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Harpreet
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Asked: 2025-08-18T09:27:34+05:30 In:

In Java, consider the following code snippet: Scanner sc = …

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In Java, consider the following code snippet:

Scanner sc = new Scanner(System.in);

System.out.print("Enter your age: ");
int age = sc.nextInt();

System.out.print("Enter your full name: ");
String name = sc.nextLine();

System.out.println("Age: " + age);
System.out.println("Name: " + name);

When the input is:

20
Rahul Sharma

The output is:

Age: 20
Name:

Explain why the nextLine() method appears to “skip” input in this case.

javanextline()scanner
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    1 Answer

    1. Pankaj Gupta
      Scholar

      The nextLine() method appears to skip input because after executing nextInt(), the newline character (\n) from pressing Enter is still left in the input buffer. When nextLine() is called immediately after, it reads this leftover newline character instead of waiting for new user input. As a result, iRead more

      The nextLine() method appears to skip input because after executing nextInt(), the newline character (\n) from pressing Enter is still left in the input buffer.

      When nextLine() is called immediately after, it reads this leftover newline character instead of waiting for new user input. As a result, it returns an empty string and seems to “skip” the input.

      To fix the issue, insert an extra sc.nextLine(); after nextInt() to consume the leftover newline character.

      Scanner sc = new Scanner(System.in);

      System.out.print(“Enter your age: “);
      int age = sc.nextInt();
      sc.nextLine(); // consume the leftover newline

      System.out.print(“Enter your full name: “);
      String name = sc.nextLine();

      System.out.println(“Age: ” + age);
      System.out.println(“Name: ” + name);

      Now, if the input is:

      20
      Rahul Sharma

      The output will be:

      Age: 20
      Name: Rahul Sharma

      See less

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