I have specified the size of the array using user input but my for loop is taking input only size-1 time.
import java.util.*;
public class Main{
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int time=sc.nextInt();
String input[]=new String[time];
for(int i=0;i<time;i++)
{
input[i]=sc.nextLine();
}
for(int i=0;i<time;i++)
{
int len=input[i].length();
if(len>4)
{
System.out.println(input[i].charAt(0)+ Integer.toString(len-2)+input[i].charAt(len-1));
}
else
System.out.println(input[i]);
}
}
}
i changed my code and it is working fine
changed
int time=sc.nextInt();
with
int time=Integer.parseInt(sc.nextLine());
but i don't know the reason behind this . Please can anyone explain me
The Scanner.nextInt() method scans the next token of the input as an int, not the line. For example, if you give an int input and then hit enter, then it takes only the int, not the carriage return.
If you give a sample input like this:
2 xyz //hit enter and give the next input
abc
You'll see the nextInt() will take the 2 as input from that line and the upcoming first iteration for Scanner.nextLine() will consider the xyz as first input and in the next iteration, as we gave abc, it will be considered as the second. All these time you're code was working, but you couldn't see as it was taking the empty string as the first input due to the carriage return from the previous line.
However, The Scanner.nextLine() takes the whole line as input, along with the carriage return and then parses the int to the integer, so, you get the next lines for the string input for your array.
Hope that makes everything clear.
The problem is with the nextLine() method used in the first for loop. Because the method advances the scanner to the next line and returns the input that was skipped, it kind of "eats" one of your loop iterations and it ends up allowing you to input time - 1 elements into the array instead of time amount of elements. If you just use sc.next() instead, the program works perfectly fine, so you don't need to use
int time=Integer.parseInt(sc.nextLine());
as it may be a bit more complicated (in my opinion) than just replacing nextLine() with next(). Here is the code:
import java.util.*;
public class Main{
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int time = sc.nextInt();
String input[] = new String[time];
for(int i = 0;i < time;i++)
{
input[i] = sc.next();
}
for(int i = 0;i < time;i++)
{
int len = input[i].length();
if(len > 4)
{
System.out.println(input[i].charAt(0) + Integer.toString(len - 2) + input[i].charAt(len - 1));
}
else
System.out.println(input[i]);
}
sc.close();
}
}
Okay so I'm having a slight problem with scanner advancing an extra line. I have a file that has many lines containing integers each separated by one space. Somewhere in the file there is a line with no integers and just the word "done".
When done is found we exit the loop and print out the largest prime integer that is less than each given integer in each line(if integer is already prime do nothing to it). We do this all the way up until the line with "done".
My problem: lets say the file contains 6 lines and on the 6th line is the word done. My output would skip lines 1, 3 and 5. It would only return the correct values for line 2 and 4.
Here's a snippet of code where I read the values in:
Scanner in = new Scanner(
new InputStreamReader(socket.getInputStream()));
PrintStream out = new PrintStream(socket.getOutputStream());
while(in.nextLine() != "done"){
String[] arr = in.nextLine().split(" ");
Now I sense the problem is that the nextLine call in my loop advances the line and then the nextline.split call also advances the line. Thus, all odd number lines will be lost. Would there be another way to check for "done" without advancing a line or is there a possible command I could call to somehow reset the scanner back to the start of the loop?
The problem is you have 2 calls to nextLine() try something like this
String line = in.nextLine();
while (!"done".equals(line)) {
String[] arr = line.split(" ");
// Process the line
if (!in.hasNextLine()) {
// Error reached end of file without finding done
}
line = in.nextLine();
}
Also note I fixed the check for "done" you should be using equals().
I think you are looking for this
while(in.hasNextLine()){
String str = in.nextLine();
if(str.trim().equals("done"){
break;
}else{
String[] arr = str.split("\\s+");
//then do whatever you want to do
}
}
My program seems stuck in the middle of my while loop, without crashing and without looping infinitely. It just stops.
The loop runs for as many input as the user provides, but then does not move on the the next line of code.
It is my first time using the hasNextDouble() in java. Am I doing it right?
Here is the while loop in question:
System.out.print("Grades (separated by a space)");
while(in.hasNextDouble())
{
student1.addGrade(in.nextDouble());
}
And here is a little bit more of my code:
Scanner in = new Scanner(System.in);
String input = "";
Student student1 = new Student();
GradeBook book = new GradeBook();
// Sets the name of the first student
System.out.print("Name: ");
input = in.nextLine();
student1.setNames(input);
// Sets the grades of the first student
System.out.print("Grades (separated by a space)");
while(in.hasNextDouble()){
student1.addGrade(in.nextDouble());
}
// Put the student into the GradeBook
book.addStudent(student1);
// Prints the report
System.out.print(book.reportGrades());
You state you want space separated input, in a single line. I would suggest taking the input as a String, and then splitting it up, like
Scanner in = new Scanner(System.in);
String line = in.nextLine();
for(String s : line.split(" ")){
student1.addGrade(Double.parseDouble(s)); //gives exception if value is not double
}
Scanner.hasNextDouble will continue to return true, until you enter a non-Double value.
With hasNext() you check if there is anything, then with hasNextDouble() check if the next input can be cast to double. You read the value with next(), but the value is still a string, so you need to parse it to double.
Also, you need a way to get out of the loop with break when input is no longer a number.
while (in.hasNext()) {
if (in.hasNextDouble()) {
student1.add(Double.parseDouble(in.next()));
} else {
break;
}
}
What is the main difference between next() and nextLine()?
My main goal is to read the all text using a Scanner which may be "connected" to any source (file for example).
Which one should I choose and why?
I always prefer to read input using nextLine() and then parse the string.
Using next() will only return what comes before the delimiter (defaults to whitespace). nextLine() automatically moves the scanner down after returning the current line.
A useful tool for parsing data from nextLine() would be str.split("\\s+").
String data = scanner.nextLine();
String[] pieces = data.split("\\s+");
// Parse the pieces
For more information regarding the Scanner class or String class refer to the following links.
Scanner: http://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html
String: http://docs.oracle.com/javase/7/docs/api/java/lang/String.html
next() can read the input only till the space. It can't read two words separated by a space. Also, next() places the cursor in the same line after reading the input.
nextLine() reads input including space between the words (that is, it reads till the end of line \n). Once the input is read, nextLine() positions the cursor in the next line.
For reading the entire line you can use nextLine().
From JavaDoc:
A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace.
next(): Finds and returns the next complete token from this scanner.
nextLine(): Advances this scanner past the current line and returns the input that was skipped.
So in case of "small example<eol>text" next() should return "small" and nextLine() should return "small example"
The key point is to find where the method will stop and where the cursor will be after calling the methods.
All methods will read information which does not include whitespace between the cursor position and the next default delimiters(whitespace, tab, \n--created by pressing Enter). The cursor stops before the delimiters except for nextLine(), which reads information (including whitespace created by delimiters) between the cursor position and \n, and the cursor stops behind \n.
For example, consider the following illustration:
|23_24_25_26_27\n
| -> the current cursor position
_ -> whitespace
stream -> Bold (the information got by the calling method)
See what happens when you call these methods:
nextInt()
read 23|_24_25_26_27\n
nextDouble()
read 23_24|_25_26_27\n
next()
read 23_24_25|_26_27\n
nextLine()
read 23_24_25_26_27\n|
After this, the method should be called depending on your requirement.
What I have noticed apart from next() scans only upto space where as nextLine() scans the entire line is that next waits till it gets a complete token where as nextLine() does not wait for complete token, when ever '\n' is obtained(i.e when you press enter key) the scanner cursor moves to the next line and returns the previous line skipped. It does not check for the whether you have given complete input or not, even it will take an empty string where as next() does not take empty string
public class ScannerTest {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int cases = sc.nextInt();
String []str = new String[cases];
for(int i=0;i<cases;i++){
str[i]=sc.next();
}
}
}
Try this program by changing the next() and nextLine() in for loop, go on pressing '\n' that is enter key without any input, you can find that using nextLine() method it terminates after pressing given number of cases where as next() doesnot terminate untill you provide and input to it for the given number of cases.
next() and nextLine() methods are associated with Scanner and is used for getting String inputs. Their differences are...
next() can read the input only till the space. It can't read two words separated by space. Also, next() places the cursor in the same line after reading the input.
nextLine() reads input including space between the words (that is, it reads till the end of line \n). Once the input is read, nextLine() positions the cursor in the next line.
import java.util.Scanner;
public class temp
{
public static void main(String arg[])
{
Scanner sc=new Scanner(System.in);
System.out.println("enter string for c");
String c=sc.next();
System.out.println("c is "+c);
System.out.println("enter string for d");
String d=sc.next();
System.out.println("d is "+d);
}
}
Output:
enter string for c
abc def
c is abc
enter string for d
d is def
If you use nextLine() instead of next() then
Output:
enter string for c
ABC DEF
c is ABC DEF
enter string for d
GHI
d is GHI
In short: if you are inputting a string array of length t, then Scanner#nextLine() expects t lines, each entry in the string array is differentiated from the other by enter key.And Scanner#next() will keep taking inputs till you press enter but stores string(word) inside the array, which is separated by whitespace.
Lets have a look at following snippet of code
Scanner in = new Scanner(System.in);
int t = in.nextInt();
String[] s = new String[t];
for (int i = 0; i < t; i++) {
s[i] = in.next();
}
when I run above snippet of code in my IDE (lets say for string length 2),it does not matter whether I enter my string as
Input as :- abcd abcd or
Input as :-
abcd
abcd
Output will be like
abcd
abcd
But if in same code we replace next() method by nextLine()
Scanner in = new Scanner(System.in);
int t = in.nextInt();
String[] s = new String[t];
for (int i = 0; i < t; i++) {
s[i] = in.nextLine();
}
Then if you enter input on prompt as -
abcd abcd
Output is :-
abcd abcd
and if you enter the input on prompt as
abcd (and if you press enter to enter next abcd in another line, the input prompt will just exit and you will get the output)
Output is:-
abcd
From javadocs
next() Returns the next token if it matches the pattern constructed from the specified string.
nextLine() Advances this scanner past the current line and returns the input that was skipped.
Which one you choose depends which suits your needs best. If it were me reading a whole file I would go for nextLine until I had all the file.
From the documentation for Scanner:
A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace.
From the documentation for next():
A complete token is preceded and followed by input that matches the delimiter pattern.
Just for another example of Scanner.next() and nextLine() is that like below :
nextLine() does not let user type while next() makes Scanner wait and read the input.
Scanner sc = new Scanner(System.in);
do {
System.out.println("The values on dice are :");
for(int i = 0; i < n; i++) {
System.out.println(ran.nextInt(6) + 1);
}
System.out.println("Continue : yes or no");
} while(sc.next().equals("yes"));
// while(sc.nextLine().equals("yes"));
Both functions are used to move to the next Scanner token.
The difference lies in how the scanner token is generated
next() generates scanner tokens using delimiter as White Space
nextLine() generates scanner tokens using delimiter as '\n' (i.e Enter
key presses)
A scanner breaks its input into tokens using a delimiter pattern, which is by default known the Whitespaces.
Next() uses to read a single word and when it gets a white space,it stops reading and the cursor back to its original position.
NextLine() while this one reads a whole word even when it meets a whitespace.the cursor stops when it finished reading and cursor backs to the end of the line.
so u don't need to use a delimeter when you want to read a full word as a sentence.you just need to use NextLine().
public static void main(String[] args) {
// TODO code application logic here
String str;
Scanner input = new Scanner( System.in );
str=input.nextLine();
System.out.println(str);
}
I also got a problem concerning a delimiter.
the question was all about inputs of
enter your name.
enter your age.
enter your email.
enter your address.
The problem
I finished successfully with name, age, and email.
When I came up with the address of two words having a whitespace (Harnet street) I just got the first one "harnet".
The solution
I used the delimiter for my scanner and went out successful.
Example
public static void main (String args[]){
//Initialize the Scanner this way so that it delimits input using a new line character.
Scanner s = new Scanner(System.in).useDelimiter("\n");
System.out.println("Enter Your Name: ");
String name = s.next();
System.out.println("Enter Your Age: ");
int age = s.nextInt();
System.out.println("Enter Your E-mail: ");
String email = s.next();
System.out.println("Enter Your Address: ");
String address = s.next();
System.out.println("Name: "+name);
System.out.println("Age: "+age);
System.out.println("E-mail: "+email);
System.out.println("Address: "+address);
}
The basic difference is next() is used for gettting the input till the delimiter is encountered(By default it is whitespace,but you can also change it) and return the token which you have entered.The cursor then remains on the Same line.Whereas in nextLine() it scans the input till we hit enter button and return the whole thing and places the cursor in the next line.
**
Scanner sc=new Scanner(System.in);
String s[]=new String[2];
for(int i=0;i<2;i++){
s[i]=sc.next();
}
for(int j=0;j<2;j++)
{
System.out.println("The string at position "+j+ " is "+s[j]);
}
**
Try running this code by giving Input as "Hello World".The scanner reads the input till 'o' and then a delimiter occurs.so s[0] will be "Hello" and cursor will be pointing to the next position after delimiter(that is 'W' in our case),and when s[1] is read it scans the "World" and return it to s[1] as the next complete token(by definition of Scanner).If we use nextLine() instead,it will read the "Hello World" fully and also more till we hit the enter button and store it in s[0].
We may give another string also by using nextLine(). I recommend you to try using this example and more and ask for any clarification.
The difference can be very clear with the code below and its output.
public static void main(String[] args) {
List<String> arrayList = new ArrayList<>();
List<String> arrayList2 = new ArrayList<>();
Scanner input = new Scanner(System.in);
String product = input.next();
while(!product.equalsIgnoreCase("q")) {
arrayList.add(product);
product = input.next();
}
System.out.println("You wrote the following products \n");
for (String naam : arrayList) {
System.out.println(naam);
}
product = input.nextLine();
System.out.println("Enter a product");
while (!product.equalsIgnoreCase("q")) {
arrayList2.add(product);
System.out.println("Enter a product");
product = input.nextLine();
}
System.out.println();
System.out.println();
System.out.println();
System.out.println();
System.out.println("You wrote the following products \n");
for (String naam : arrayList2) {
System.out.println(naam);
}
}
Output:
Enter a product
aaa aaa
Enter a product
Enter a product
bb
Enter a product
ccc cccc ccccc
Enter a product
Enter a product
Enter a product
q
You wrote the following products
aaa
aaa
bb
ccc
cccc
ccccc
Enter a product
Enter a product
aaaa aaaa aaaa
Enter a product
bb
Enter a product
q
You wrote the following products
aaaa aaaa aaaa
bb
Quite clear that the default delimiter space is adding the products separated by space to the list when next is used, so each time space separated strings are entered on a line, they are different strings.
With nextLine, space has no significance and the whole line is one string.
import java.util.Scanner;
public class ClassList {
static final int SIZE = 10;
public static void main(String Args[]) {
Student[] students = new Student[SIZE];
Scanner s = new Scanner(System.in);
String name, id;
float gpa;
do {
System.out.println("Enter a Student");
name = s.nextLine();
System.out.println("Enter a ID");
id = s.nextLine();
System.out.println("Enter a GPA");
gpa = s.nextFloat();
} while(!name.equals("quit"));
}
}
basically, what happened when you run this program is that it will run one fine, then on the 2nd iteration, it will ask for a student name. but it will just skip over where the user should input it and go onto "get ID". How do i stop that.
In the first iteration, nextFloat() doesn't consume any new line characters. In the second iteration, the nextLine() method will consume the newline characters left from the first iteration. As a result, it appears that the nextLine() method in the second iteration skips over user input, while it's just consuming the newline characters left from the previous iteration.
One way to fix this problem is to call nextLine() after nextFloat() to consume the leftover new line character.
You can place
s.nextLine();
at the end of the loop to consume the newline character. Currently Scanner.getFloat() is passing the newline character through to your statement:
name = s.nextLine();
which is not blocking as it has been supplied with this character.