wrote too many java thing today.. just want to check with you guys if anyone can help as I don't have someone that know java.. Thank you

To get out put you need to run the program. Right now main is empty and does nothing.
Here is an example of using Toing object:
public static void main(String [] args){
//create a Toing object
Toing t = new Toing();
//use it to calculate distance
int distance = t.distance(500,1200);
System.out.println("distance = "+ distance);
}

your main method is empty therefore the app is good design but doing nothing..
if you want to see some output in the console use System.out.println() method
in your case you need an instance of the class Toing
so do this:
public static void main(String [] args){
Toing myToingInstance = new Toing();
int distance = myToingInstance.distance(0,63);
System.out.println(" Tchebychev distance from 0 to 63 is: "+ distance);
}

Related

Issue with narrowing Number type to UInt32

So I have a code I previously wrote in C# that I am trying to convert to JS.
However I just encountered a strange issue while narrowing down my Number type to UInt32 Range in JS.
However, If I implement the same code in C# or Object Pascal (languages which have native UInt32 types), I get a different result.
Can someone please show me how I can get results that are consistent with the C# code in JS.
For simplicity sake, I am using defined constants to show the problem.
The result of my Code is
JS: 2444377276
C# and Object Pascal: 2444377275
function modulo(a, b) {
return a - Math.floor(a / b) * b;
}
function ToInteger(x) {
x = Number(x);
return x < 0 ? Math.ceil(x) : Math.floor(x);
}
function ToUint32(x) {
return modulo(ToInteger(x), Math.pow(2, 32));
}
let a = (1207988537 * 16777619);
console.log(a >>> 0);
let uu = new Uint32Array([a]);
console.log(uu[0]);
console.log(ToUint32(a));
// will only work in ES2020 upwards but produces same result as other JS solutions above
//console.log(BigInt.asUintN(32, BigInt(a)));
C# code
using System;
public class Test
{
public static void Main()
{
// your code goes here
unchecked
{
uint a = (uint)(1207988537 * 16777619);
Console.WriteLine(a);
}
}
}
Object Pascal Code
{$MODE DELPHI}
program ideone;
var
a: UInt32;
begin
(* your code goes here *)
a := UInt32(1207988537 * 16777619);
Writeln(a);
end.

What does CryptoJS.enc.Hex.parse(hash) do, and how to replicate it in Java?

Basically, I want to replicate the following javascript code in java:
let h = calculateHmac(CryptoJS.enc.Hex.parse(hash), key);
const calculateHmac = (seed, salt) => {
const hmac = CryptoJS.HmacSHA256(seed, salt);
return hmac;
}
Here's my Java code so far:
public static String encode(String key, String hash) throws Exception {
Mac sha256_HMAC = Mac.getInstance("HmacSHA256");
SecretKeySpec secret_key = new SecretKeySpec(key.getBytes("UTF-8"), "HmacSHA256");
sha256_HMAC.init(secret_key);
return bytesToHex(sha256_HMAC.doFinal(hash.getBytes("UTF-8")));
}
My java code correctly gets the hmac, but my result is different from the js due to the CryptoJS.enc.Hex.parse(hash) part (I'm using just hash in java). What does CryptoJS.enc.Hex.parse(hash) actually do? When I print it, it seems to print the same as just hash. And how do I replicate that line in java?
For reference, I'm using the key 0000000000000000004d6ec16dafe9d8370958664c1dc422f452892264c59526 and hash 71b12b1ee6aad3763a3f844bec2512e6985f727de60224237552df77aa2d1d12. When I run it through an online hmac generator tool (for example https://www.freeformatter.com/hmac-generator.html#ad-output) I get 2d1044151d96fadd070261880e6564f7a52a33af293b57cc40fb75e099d1e232 as the output, which my java code correctly does, but is different from the js code because of the CryptoJS.enc.Hex.parse(hash) line. Please let me know what I'm missing.
EDIT: edited for clarity
Assuming that hash-value in JS is a String var we do need a way to get a byte[] out of it. The simple way of
byte[] hashValue = hash.getBytes("UTF-8");
does not work as we need the value and not the ASCII-code of the characters. Simply use the hexStringToByteArray method to
transform a hex-string representation to a byte array ("bytesToHex" is a service method to print out the byte array):
import java.io.UnsupportedEncodingException;
public class ParseHexstring {
public static void main(String[] args) throws UnsupportedEncodingException {
System.out.println("https://stackoverflow.com/questions/62497931/what-does-cryptojs-enc-hex-parsehash-do-and-how-to-replicate-it-in-java");
String hash = "124d45ff6789ac01234543789012";
byte[] hashGetBytes = hash.getBytes("UTF-8");
byte[] hashValue = hexStringToByteArray(hash);
System.out.println("hash String : " + hash);
System.out.println("hashGetBytes: " + bytesToHex(hashGetBytes));
System.out.println("hashValue : " + bytesToHex(hashValue));
}
public static byte[] hexStringToByteArray(String s) {
int len = s.length();
byte[] data = new byte[len / 2];
for (int i = 0; i < len; i += 2) {
data[i / 2] = (byte) ((Character.digit(s.charAt(i), 16) << 4)
+ Character.digit(s.charAt(i + 1), 16));
}
return data;
}
private static String bytesToHex(byte[] bytes) {
StringBuffer result = new StringBuffer();
for (byte b : bytes) result.append(Integer.toString((b & 0xff) + 0x100, 16).substring(1));
return result.toString();
}
}
That the output:
https://stackoverflow.com/questions/62497931/what-does-cryptojs-enc-hex-parsehash-do-and-how-to-replicate-it-in-java
hash String : 124d45ff6789ac01234543789012
hashGetBytes: 31323464343566663637383961633031323334353433373839303132
hashValue : 124d45ff6789ac01234543789012

How to get the real value in a mathematic function

hello everyone i have this code that makes functions in mathematics but with this function the result is wrong o.O
x=0
"-3*X^2-16*X+2"
Code:
public static void main(String[] args) throws Exception {
ScriptEngineManager manager = new ScriptEngineManager();
ScriptEngine engine = manager.getEngineByName("js");
engine.put("X", 0);
Object operation = engine.eval("-3*X^2-16*X+2");
//Object operation2 = engine.eval("(X+3)");
System.out.println("Evaluado operacion 1: " + operation);
//System.out.println("Evaluado operacion 2: " + operation2);
}
the result is 2 but i get 4
Evaluado operacion 1: 4
i have other code that i made
/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package gustavo_santa;
import javax.script.ScriptEngine;
import javax.script.ScriptEngineManager;
import javax.script.ScriptException;
import javax.swing.JOptionPane;
/**
*
* #author osmarvirux
*/
public class SerieB {
//xi and x4
int xi;
int x4;
//f(x) function variables
String positive_negative;
int num_one;
int elevation_one;
String add_subtract_one;
int num_two;
int elevation_two;
String add_subtract_two;
int num_three;
//results
String xi_result;
String x4_result;
public SerieB(int xi, int x4, String positive_negative, int num_one, int elevation_one, String add_subtract_one, int num_two, int elevation_two, String add_subtract_two, int num_three) {
this.xi = xi;
this.x4 = x4;
this.positive_negative = positive_negative;
this.num_one = num_one;
this.elevation_one = elevation_one;
this.add_subtract_one = add_subtract_one;
this.num_two = num_two;
this.elevation_two = elevation_two;
this.add_subtract_two = add_subtract_two;
this.num_three = num_three;
}
public void Procedure_xi(){
ScriptEngineManager manager = new ScriptEngineManager();
ScriptEngine engine = manager.getEngineByName("js");
if (positive_negative == "-"){
try {
xi_result=(num_one*(Math.pow(xi, elevation_one)))+add_subtract_one+(num_two*(Math.pow(xi, elevation_two)))
+add_subtract_two+num_three;
Object result = engine.eval(xi_result);
System.out.println(xi_result+" = "+result);
} catch(ScriptException se) {
se.printStackTrace();
}
}else{
try {
xi_result=((-num_one*(Math.pow(xi, elevation_one)))+add_subtract_one+(num_two*(Math.pow(xi, elevation_two)))
+add_subtract_two+num_three);
Object result = engine.eval(xi_result);
System.out.println(xi_result+" = "+result);
} catch(ScriptException se) {
se.printStackTrace();
}
}
}
public void Procedure_x4(){
ScriptEngineManager manager = new ScriptEngineManager();
ScriptEngine engine = manager.getEngineByName("js");
if (positive_negative == "-"){
try {
x4_result=(num_one*(Math.pow(x4, elevation_one)))+add_subtract_one+(num_two*(Math.pow(x4, elevation_two)))
+add_subtract_two+num_three;
Object result = engine.eval(x4_result);
System.out.println(x4_result+" = "+result);
} catch(ScriptException se) {
se.printStackTrace();
}
}else{
try {
x4_result=((-num_one*(Math.pow(x4, elevation_one)))+add_subtract_one+(num_two*(Math.pow(x4, elevation_two)))
+add_subtract_two+num_three);
Object result = engine.eval(x4_result);
System.out.println(x4_result+" = "+result);
} catch(ScriptException se) {
se.printStackTrace();
}
}
}
public static void main(String[] args){
//-3x^2-16x+2
SerieB obj = new SerieB(0, 1, "+", -3, 2, "-", 16, 1, "+", 2);
obj.Procedure_xi();
obj.Procedure_x4();
}
}
the result with this code is 2 but i wanna use
ScriptEngineManager manager = new ScriptEngineManager();ScriptEngineManager manager = new ScriptEngineManager();
because is a library and i think is more precise and i dont wanna use my code because there are many lines and i dont know if is 100% efficient. someone can help me? or give me a recomendation to resolve this mathematic functions? thanks a lot
The result you're getting is correct.
The confusion arises from the fact that what you're assuming to be the power operator (^) is actually the bitwise XOR operator in JavaScript (you're using a JavaScript script engine).
So, evaluating 0 ^ 2 yields 2, while evaluating Math.pow(0, 2) yields 0, hence the difference.
To get the result you expect, the expression would have to read:
-3*Math.pow(X,2)-16*X+2
You could pre-process the expression to replace the exponential operations with invocations of Math.pow():
let X = 0;
let expression = "-3*X^2-16*X+2"
let processed = expression.replace(/(\w+)\^(\w+)/g, 'Math.pow($1,$2)');
console.log(processed); // prints "-3*Math.pow(X,2)-16*X+2"
console.log(eval(processed)); // prints "2"
Using the script engine, that could look like:
ScriptEngineManager manager = new ScriptEngineManager();
ScriptEngine engine = manager.getEngineByName("js");
engine.put("X", 0);
engine.put("expression", "-3*X^2-16*X+2");
engine.put("processed", engine.eval("expression.replace(/(\\w+)\\^(\\w+)/g, 'Math.pow($1,$2)')"));
System.out.println(engine.eval("eval(processed)")); // 2.0
Or, if you prefer to do the regular expression replacement in Java:
String expression = "-3*X^2-16*X+2";
String processed = expression.replaceAll("(\\w+)\\^(\\w+)", "Math.pow($1,$2)");
ScriptEngineManager manager = new ScriptEngineManager();
ScriptEngine engine = manager.getEngineByName("js");
engine.put("X", 0);
System.out.println(engine.eval(processed)); // 2.0
You have incorrect syntax for power. Replace -3*X^2-16*X+2 with -3*Math.pow(X,2)-16*X+2. See Javascript, What does the ^ (caret) operator do?
I think what's happening is your program is evaluating
(-3*0)^2 - 16*0 + 2 = 0^2 +2 = 2+2 =4
because the operator ^ in computer science mean bitwise exclusive or,
which basically means if the bits are both 1 or 0 then change them to 0, else 1. And 2 is represented by 10 while 0 is 0 so 2^0 = 2
Try replacing ^2 with *X
alternatively you can use Math.pow(X,n) , if you need exponentiation to some power n, but for squaring it's better to just write it out as X*X
The Below is Outdated Due to Edits to Question, but is still functional:
In the second part of your question you wrote
Object operation = engine.eval("-3*(X^2)-16*X +2");
String processed = operation.replace(/(\w+)\^(\w+)/g, 'Math.pow($1,$2)');
based on an answer by another user. You should have written
String expr = "-3*(X^2)-16*X +2";
String processed = expr.replaceAll("(\\w+)(\\^)(\\w+)", 'Math.pow($1,$2'));
Object operation = engine.eval(processed);

PhoneBookEntry Driver Program error [duplicate]

I am using the Scanner methods nextInt() and nextLine() for reading input.
It looks like this:
System.out.println("Enter numerical value");
int option;
option = input.nextInt(); // Read numerical value from input
System.out.println("Enter 1st string");
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)
The problem is that after entering the numerical value, the first input.nextLine() is skipped and the second input.nextLine() is executed, so that my output looks like this:
Enter numerical value
3 // This is my input
Enter 1st string // The program is supposed to stop here and wait for my input, but is skipped
Enter 2nd string // ...and this line is executed and waits for my input
I tested my application and it looks like the problem lies in using input.nextInt(). If I delete it, then both string1 = input.nextLine() and string2 = input.nextLine() are executed as I want them to be.
That's because the Scanner.nextInt method does not read the newline character in your input created by hitting "Enter," and so the call to Scanner.nextLine returns after reading that newline.
You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself).
Workaround:
Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline
int option = input.nextInt();
input.nextLine(); // Consume newline left-over
String str1 = input.nextLine();
Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.
int option = 0;
try {
option = Integer.parseInt(input.nextLine());
} catch (NumberFormatException e) {
e.printStackTrace();
}
String str1 = input.nextLine();
The problem is with the input.nextInt() method; it only reads the int value. So when you continue reading with input.nextLine() you receive the "\n" Enter key. So to skip this you have to add the input.nextLine().
Try it like this, instead:
System.out.print("Insert a number: ");
int number = input.nextInt();
input.nextLine(); // This line you have to add (It consumes the \n character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();
It's because when you enter a number then press Enter, input.nextInt() consumes only the number, not the "end of line". When input.nextLine() executes, it consumes the "end of line" still in the buffer from the first input.
Instead, use input.nextLine() immediately after input.nextInt()
There seem to be many questions about this issue with java.util.Scanner. I think a more readable/idiomatic solution would be to call scanner.skip("[\r\n]+") to drop any newline characters after calling nextInt().
EDIT: as #PatrickParker noted below, this will cause an infinite loop if user inputs any whitespace after the number. See their answer for a better pattern to use with skip: https://stackoverflow.com/a/42471816/143585
TL;DR
nextLine() is safe to call when (a) it is first reading instruction, (b) previous reading instruction was also nextLine().
If you are not sure that either of above is true you can use scanner.skip("\\R?") before calling scanner.nextLine() since calls like next() nextInt() will leave potential line separator - created by return key which will affect result of nextLine(). The .skip("\\R?") will let us consume this unnecessary line separator.
skip uses regex where
\R represents line separators
? will make \R optional - which will prevent skip method from:
waiting for matching sequence
in case of reaching end of still opened source of data like System.in, input stream from socket, etc.
throwing java.util.NoSuchElementException in case of
terminated/closed source of data,
or when existing data doesn't match what we want to skip
Things you need to know:
text which represents few lines also contains non-printable characters between lines (we call them line separators) like
carriage return (CR - in String literals represented as "\r")
line feed (LF - in String literals represented as "\n")
when you are reading data from the console, it allows the user to type his response and when he is done he needs to somehow confirm that fact. To do so, the user is required to press "enter"/"return" key on the keyboard.
What is important is that this key beside ensuring placing user data to standard input (represented by System.in which is read by Scanner) also sends OS dependant line separators (like for Windows \r\n) after it.
So when you are asking the user for value like age, and user types 42 and presses enter, standard input will contain "42\r\n".
Problem
Scanner#nextInt (and other Scanner#nextType methods) doesn't allow Scanner to consume these line separators. It will read them from System.in (how else Scanner would know that there are no more digits from the user which represent age value than facing whitespace?) which will remove them from standard input, but it will also cache those line separators internally. What we need to remember, is that all of the Scanner methods are always scanning starting from the cached text.
Now Scanner#nextLine() simply collects and returns all characters until it finds line separators (or end of stream). But since line separators after reading the number from the console are found immediately in Scanner's cache, it returns empty String, meaning that Scanner was not able to find any character before those line separators (or end of stream).
BTW nextLine also consumes those line separators.
Solution
So when you want to ask for number and then for entire line while avoiding that empty string as result of nextLine, either
consume line separator left by nextInt from Scanners cache by
calling nextLine,
or IMO more readable way would be by calling skip("\\R") or skip("\r\n|\r|\n") to let Scanner skip part matched by line separator (more info about \R: https://stackoverflow.com/a/31060125)
don't use nextInt (nor next, or any nextTYPE methods) at all. Instead read entire data line-by-line using nextLine and parse numbers from each line (assuming one line contains only one number) to proper type like int via Integer.parseInt.
BTW: Scanner#nextType methods can skip delimiters (by default all whitespaces like tabs, line separators) including those cached by scanner, until they will find next non-delimiter value (token). Thanks to that for input like "42\r\n\r\n321\r\n\r\n\r\nfoobar" code
int num1 = sc.nextInt();
int num2 = sc.nextInt();
String name = sc.next();
will be able to properly assign num1=42 num2=321 name=foobar.
It does that because input.nextInt(); doesn't capture the newline. you could do like the others proposed by adding an input.nextLine(); underneath.
Alternatively you can do it C# style and parse a nextLine to an integer like so:
int number = Integer.parseInt(input.nextLine());
Doing this works just as well, and it saves you a line of code.
Instead of input.nextLine() use input.next(), that should solve the problem.
Modified code:
public static Scanner input = new Scanner(System.in);
public static void main(String[] args)
{
System.out.print("Insert a number: ");
int number = input.nextInt();
System.out.print("Text1: ");
String text1 = input.next();
System.out.print("Text2: ");
String text2 = input.next();
}
If you want to read both strings and ints, a solution is to use two Scanners:
Scanner stringScanner = new Scanner(System.in);
Scanner intScanner = new Scanner(System.in);
intScanner.nextInt();
String s = stringScanner.nextLine(); // unaffected by previous nextInt()
System.out.println(s);
intScanner.close();
stringScanner.close();
In order to avoid the issue, use nextLine(); immediately after nextInt(); as it helps in clearing out the buffer. When you press ENTER the nextInt(); does not capture the new line and hence, skips the Scanner code later.
Scanner scanner = new Scanner(System.in);
int option = scanner.nextInt();
scanner.nextLine(); //clearing the buffer
If you want to scan input fast without getting confused into Scanner class nextLine() method , Use Custom Input Scanner for it .
Code :
class ScanReader {
/**
* #author Nikunj Khokhar
*/
private byte[] buf = new byte[4 * 1024];
private int index;
private BufferedInputStream in;
private int total;
public ScanReader(InputStream inputStream) {
in = new BufferedInputStream(inputStream);
}
private int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) return -1;
}
return buf[index++];
}
public char scanChar(){
int c=scan();
while (isWhiteSpace(c))c=scan();
return (char)c;
}
public int scanInt() throws IOException {
int integer = 0;
int n = scan();
while (isWhiteSpace(n)) n = scan();
int neg = 1;
if (n == '-') {
neg = -1;
n = scan();
}
while (!isWhiteSpace(n)) {
if (n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = scan();
}
}
return neg * integer;
}
public String scanString() throws IOException {
int c = scan();
while (isWhiteSpace(c)) c = scan();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = scan();
} while (!isWhiteSpace(c));
return res.toString();
}
private boolean isWhiteSpace(int n) {
if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true;
else return false;
}
public long scanLong() throws IOException {
long integer = 0;
int n = scan();
while (isWhiteSpace(n)) n = scan();
int neg = 1;
if (n == '-') {
neg = -1;
n = scan();
}
while (!isWhiteSpace(n)) {
if (n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = scan();
}
}
return neg * integer;
}
public void scanLong(long[] A) throws IOException {
for (int i = 0; i < A.length; i++) A[i] = scanLong();
}
public void scanInt(int[] A) throws IOException {
for (int i = 0; i < A.length; i++) A[i] = scanInt();
}
public double scanDouble() throws IOException {
int c = scan();
while (isWhiteSpace(c)) c = scan();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = scan();
}
double res = 0;
while (!isWhiteSpace(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, scanInt());
}
res *= 10;
res += c - '0';
c = scan();
}
if (c == '.') {
c = scan();
double m = 1;
while (!isWhiteSpace(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, scanInt());
}
m /= 10;
res += (c - '0') * m;
c = scan();
}
}
return res * sgn;
}
}
Advantages :
Scans Input faster than BufferReader
Reduces Time Complexity
Flushes Buffer for every next input
Methods :
scanChar() - scan single character
scanInt() - scan Integer value
scanLong() - scan Long value
scanString() - scan String value
scanDouble() - scan Double value
scanInt(int[] array) - scans complete Array(Integer)
scanLong(long[] array) - scans complete Array(Long)
Usage :
Copy the Given Code below your java code.
Initialise Object for Given Class
ScanReader sc = new ScanReader(System.in);
3. Import necessary Classes :
import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
4. Throw IOException from your main method to handle Exception
5. Use Provided Methods.
6. Enjoy
Example :
import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
class Main{
public static void main(String... as) throws IOException{
ScanReader sc = new ScanReader(System.in);
int a=sc.scanInt();
System.out.println(a);
}
}
class ScanReader....
sc.nextLine() is better as compared to parsing the input.
Because performance wise it will be good.
I guess I'm pretty late to the party..
As previously stated, calling input.nextLine() after getting your int value will solve your problem. The reason why your code didn't work was because there was nothing else to store from your input (where you inputted the int) into string1. I'll just shed a little more light to the entire topic.
Consider nextLine() as the odd one out among the nextFoo() methods in the Scanner class. Let's take a quick example.. Let's say we have two lines of code like the ones below:
int firstNumber = input.nextInt();
int secondNumber = input.nextInt();
If we input the value below (as a single line of input)
54 234
The value of our firstNumber and secondNumber variable become 54 and 234 respectively. The reason why this works this way is because a new line feed (i.e \n) IS NOT automatically generated when the nextInt() method takes in the values. It simply takes the "next int" and moves on. This is the same for the rest of the nextFoo() methods except nextLine().
nextLine() generates a new line feed immediately after taking a value; this is what #RohitJain means by saying the new line feed is "consumed".
Lastly, the next() method simply takes the nearest String without generating a new line; this makes this the preferential method for taking separate Strings within the same single line.
I hope this helps.. Merry coding!
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
scan.nextLine();
double d = scan.nextDouble();
scan.nextLine();
String s = scan.nextLine();
System.out.println("String: " + s);
System.out.println("Double: " + d);
System.out.println("Int: " + i);
}
if I expect a non-empty input
avoids:
–  loss of data if the following input is eaten by an unchecked scan.nextLine() as workaround
–  loss of data due to only partially read lines because scan.nextLine() was replaced by scan.next() (enter: "yippie ya yeah")
–  Exceptions that are thrown when parsing input with Scanner methods (read first, parse afterwards)
public static Function<Scanner,String> scanLine = (scan -> {
String s = scan.nextLine();
return( s.length() == 0 ? scan.nextLine() : s );
});
used in above example:
System.out.println("Enter numerical value");
int option = input.nextInt(); // read numerical value from input
System.out.println("Enter 1st string");
String string1 = scanLine.apply( input ); // read 1st string
System.out.println("Enter 2nd string");
String string2 = scanLine.apply( input ); // read 2nd string
Use 2 scanner objects instead of one
Scanner input = new Scanner(System.in);
System.out.println("Enter numerical value");
int option;
Scanner input2 = new Scanner(System.in);
option = input2.nextInt();
In one of my usecase, I had the scenario of reading a string value preceded by a couple of integer values. I had to use a "for / while loop" to read the values. And none of the above suggestions worked in this case.
Using input.next() instead of input.nextLine() fixed the issue. Hope this might be helpful for those dealing with similar scenario.
As nextXXX() methods don't read newline, except nextLine(). We can skip the newline after reading any non-string value (int in this case) by using scanner.skip() as below:
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(x);
double y = sc.nextDouble();
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(y);
char z = sc.next().charAt(0);
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(z);
String hello = sc.nextLine();
System.out.println(hello);
float tt = sc.nextFloat();
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(tt);
Use this code it will fix your problem.
System.out.println("Enter numerical value");
int option;
option = input.nextInt(); // Read numerical value from input
input.nextLine();
System.out.println("Enter 1st string");
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)
To resolve this problem just make a scan.nextLine(), where scan is an instance of the Scanner object. For example, I am using a simple HackerRank Problem for the explanation.
package com.company;
import java.util.Scanner;
public class hackerrank {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
double d = scan.nextDouble();
scan.nextLine(); // This line shall stop the skipping the nextLine()
String s = scan.nextLine();
scan.close();
// Write your code here.
System.out.println("String: " + s);
System.out.println("Double: " + d);
System.out.println("Int: " + i);
}
}
The nextLine() will read enter directly as an empty line without waiting for the text.
Simple solution by adding an extra scanner to consume the empty line:
System.out.println("Enter numerical value");
int option;
option = input.nextInt(); // Read numerical value from input
input.nextLine();
System.out.println("Enter 1st string");
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)
This is a very basic problem for beginner coders in java. The same problem I also have faced when I started java (Self Taught).
Actually, when we take an input of integer dataType, it reads only integer value and leaves the newLine(\n) character and this line(i.e. leaved new line by integer dynamic input )creates the problem when we try to take a new input.
eg. Like if we take the integer input and then after try to take an String input.
value1=sc.nextInt();
value2=sc.nextLine();
the value2 will auto read the newLine character and will not take the user input.
Solution:
just we need to add one line of code before taking the next user input i.e.
sc.nextLine();
or
value1=sc.nextInt();
sc.nextLine();
value2=sc.nextLine();
Note: don't forget to close the Scanner to prevent memory leak;
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
scan.nextLine();//to Ignore the rest of the line after (integer input)nextInt()
double d=scan.nextDouble();
scan.nextLine();
String s=scan.nextLine();
scan.close();
System.out.println("String: " + s);
System.out.println("Double: " + d);
System.out.println("Int: " + i);
Why not use a new Scanner for every reading? Like below. With this approach you will not confront your problem.
int i = new Scanner(System.in).nextInt();
The problem is with the input.nextInt() method - it only reads the int value. So when you continue reading with input.nextLine() you receive the "\n" Enter key. So to skip this you have to add the input.nextLine(). Hope this should be clear now.
Try it like that:
System.out.print("Insert a number: ");
int number = input.nextInt();
input.nextLine(); // This line you have to add (It consumes the \n character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();

Android Collections.min NoSuchElementException

Right now, I have a program that contains this piece of code:
if(Highest6.y == 0){
if(G.y == Collections.min(YUpper) && !notdone){ Highest6 = G; YUpper.remove(Integer.valueOf(G.y)); notdone = true;}
}
When I run it, I get this error:
The most interesting thing that I have identical snippets, which appear in different HighestX.y statements(I have six of them). And this error occurs only in the last one. Anybody knows why this keeps happening? Thanks in advance.
Here's code for my list:
List<Integer> YPoint = new java.util.ArrayList(Arrays.asList(A.y, B.y, C.y, D.y, E.y, F.y, G.y, K.y, Q.y, L.y, M.y, N.y));
List<Integer> YUpper = new java.util.ArrayList(Arrays.asList());
int Classified = 0;
int Highest = 0;
while(Classified != 6){
Highest = Collections.min(YPoint);
YPoint.remove(Integer.valueOf(Highest));
YUpper.add(Integer.valueOf(Highest));
Classified++;
}
I think the problem is that your collection is empty. From the documentation of Collections min returns:
NoSuchElementException - if the collection is empty
That is: your yUpper Arraylist is empty:
Collections.min(YUpper)
And it's empty because you never enter that while loop:
while(Classified != 6){
As classified is 0
Style note: use camelCase for variables & methods. That's the way Java code is meant to be written. Not MyVar but myVar. It's easier to read to Java people.

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