Next keyword in java

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This post eradicates the confusion between next() and nextLine() and gives you a clear idea of how they work.

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.

Look at the output keenly. When the input for c is given as abc def, it will not read the entire input abc def. next() reads only abc and places the cursor after the alphabet c. Now, when value for d is got using next(), it will not get new value. Rather, it takes the value after the current cursor position(since the cursor is after c, def is assigned to d and the cursor stays after f).

Since nextLine() is used for reading input for c, the entire line is read and hence ABC DEF is stored in c and the cursor is placed in the next line. Now when d is read using next() or nextLine(), the cursor will be in the next line and hence the input for d is read and is placed in d (here d gets the input GHI).

So whenever next() is used, make sure nextLine() used to place the cursor in the next line. Hence the above program is also written as,

39 comments:

Hi, thank you for writing such a useful blog, but im a java beginner and i dont know what does the word ‘static’ mean. I mean i can define it (attached to the class instead of the object) but i dont know how to use it.. can u write a blog about it please? thank you!(: Reply Delete

hi buddy! thnx for ur comment. 🙂 as u suggested me to write a blog about ‘static’, i’ve published a post about it! 😉 read it! and i hope you’ll understand. 🙂 comment on my blog about it and for further queries 🙂 Delete

here is a problem?Can anybody explain it.

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public static void main(String[] args) // write your code here
Scanner scanner = new Scanner(System.in);
int numOfPriceTags = scanner.nextInt();
int numOfFruits = scanner.nextInt();
int tagList[] = new int[102];

int count[] = new int[102];
List fruitList = new ArrayList();

String myFruit = scanner.next();//hre is the problem if I use here nextLine then myFruit takes «» for the first time latter it works for the latter iterations

int p = fruitList.indexOf(myFruit);
if (p >= 0)
count[p]++;
else fruitList.add(myFruit);
p = fruitList.indexOf(myFruit);
count[p]++;
>
>
Arrays.sort(count,0,(fruitList.size()));
Arrays.sort(tagList,0,numOfPriceTags);
int maximum = 0;
int minimum = 0;
int k = 0;
for (int j = fruitList.size() — 1; j >= 0; j—) minimum += count[j] * tagList[k];
k++;
maximum += count[j] * tagList[numOfPriceTags — k];

Just like your man explained there the nextInt() does not advance to the next line but rather it stays in as much as next() does not advance. So to make it advance put scanner.nextLine() after the nextInt(), this should work. worked like a charm on my project as well. BTW, thatnks dude for explaining the difference in plain English 🙂 Delete

do
System.out.println(«Enter the directory:»);
p=ob.nextLine();
System.out.println(p);
tester[check]=p;
check++;
for(int i=0;i if(tester[i].equals(p))
System.out.println(«The directory is already present»);
positive=1;
break;
>
else

>
>while(positive!=0);

in the next entry it is showing null pointer exceptio
but if use next(); it executes properly but didnt read the directory with sapces in name. what to do..please help me asap..
thanku Reply Delete

Did you initialise the tester array with right amount of array size? Before assigning something to the array you need to set aside memory using new keyword. Hard to say more without seeing much of the code. Delete

public static void searchString(String[] str,int i)
<
//String search;
System.out.println(«\n\nSearch word: «);
String search=kb.next();
for(int k=0; k if(search.equals(str[k]))
System.out.println(str[k]+» is found at line «+ (k+1));
break;

>else//(!search.equals(str[k]))
System.out.println(search+» is not in the list!please try again: «);
break;// System.out.println(«\nSearch word: «);
> // search=kb.next();

>
im using input=scanner.nextLine() in my input Array of string,problem is
i have to input again a string to find and compare it with my array and im using search=scanner.next()..the prob is my next() cannot read my nextLine in the array which contain a two words. i have to read my input nextLine in my array,what should i do?do i need to replace my second string next() so that it will read?ive tried it but its not working. plss help Reply Delete

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Next keyword in java

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Difference between next() and nextline() in Java

One of the simplest methods for receiving input of the basic data types, also including int, double, and strings, in Java, is to use the Scanner class, which is part of the java.util.package. Time is a limited resource in competitive programming, thus employing the Scanner class is not the most effective solution in these circumstances.

Let’s examine each approach separately using the following example:

next() Method

Input from the user is obtained using the next() function of the Scanner class. A Scanner class instance must be created before the next() method of both the Scanner class can be used. The next() method doesn’t let the user’s input through until it sees a white space (» «). The Scanner’s subsequent entire token is returned as a result.

To comprehend how the next() technique utilizes user input, let’s look at an example.

NextExpl.java

// importing the required classes and packages import java.util.Scanner; // creating the class NextExpl to understand nextLine() method of the Scanner class public class NextExpl < // Driver code public static void main(String args[]) < String input; // create an object for the Scanner class Scanner s = new Scanner(System.in); System.out.println(" Enter any string of your choice : "); // taking the input from the user and storing it in the input input = s.next(); System.out.println(" User input string : \n"+input); // closing the Scanner class object s.close(); >> 

nextLine() Method

Input from the user is also taken using the nextLine() function of the Scanner class. Additionally, we must make an instance of both the Scanner class before we can utilize the nextLine() method. The nextLine() method has the ability to read input all the way across a line. When the user pushes the enter button or line change, it so halts reading the user’s input.

Let’s look at an illustration to understand better how well the nextLine() method works to capture user input.

// importing the required classes and packages import java.util.Scanner; // creating the class NextLineExpl to understand nextLine() method of the Scanner class public class NextLineExpl < // Driver code public static void main(String args[]) < String input; // create an object for the Scanner class Scanner s = new Scanner(System.in); System.out.println(" Enter any string of your choice : "); // taking the input from the user and storing it in the input input = s.nextLine(); System.out.println(" User input string : \n"+input); // closing the Scanner class object s.close(); >> 

Difference between the next() and the nextLine()

The following table compares the next() and nextLine() functions. It describes the differences between them:

S.no	next()	nextLine()
1	Up until the space character, it reads the input.	Up to the line change, it reads input.
2	It cannot read words with spaces between them.	This technique can be used to read words with spaces in them.
3	When it encounters a blank space, it stops reading its input.	When it encounters the character «n» or is pressed, it stops reading the input.
4	The technique moves the pointer to the same line after reading the input.	The procedure moves the pointer to the following line after reading the input.
5	The next() method’s escape sequence is space.	The nextLine() method’s escape sequence is «n.»
6	Syntax: Scanner s = new Scanner(System.in); s.next();	Syntax: Scanner s = new Scanner(System.in); s.nextLine();

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Difference between next() and hasNext() in java collections?

Java provides Iterator and ListIterator classes to retrieve the elements of the collection objects.

The hasNext() method

The hasNext() method of these interfaces returns true if the collection object has the next element else it returns false.

Example

import java.util.ArrayList; import java.util.Iterator; public class hasNextExample < public static void main(String[] args)< ArrayList list = new ArrayList(); //Instantiating an ArrayList object list.add("JavaFX"); list.add("Java"); Iterator it = list.iterator(); System.out.println(it.hasNext()); it.next(); System.out.println(it.hasNext()); it.next(); System.out.println(it.hasNext()); > >

Output

The next() method

The next() methods of the Iterator and ListIterator returns the next element of the collection.

Example

import java.util.ArrayList; import java.util.Iterator; public class nextExample < public static void main(String[] args)< ArrayList list = new ArrayList(); //Instantiating an ArrayList object list.add("JavaFX"); list.add("Java"); list.add("WebGL"); list.add("OpenCV"); list.add("OpenNLP"); list.add("JOGL"); list.add("Hadoop"); list.add("HBase"); list.add("Flume"); list.add("Mahout"); list.add("Impala"); System.out.println("Contents of the array list (first to last): "); Iterator it = list.iterator(); while(it.hasNext()) < System.out.println(it.next()); >> >

Output

Contents of the array list (first to last): JavaFX Java WebGL OpenCV OpenNLP JOGL Hadoop HBase Flume Mahout Impala

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