A simple example to illustrate how java.util.Scanner works would be reading a single integer from System.in. It’s really quite simple. Scanner sc = new Scanner(System.in); int i = sc.nextInt(); To retrieve a username I would probably use sc.nextLine(). System.out.println(“Enter your username: “); Scanner scanner = new Scanner(System.in); String username = scanner.nextLine(); System.out.println(“Your username is ” … Read more
As per the javadoc for Scanner: When a scanner throws an InputMismatchException, the scanner will not pass the token that caused the exception, so that it may be retrieved or skipped via some other method. That means that if the next token is not an int, it throws the InputMismatchException, but the token stays there. … Read more
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 = … Read more
This has really puzzled me for a while but this is what I found in the end. When you call, sc.close() in first method, it not only closes your scanner but closes your System.in input stream as well. You can verify it by printing its status at very top of the second method as : … Read more
Overview of Scanner.hasNextXXX methods java.util.Scanner has many hasNextXXX methods that can be used to validate input. Here’s a brief overview of all of them: hasNext() – does it have any token at all? hasNextLine() – does it have another line of input? For Java primitives hasNextInt() – does it have a token that can be … Read more
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 … Read more
Idiomatic Example: The following is how to properly use the java.util.Scanner class to interactively read user input from System.in correctly( sometimes referred to as stdin, especially in C, C++ and other languages as well as in Unix and Linux). It idiomatically demonstrates the most common things that are requested to be done. package com.stackoverflow.scanner; import … Read more
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 … Read more
You’re mixing up && and ||, so your first if will never run as mentioned in the comments. So it looks like the only “Not Weird” print out is 2, 4 and even numbers > 20. So use this to your advantage to just check for the “Weird” outputs, otherwise print “Not Weird”. if (n … Read more