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

Mary-Kate Olsen
Release: 2024-12-27 07:21:09
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Why Does `scanner.nextLine()` Behave Differently After `scanner.nextInt()` in Java?

Using Scanner.nextLine() for User Input

Question:

Why does the scanner.nextLine() method behave differently in the following two Java code examples?

// Working Example
Scanner scanner = new Scanner(System.in);
System.out.print("Enter a sentence: ");
String sentence = scanner.nextLine();
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// Not Working Example
while (true) {
    System.out.print("Enter a sentence: ");
    int selection = scanner.nextInt();
    String sentence = scanner.nextLine();
}
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Answer:

The difference in behavior arises from how scanner.nextInt() consumes input compared to scanner.nextLine().

In the working example, scanner.nextLine() reads an entire line of input, including spaces, until it encounters a newline character. In contrast, scanner.nextInt() reads only the integer portion of the input, leaving any remaining characters in the input buffer.

When the nextInt() call is used without any subsequent calls to nextLine(), any characters remaining in the input buffer (e.g., a newline character) are not consumed, which can cause issues for subsequent nextLine() calls.

In the non-working example, after the user enters a number, the remaining newline character is not consumed by nextInt(). Consequently, the subsequent call to nextLine() immediately reads the newline character, resulting in an empty string being assigned to sentence.

To resolve this issue, an additional call to scanner.nextLine() can be placed after each nextInt() call to consume any remaining characters in the input buffer.

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