Why Does Java's `nextLine()` Behave Unexpectedly After `nextInt()`?

nextLine() Comprehension

In Java's java.util.Scanner API, the nextLine() method gives developers the ability to parse string input from the user. However, it can behave differently when used within loops or after other input parsing methods like nextInt().

Scenario #1:

A code snippet that uses nextLine() successfully:

Scanner scanner = new Scanner(System.in);

// Wait for user input before continuing
String sentence = scanner.nextLine();

// Continue with more input
int index = scanner.nextInt();

In this scenario, nextLine() waits for user input before moving on to nextInt().

Scenario #2:

Now consider a code snippet that behaves unexpectedly:

Scanner scanner = new Scanner(System.in);

// Infinite loop
while (true) {
    // Menu options
    int selection = scanner.nextInt();

    // Input only works after reading the number
    if (selection == 1) {
        String sentence = scanner.nextLine();
        int index = scanner.nextInt();
    }
}

Here, nextLine() doesn't wait for user input after nextInt(). It skips to the next line, making it impossible to enter a sentence.

Explanation:

The difference between these examples lies in the nextInt() method. It reads numerical input, but it doesn't consume the newline character (n) that follows. As a result, when nextLine() is called after nextInt(), it tries to read the already-consumed newline character, leading to the unexpected behavior in Scenario #2.

Solution:

To resolve this, insert an additional nextLine() call after each nextInt() to consume the remaining input on the line before moving on to the next input request. This ensures that nextLine() always waits for user input.

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