This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Scanner issue when using nextLine after nextXXX [duplicate]
Closed 6 years ago.
In this snippet what I am doing taking three different types of variables and adding them and then printing then down.
public static void main(String[] args) {
int i = 4;
double d = 4.0;
String s = "HackerRank ";
Scanner scan = new Scanner(System.in);
int firstVariable = 0;
double secondVariable = 0.0;
String theString = "";
firstVariable = scan.nextInt();
secondVariable = scan.nextDouble();
theString = scan.nextLine();
System.out.println(firstVariable+i);
System.out.println(secondVariable+d);
System.out.println(s+""+theString);
}
I am providing input for firstVariable hitting enter and then providing the input for secondVariable and now as soon as I hit enter theString is capturing that value(I know it should capture it).
EDIT: in this case how should I provide the input to theString without as well as with space ?
I did try something like this,
while(scan.hasNext())
theString = scan.nextLine();
But it didn't work either.
Simply check scan.nextLine() is empty if so call scan.nextLine() again as next:
secondVariable = scan.nextDouble();
theString = scan.nextLine().trim();
if (theString.isEmpty()) {
theString = scan.nextLine();
}
Another approach with a pattern:
firstVariable = scan.nextInt();
secondVariable = scan.nextDouble();
theString = scan.next("\\w+");
There are two ways to solve your problem:
The first one is to use
nextLine()
each time you read from the user like this:
int firstVariable = scan.nextInt();
scan.nextLine();
double secondVariable = scan.nextDouble();
scan.nextLine();
String theString = scan.nextLine();
The second one is to parse the integer and double value from nextLine() like this:
int firstVariable = Integer.parseInt(scan.nextLine());
Double secondVariable = Double.parseDouble(scan.nextLine());
String theString = scan.nextLine();
You need to add an extra scan.nextLine(); before your String theString, to capture the Enter after the double. Like this:
firstVariable = scan.nextInt();
secondVariable = scan.nextDouble();
scan.nextLine();
theString = scan.nextLine();
Just as an advice to reduce your code, there's no need to add these:
int firstVariable = 0;
double secondVariable = 0.0;
String theString = "";
Just add the type to the variables to capture the scan:
int firstVariable = scan.nextInt();
double secondVariable = scan.nextDouble();
scan.nextLine();
String theString = scan.nextLine();
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 7 years ago.
I don't know why, but the below code makes the user run the code again, whether they choose to or not. I've tried many things, but it doesn't work correctly.
Thanks!
public static void main (String [ ] args)
{
boolean a = true;
while (a)
{
Scanner scan = new Scanner(System.in);
System.out.print("Enter an integer: ");
int x = scan.nextInt();
System.out.print("\n\nEnter a second integer: ");
int z = scan.nextInt();
System.out.println();
System.out.println();
binaryConvert1(x, z);
System.out.println("\n\nWould you like to run this code again? Enter \"Y\" or \"N\".");
System.out.print("Enter your response here: ");
String RUN = scan.nextLine();
String run = RUN.toLowerCase();
if (run.equals("n"))
{
a = false;
}
System.out.println();
System.out.println();
}
System.out.println("Goodbye.");
}
Scanner.nextInt() doesn't consume the line ending characters from the buffer, which is why when you read the value of the "yes/no" question with scan.nextLine(), you'll receive an empty string instead of the value the user entered.
A simple way to fix this is to explicitly parse the integer from raw lines using Integer.parseInt():
System.out.print("Enter an integer: ");
int x = Integer.parseInt(scan.nextLine());
System.out.print("\n\nEnter a second integer: ");
int z = Integer.parseInt(scan.nextLine());
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)
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();
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();
Use BufferedReader class to input string, this will not create problems
This question already has an answer here:
How to use java.util.Scanner to correctly read user input from System.in and act on it?
(1 answer)
Closed 7 years ago.
I'm trying to teach my friend Java. I tried this simple calculator.
public static void main(String[] args) {
boolean powerOn = true;
while(powerOn) {
#SuppressWarnings("resource")
Scanner userInput = new Scanner(System.in);
System.out.println("Welcome to the calculator\nPlease enter a number (Enter 3.14 for PI)");
double firstNumber = userInput.nextDouble();
if(firstNumber == 3.14) firstNumber = Math.PI;
System.out.println("Please enter an operation(+,-,*,/, Square Root)");
String operation = userInput.next();
if(operation.equalsIgnoreCase("Square Root")) System.out.println(Math.sqrt(firstNumber));
else {
System.out.println("Please enter another number");
double secondNumber = userInput.nextDouble();
if(secondNumber == 3.14) secondNumber = Math.PI;
if (operation.equals("+")) {
if(firstNumber == 9 && secondNumber == 10) System.out.println("21");
else System.out.println(firstNumber+secondNumber);
}
else if (operation.equals("-")) System.out.println(firstNumber-secondNumber);
else if (operation.equals("*")) System.out.println(firstNumber*secondNumber);
else if (operation.equals("/")) System.out.println(firstNumber/secondNumber);
}
System.out.println("Power off?");
String off = userInput.next();
if(off.contains("y")) System.exit(1);
}
}
If you do square root, it prints enter another number and then throws an exception... I know what's happening, but why and how do I prevent it?
And just a side note can someone explain to me the difference between Scanner#next() and Scanner#nextLine?
next() returns the string before the space while nextLine() as you might have guessed returns the whole line .
You are probably also getting the error because of the same reason as your string "Square Root" has a space and hence operation variable always have "Sqaure" value assigned to it rather than "Square Root".
This pushes it to the else block instead of if.
To fix your issue add a scan.nextLine();
Before you scan in the operation. And while scanning in the operation use a scan.nextLine(); as well instead of the scan.next();
Think of the scan as the location of the cursor. A scan.nextDouble(); will place the cursor after the last digit of the number, and a scan.nextLine(); will scan in what ever is inbetween the cursor's location and the beginning of the next line. scan.next(); won't scan in the entire line.
You need to update the following code
String operation = userInput.next();
with following
String operation = userInput.nextLine();
The reason being the next() will only return what comes before a space. nextLine() automatically moves the scanner down after returning the current line.
In your current code, when you try to get the square root, the operation will just store "Square" instead of "Square Root" and therefore the following if condition isn't satisfied
if(operation.equalsIgnoreCase("Square Root"))
and the code goes into else condition and asks for another number and tries to process it and in the else block it throws error on this line
double secondNumber = userInput.nextDouble();
Apart from changing the next() to nextLine() in your code, you need to update following
double firstNumber = userInput.nextDouble();
with
double firstNumber = userInput.nextDouble();
userInput.nextLine();
to ensure that the scanner doesn't skip your nextLine() because in nextDouble() method only the double is consumed and the next line characters (\n) aren't.
Similarly update
double secondNumber = userInput.nextDouble();
with
double secondNumber = userInput.nextDouble();
userInput.nextLine();
For more info about skipping of nextLine() method while using nextDouble() refer this link.
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 9 years ago.
When I execute my code my program is terminated without scanning the string.
double x, y;
String s;
Scanner scan = new Scanner(System.in);
System.out.println("Enter Number: ");
x = scan.nextDouble();
System.out.println("Enter Number 2: ");
y = scan.nextDouble();
System.out.println("Enter Operater: x,+,/,-");
s = scan.nextLine();
if(s.equals("x"))
{
System.out.print(x * y);
}
else if(s.equals("+"))
{
System.out.print(x + y);
}
else if(s.equals("/"))
{
System.out.print(x / y);
}
else if(s.equals("-"))
{
System.out.print(x - y);
}
scan.close();
my program ends before s = scan.nextline();
How come it ends before?
End of line you leave in the buffer.
next( ) reads a token from the buffer until the next white space, while nextLine( ) reads up to \n
...
System.out.print("Enter Number 2: ");
y = scan.nextDouble();
System.out.print("Enter Operater: x,+,/,-");
s = scan.next();
...
Enter Number: 1
Enter Number 2: 2
Enter Operater: x,+,/,--
-1.0
The user input, optimally, would look like this:
-CURSOR HERE- num1 NEWLINE
num2 NEWLINE
operator NEWLINE
If you do several nextDouble() calls, the program will read one double first.
num1 -CURSOR HERE- NEWLINE
num2 NEWLINE
operator NEWLINE
Then, the user must type an enter to submit the input, so the second nextDouble() can't find anything to read, since there is no number directly after the cursor. It needs a nextLine() to absorb the newline.
Unfortunately, you have a nextLine() in the wrong place, which absorbs the newline.
num1 NEWLINE
-CURSOR HERE- num2 NEWLINE
operator NEWLINE
So, your program absorbs the double in the first nextDouble(), nothing in the second, and the newline in nextLine().
To fix this, put a scan.nextLine() right after each nextDouble(). You don't have to read the nextLine() calls into anything, other than the one that has the operator.