Reading string input in Java - Step-by-Step Execution

Concept Flow - Reading string input

Start Program

↓

Create Scanner

↓

Prompt User

↓

Wait for Input

↓

Read Input String

↓

Store in Variable

↓

Use or Print String

↓

End Program

The program starts, creates a Scanner to read input, waits for the user to type a string, stores it, then uses or prints it.

Execution Sample

Java

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter your name: ");
        String name = sc.nextLine();
        System.out.println("Hello, " + name + "!");
        sc.close();
    }
}

This code asks the user to enter their name, reads the full line as a string, then greets the user.

Execution Table

Step	Action	Input/Condition	Variable State	Output
1	Create Scanner object	N/A	sc = Scanner(System.in)
2	Print prompt	N/A	sc = Scanner(System.in)	Enter your name:
3	Wait for user input	User types: Alice	sc = Scanner(System.in)
4	Read input line	sc.nextLine()	name = "Alice"
5	Print greeting	N/A	name = "Alice"	Hello, Alice!
6	Program ends	N/A	name = "Alice"

💡 Program ends after printing greeting.

Variable Tracker

Variable	Start	After Step 4	Final
sc	null	Scanner(System.in)	Scanner(System.in)
name	null	"Alice"	"Alice"

Key Moments - 2 Insights

Why do we use sc.nextLine() instead of sc.next()?

sc.nextLine() reads the entire line including spaces, while sc.next() reads only up to the first space. See execution_table step 4 where the full input "Alice" is read.

What happens if the user just presses Enter without typing anything?

sc.nextLine() reads an empty string "" and stores it in name. The program still continues normally as shown in execution_table step 4.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of 'name' after step 4?

A "Alice"

B null

C "Enter your name:"

D Scanner(System.in)

💡 Hint

Check the 'Variable State' column at step 4 in the execution_table.

At which step does the program print the greeting message?

A Step 2

B Step 4

C Step 5

D Step 6

💡 Hint

Look at the 'Output' column in the execution_table to find when greeting is printed.

If the user input was "Bob Smith", which method ensures the full name is read?

A sc.next()

B sc.nextLine()

C System.out.print()

D Scanner(System.in)

💡 Hint

Refer to key_moments about difference between next() and nextLine().

Concept Snapshot

Reading string input in Java:
- Create Scanner: Scanner sc = new Scanner(System.in);
- Prompt user: System.out.print("Enter: ");
- Read full line: String s = sc.nextLine();
- Use the string as needed
- nextLine() reads entire line including spaces
- Always close Scanner when done (not shown here)

Full Transcript

This example shows how Java reads string input from the user. First, a Scanner object is created to read from the keyboard. Then the program prints a prompt asking the user to enter their name. It waits for the user to type and press Enter. The method nextLine() reads the entire line typed, including spaces, and stores it in the variable 'name'. Finally, the program prints a greeting using the entered name. The execution table traces each step, showing variable changes and output. Key points include why nextLine() is used instead of next(), and what happens if the user inputs an empty line.

Practice

1. What is the correct way to read a full line of text input from the user in Java?

easy

A. Use Scanner.nextLine() method

B. Use Scanner.nextInt() method

C. Use System.out.println() method

D. Use Scanner.read() method

Solution

1. Step 1: Understand Scanner methods

   The Scanner class has different methods to read input. nextLine() reads the entire line as a string.

2. Step 2: Identify the correct method for full line input

   nextInt() reads only integers, System.out.println() prints output, and Scanner.read() does not exist. So, nextLine() is correct.

3. Final Answer:

   Use Scanner.nextLine() method -> Option A

4. Quick Check:

   Read full line = nextLine() [OK]

Hint: Use nextLine() to read full text line input [OK]

Common Mistakes:

• Using nextInt() to read strings
• Confusing output methods with input
• Forgetting to import Scanner

2. Which of the following is the correct syntax to create a Scanner object to read input from the keyboard?

easy

A. Scanner input = new Scanner(System.out);

B. Scanner input = new Scanner(System.file);

C. Scanner input = new Scanner(System.err);

D. Scanner input = new Scanner(System.in);

Solution

1. Step 1: Understand Scanner constructor

   The Scanner constructor takes an InputStream to read from. System.in is the standard input stream (keyboard).

2. Step 2: Identify correct stream for keyboard input

   System.out and System.err are output streams, and System.file does not exist. So, new Scanner(System.in) is correct.

3. Final Answer:

   Scanner input = new Scanner(System.in); -> Option D

4. Quick Check:

   Keyboard input = System.in [OK]

Hint: Use System.in to read keyboard input with Scanner [OK]

Common Mistakes:

• Using System.out or System.err instead of System.in
• Trying to read from non-existent System.file
• Missing parentheses in Scanner constructor

3. What will be the output of the following Java code if the user inputs: Hello World

import java.util.Scanner;
public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String text = sc.nextLine();
    System.out.println(text);
    sc.close();
  }
}

medium

A. Hello

B. Hello World

C. World

D. Compilation error

Solution

1. Step 1: Understand nextLine() behavior

   The nextLine() method reads the entire line including spaces until the user presses Enter.

2. Step 2: Analyze the input and output

   User inputs "Hello World" and presses Enter. The program reads the full line and prints it exactly.

3. Final Answer:

   Hello World -> Option B

4. Quick Check:

   nextLine() reads full line = Hello World [OK]

Hint: nextLine() reads entire line including spaces [OK]

Common Mistakes:

• Assuming nextLine() reads only first word
• Confusing next() with nextLine()
• Forgetting to close Scanner

4. Identify the error in the following code snippet that reads a string input from the user:

import java.util.Scanner;
public class Test {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String name = sc.next();
    System.out.println(name);
    sc.close();
  }
}

What is the problem if the user inputs "John Doe"?

medium

A. The program does not compile

B. The program throws a runtime exception

C. The program reads only "John" and ignores "Doe"

D. The program reads the full line "John Doe" correctly

Solution

1. Step 1: Understand difference between next() and nextLine()

   next() reads input only until the first space, so it reads "John" only.

2. Step 2: Analyze input and output behavior

   User inputs "John Doe" but next() stops at space, so "Doe" is ignored in this read.

3. Final Answer:

   The program reads only "John" and ignores "Doe" -> Option C

4. Quick Check:

   next() reads until space = partial input [OK]

Hint: Use nextLine() to read full input with spaces [OK]

Common Mistakes:

• Using next() when full line is needed
• Expecting next() to read multiple words
• Not testing with multi-word input

5. You want to read multiple lines of user input until the user types "exit". Which code snippet correctly implements this using Scanner and nextLine()?

hard

A. Scanner sc = new Scanner(System.in);
String line;
do {
line = sc.nextLine();
System.out.println(line);
} while (!line.equals("exit"));
sc.close();

B. Scanner sc = new Scanner(System.in);
while (sc.nextLine() != "exit") {
System.out.println(sc.nextLine());
}
sc.close();

C. Scanner sc = new Scanner(System.in);
String line = sc.next();
while (!line.equals("exit")) {
System.out.println(line);
line = sc.next();
}
sc.close();

D. Scanner sc = new Scanner(System.in);
while (true) {
String line = sc.nextLine();
if (line == "exit") break;
System.out.println(line);
}
sc.close();

Solution

1. Step 1: Understand string comparison in Java

   Use line.equals("exit") to compare strings, not ==.

2. Step 2: Analyze loop structure and input reading

   Scanner sc = new Scanner(System.in); String line; do { line = sc.nextLine(); System.out.println(line); } while (!line.equals("exit")); sc.close(); uses a do-while loop reading input once per iteration and compares correctly. Scanner sc = new Scanner(System.in); while (sc.nextLine() != "exit") { System.out.println(sc.nextLine()); } sc.close(); reads twice per loop and uses != with strings (wrong). Scanner sc = new Scanner(System.in); String line = sc.next(); while (!line.equals("exit")) { System.out.println(line); line = sc.next(); } sc.close(); uses next() which reads only one word. Scanner sc = new Scanner(System.in); while (true) { String line = sc.nextLine(); if (line == "exit") break; System.out.println(line); } sc.close(); uses == for string comparison (wrong).

3. Final Answer:

   The do-while loop with equals("exit") -> Option A

4. Quick Check:

   Use equals() and read once per loop [OK]

Hint: Use equals() for string compare and read input once per loop [OK]

Common Mistakes:

• Using == instead of equals() for strings
• Reading input multiple times per loop
• Using next() instead of nextLine() for full lines