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Scanner scanner = new Scanner(System.in);
System.out.println("Enter first number");
int FirstNumber = scanner.nextInt();
System.out.println("Entered First Number is" + FirstNumber);
System.out.println("Enter Second Number");
int SecondNumber = scanner.nextInt();
System.out.println("Entered Second Number is" + SecondNumber);

There is no error in the mentioned code everything is perfect but I have a doubt about why we didn't write a scanner.next line() method to handle enter key being pressed after the first number is entered on the console.

Pratik Bankar
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Jignesh Nayak
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2 Answers2

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nextInt reads the next token in the scanner.

According to the documentation

A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace.

And further down

The default whitespace delimiter used by a scanner is as recognized by Character.isWhitespace().

The documentation for Character.isWhitespace() says that it returns true if the given codepoint

[...] [...] is '\n', U+000A LINE FEED. [...]

Which means that \n is a separator, thus it's not part of any token, so you don't need to skip it with a dummy call to nextLine()

Federico klez Culloca
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Scanner.nextInt() does not read new lines, it only progresses in the current line. See: https://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#nextInt()

Alex_X1
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  • That is not correct, it does progress to a next line (because scanner by default tokenizes on whitespace). However, say it is consuming `1\n2\n`, after two `nextInt()` calls the remaining buffer is `\n`. If you then ask for `nextLine()`, you get an empty line (and the remaining buffer is then empty). – Mark Rotteveel Dec 21 '22 at 16:34