Why Does `scanner.nextLine()` Fail After `scanner.nextInt()` in Java?

Understanding scanner.nextLine() and Its Impact on Subsequent Input

Java's nextLine() method, defined in the java.util.Scanner class, is commonly used to read lines of text from standard input (the user's keyboard). However, its usage in certain scenarios can lead to unexpected behavior.

Consider the following example where nextLine() is used within a menu loop:

Scanner scanner = new Scanner(System.in);

while (true) {
    // Display menu options
    // ...

    System.out.print("Please enter your selection: ");
    int selection = scanner.nextInt();

    if (selection == 1) {
        // This issue arises here when reading a sentence
        System.out.print("Enter a sentence: ");
        String sentence = scanner.nextLine();
    }
    // ...
}

When this code executes, it fails to prompt the user for a sentence when the user enters the value 1. Instead, the code proceeds directly to the next input prompt for an index.

The reason for this behavior lies in the nature of nextLine(). When used after nextInt(), it reads and consumes the remaining newline character (n) in the input stream. Consequently, when nextLine() is invoked to read the sentence, it encounters an empty input buffer and returns an empty string.

To resolve this issue, it is necessary to consume the newline character explicitly using scanner.nextLine() after each scanner.nextInt(). This ensures that the input buffer is cleared and nextLine() can correctly read the user's input for the sentence.

By following this guideline, you can avoid inconsistencies when using nextLine() and ensure that your code accurately handles user input in both single- and multi-pass scenarios.

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