Challenge of Using nextLine() after Calling nextInt() or next() Method in Java

Many Java developers encounter this issue when working with the Scanner class. In this post, we’ll be understanding the issue and we’ll also look into a way to handle it. So without further ado, let’s get started.

Understanding Issue

When working with the Scanner class in Java, it’s always a good idea to understand how nextInt(), next(), and nextLine() methods behave. The nextInt() method is commonly used to read integer input from the user.

The next() method can’t read the words with blank space(or white space), whereas nextLine() can read the words with blank space. For String “Hello World” the next() method will only read the “Hello” word, whereas the nextLine() method will read the “Hello World” word. The next() method can read the input only until a space(” “) is encountered so it keeps the cursor in the same line. While nextLine() method reads the entire text up-to the new line(or return line).

The newline character issue

After calling nextInt() method, the scanner reads the integer value but leaves the newline character (represented as “\n”) in the input stream.
Eg:

System.out.print(“Enter an integer: “);
int num = scanner.nextInt();
System.out.print(“Enter a string: “);
String str = scanner.nextLine();
System.out.println(“Integer: ” + num);
System.out.println(“String: ” + str);

This newline character is not consumed by nextInt(), and if you immediately call nextLine() to read a string input, it will capture that newline character instead of the intended input. As a result, nextLine() considers this newline character as input and returns an empty string.

Ways to handle the newline character(“\n”) issue:

1) Consuming newline character

We can use an additional nextLine() just after nextInt() to consume the leftover newline character from the input stream. This ensures that the subsequent nextLine() call captures the desired input.
eg:

System.out.print(“Enter an integer: “);
int num = scanner.nextInt();

// Consumes the newline character
scanner.nextLine();

System.out.print(“Enter a string: “);
String str = scanner.nextLine();
System.out.println(“Integer: ” + num);
System.out.println(“String: ” + str);

2) Using nextLine() Directly

We can employ the nextLine() method as an alternative to nextInt() to handle user input. Unlike nextInt(), which reads only the integer value from the input, nextLine() method reads the entire line of input as a string. This can be really useful when you need to handle input that includes spaces or characters other than integers.

To do this, simply replace the nextInt() with a nextLine() method to retrieve the input as a string. You can then parse the string as an integer using the Integer.parseInt() or Integer.valueOf() method.

Eg:

System.out.print(“Enter an integer: “);
String string = scanner.nextLine();
int num = Integer.valueOf(string);
System.out.print(“Enter a string: “);
String str = scanner.nextLine();
System.out.println(“Integer: ” + num);
System.out.println(“String: ” + str);

Conclusion

Resolving the challenges of using nextLine() after calling nextInt() or next() in Java requires a clear understanding of the Scanner class’s behavior. By employing one of the suggested two solutions, developers can ensure smooth and accurate user input in their Java applications. Happy Coding!