Long input in java

Java java take long as input code example

You can take input as text and then parse it using BigInteger. input long in java Solution 1: You can’t parse this string into a long, it’s too great (greater than Long. as given below. here we have input string contains time.

How to take a input of very big length in java

You are looking for the BigInteger class, which is an arbitrary-precision integer class.

You can create one with a String argument, like so:

String longIntegerString = "10000000000000000000000000000000000000000000000000000000000000000"; BigInteger bigInteger = new BigInteger(longIntegerString); 

The class contains add(), subtract(), multiply(), divide(), mod() and other mathematical functions.

Yes, use BigInteger to store large input.

You can take input as text and then parse it using BigInteger.

Take long input in java

input long in java

Scanner input = new Scanner(System.in); long num = input.nextLong();

Take double as input java Code Example, import java.util.Scanner; Scanner input = new Scanner(System.in); double choice = input.nextDouble();

Parsing long fails for large input in Java

You can’t parse this string into a long, it’s too great (greater than Long.MAX_VALUE), you need BigInteger :

BigInteger bi = new BigInteger(st.nextToken()); 

Following your edit :

Don’t try to iterate over a bigInteger : if it’s too great to fit in a long, the loop will be too long for your time. Compare it with a reasonable limit and if it’s smaller, then get it as int and enter your loop:

 BigInteger MAX = new BigInteger("1000000"); if (bi.compareTo(MAX) <0) < int N = bi.intValue(); for (int i=0; i> 

If throws NumberFormatException because 248614876148768372675689568324619856329856295619253291561358926935829358293587932857923857934572895729511 is greater than Long.MAX_VALUE (which is 9223372036854775807 ).

As pointed out by others the valuse is too big for long However another problem is that there can be no spaces in the string which is being converted. There is a handy post on removeing white space from a string here

«The sequence of characters following an (optional) negative sign and/or radix specifier («0x», «0X», «#», or leading zero) is parsed as by the Long.parseLong method with the indicated radix (10, 16, or 8). This sequence of characters must represent a positive value or a NumberFormatException will be thrown. The result is negated if first character of the specified String is the minus sign. No whitespace characters are permitted in the String .» From the Docs

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How to take Time Input in a line in Java?

This for the above problem , regular expression is very helpful to find and collect the value from input string/text. as given below.

public static void main(String[] args) < String input = "12:00 AM 11:42 PM"; ArrayListlist = new ArrayList(); Pattern pattern = Pattern.compile("[\\d:]*[ ](AM|PM)"); //Matching the compiled pattern in the String Matcher matcher = pattern.matcher(input); while (matcher.find()) < list.add(matcher.group()); >Iterator it = list.iterator(); System.out.println("List of matches: "); while(it.hasNext()) < System.out.println(it.next()); >> 

here we have input string contains time. then we are using regex to find the pattern one the match found we are collecting that into list.

There is lots of ways of doing this. I would use Regular Expression.

 String input = "12:00 AM 11:42 PM"; String pattern = "(\\d+:\\d+ AM) (\\d+:\\d+ PM)"; Pattern r = Pattern.compile(pattern); Matcher m = r.matcher(input); if(m.find())

This gives you two strings am and pm .

Then you can use SimpleDateFormat to parse them as Date .

SimpleDateFormat simpleDateFormat = new SimpleDateFormat("HH:mm aa"); 

The format you are looking for is: hh:mm a .

Symbol	Desc	Type	Example
h	clock-hour-of-am-pm (1-12)	number	12
m	minute-of-hour	number	30
a	am-pm-of-day	text	PM

How to Take Input From User in Java?, There are two ways by which we can take input from the user or from a file. BufferedReader Class. Scanner Class. 1. BufferedReader. It is a simple class that is used to read a sequence of characters. It has a simple function that reads a character another read which reads, an array of characters, and a readLine () …

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Java: Accepting Long Input — A Guide

Below programs illustrate the above function: Program 1: Output: Program 2:Output: Program 3: To demonstrate InputMismatchException Output: Program 4: To demonstrate NoSuchElementExceptionOutput: Program 5: To demonstrate IllegalStateException Output: Reference: https://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#nextLong(int) Solution: Just use the method instead of using . Your code will look like this after the change: Solution 1: This will never finish executing until you do Ctrl+D, because the input here is , which technically doesn’t have an end «out-of-the-box»: Solution 2: I tried your code and it is working.

How to read a long value from standard input in java?

Employ the technique identified as nextLong .

Number of Digits in an Integer in Java, int length = 0; long temp = 1; while (temp <= number) < length++; temp *= 10; >return length;. In this code, temp *= 10 is the same as writing

Scanner nextLong() method in Java with Examples

The method nextLong(radix) in the java.util.Scanner class reads the subsequent token of the input as a long. Upon successful translation, the scanner moves past the matching input. When the parameter radix is not provided, it functions similarly to nextLong(radix) with the default radix assumed.

The function’s parameter, called «radix», is utilized to interpret the token as a long value.

The output of this function is a long that has been scanned from the input.

The function has three exceptions that are outlined in the following manner:

• The InputMismatchException occurs when the subsequent token either falls outside the range or does not conform to the Integer regular expression.
• The NoSuchElementException is raised when there is no more input available.
• The IllegalStateException is thrown when attempting to use a closed scanner.

The programs below demonstrate the function mentioned above.

The following codes are listed below: — // Java program to illustrate the — // nextLong() method of Scanner class in Java — // without parameter — — import and java.util.*; — — public to GFG1 < - to main(String[] argv) - to Exception - and < - - to ; - - and // create a new scanner - and // with the specified String Object - to Scanner(s); - - to (scanner.hasNext()) < - - and // if the next is a Long, - and // print found and the Long - to (scanner.hasNextLong()) < - to "Found Long value :" - and + scanner.nextLong()); - and >— — and // if no Long is found, — and // print «Not Found:» and the token — to < - to "Not found Long value :" - and + scanner.next()); - and >— and > — and scanner.close(); — and > — >

Not found Long value :Gfg Found Long value :9 Not found Long value :+ Found Long value :6 Not found Long value := Not found Long value :12.0

The following codes are listed below: // Java program to illustrate the // nextLong() method of Scanner class in Java // with parameter import and java.util.*; public and class and GFG1 < and public and static and void and main(String[] argv) and throws and Exception and < and String s = and "Gfg 9 + 6 = 12.0" and ; and // create a new scanner and // with the specified String Object and Scanner scanner = and new and Scanner(s); and while and (scanner.hasNext()) < and // if the next is a Long, and // print found and the Long and if and (scanner.hasNextLong()) < and System.out.println( and "Found Long value :" and + scanner.nextLong( and 12 and )); and >and // if no Long is found, and // print «Not Found:» and the token and else and < and System.out.println( and "Not found Long value :" and + scanner.next()); and >and > and scanner.close(); and > >

Not found Long value :Gfg Found Long value :9 Not found Long value :+ Found Long value :6 Not found Long value := Not found Long value :12.0

Program 3 aims to showcase the occurrence of InputMismatchException.

The following codes are listed: — // Java program to illustrate the , // nextLong() method of Scanner class in Java , // InputMismatchException , , import , java.util.*; , , public , class , GFG1 < - , public , static , void , main(String[] argv) - , throws , Exception - , < - - , try , < - - , String s = , "Gfg 9 + 6 = 12.0" , ; - - , // create a new scanner - , // with the specified String Object - , Scanner scanner = , new , Scanner(s); - - , while , (scanner.hasNext()) < - - , // if the next is a Long, - , // print found and the Long - , // since the value 60 is out of range - , // it throws an exception - - , System.out.println( , "Next Long value :" - , + scanner.nextLong()); - , >— , scanner.close(); — , > — , catch , (Exception e) < - , System.out.println( , "Exception thrown: " , + e); - , >— , > — >

Exception thrown: java.util.InputMismatchException

The fourth program aims to showcase the occurrence of NoSuchElementException.

Not found Long value :Gfg Exception thrown: java.util.NoSuchElementException

Program number 5 showcases how an IllegalStateException can be triggered.

The text contains a list of 101 MSDT codes, ranging from MSDT code 1 to MSDT code 101, which can be rewritten as a list of MSDT codes numbered from 1 to 101.

Scanner Closed Trying to get next Long value Exception thrown: java.lang.IllegalStateException: Scanner closed

The documentation for the Java SE 7 API provides a reference for the Scanner class method nextLong(int). The reference can be found at the following URL: https://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#nextLong(int)

Java Program to Convert String to Long, Using the parseLong() method of the Long class; Using valueOf() method of long class; Using constructor of Long class. Illustrations: Input

How to output a long string from Scanner input in Java [duplicate]

Utilize Scanner.nextLine() as an alternative to Scanner.next() .

As per the documentation, it can be seen in Scanner.next() .

Retrieves and delivers the subsequent token that is fully formed from the scanner.

Regarding Scanner.nextLine() , it will solely analyze the initial word, which is also referred to as the token. However, according to the documentation, Scanner.nextLine() functions differently.

This function retrieves the remaining portion of the present line, without considering the line terminator located at its conclusion.

This implies that it will scan the complete sentence.

After implementing the modification, the appearance of your code will be altered in the following manner.

// . System.out.println("Enter room name " + (i + 1)); rooms[i] = input.nextLine(); // . 

Java.util.Arrays.fill(long[], long) Method, Java.util.Arrays.fill(long[], long) Method, The java.util.Arrays.fill(long[] a, long val) method assigns the specified long value to each element of the

Java code taking input is taking a long time to execute

To stop the endless execution of System.in , you need to use the Ctrl+D command since it lacks a pre-defined end point.

Your code is functional, but I am uncertain about the input you provided.

To view comments, I added another «System out».

 import java.util.*; public class Main < public static void main(String [] args)< System.out.println("Show"); Scanner input = new Scanner(System.in); System.out.println("In:"); int x = input.nextInt(); System.out.println("Out"); int sum = 0; for(int I = 1; I > System.out.println ("The sum of the factors is " + sum); > > 

The result obtained when 12345 is provided as input is as follows.

 Show In: 12345 Out The sum of the factors is 19776 

It seems like you made a mistake that I used to often make while running command line programs in Eclipse or similar software. Despite repeatedly clicking the «Run» button, I was not able to see any output and became frustrated.

After launching the debugger, I proceeded to go through the entire process, and observed that it halts at a specific line.

Following that, I came to the realization of my foolishness as the console was patiently waiting for my input.

At the specific line where the program runs, it halts and awaits your input followed by the Enter key. This is where the nextInt method comes into play, causing the program to block until input is received.

It’s typical for a command line program to pause and wait for user input. However, if the program doesn’t display any output before the prompt, it may not be immediately apparent that it’s already running.

How to solve this problem

Simply input a command into the console and let the program execute to display the accurate output.

Scanner nextLong() method in Java with Examples, The nextLong(radix) method of java.util.Scanner class scans the next token of the input as a long. If the translation is successful,

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