Syntax for command line arguments in Java - Mini Project: Build & Apply

Syntax for command line arguments

📖 Scenario: You want to create a simple Java program that reads names from the command line when you run it. This is like telling your program some information before it starts, just like giving instructions to a friend before they begin a task.

🎯 Goal: Build a Java program that takes exactly two names as command line arguments and prints a greeting message using those names.

📋 What You'll Learn

Create a Java class named Greeting

Use the main method with String[] args to receive command line arguments

Access the first and second command line arguments using args[0] and args[1]

Print a greeting message combining both names

💡 Why This Matters

🌍 Real World

Many Java programs accept information from the command line to customize their behavior without changing the code. For example, tools, scripts, and utilities use command line arguments to know what files to process or what options to use.

💼 Career

Understanding command line arguments is important for Java developers because it helps in creating flexible programs and working with automation scripts, build tools, and deployment processes.

1

Create the Greeting class with main method

Write a public class called Greeting with a public static void main method that takes String[] args as parameter.

Java

public class Greeting {
    public static void main(String[] args) {
        // Your code here
    }
}

Hint

The main method is the entry point of a Java program. It must be public static void main(String[] args).

2

Add variables to store the first and second command line arguments

Inside the main method, create two String variables named firstName and secondName. Assign args[0] to firstName and args[1] to secondName.

Java

public class Greeting {
    public static void main(String[] args) {
        // Your code here
    }
}

Hint

Command line arguments are stored in the args array. The first argument is args[0], the second is args[1].

3

Create a greeting message using the two names

Inside the main method, create a String variable named message that combines firstName and secondName into the text: "Hello, <firstName> and <secondName>!" using string concatenation.

Java

public class Greeting {
    public static void main(String[] args) {
        String firstName = args[0];
        String secondName = args[1];
        // Your code here
    }
}

Hint

Use the + operator to join strings and variables into one message.

4

Print the greeting message

Use System.out.println to print the message variable inside the main method.

Java

public class Greeting {
    public static void main(String[] args) {
        String firstName = args[0];
        String secondName = args[1];
        String message = "Hello, " + firstName + " and " + secondName + "!";
        // Your code here
    }
}

Hint

Use System.out.println(message); to show the message on the screen.

Practice

1. What is the correct way to declare the main method to accept command line arguments in Java?

easy

A. public void main(String args)

B. public static void main(String[] args)

C. static void main(String args[])

D. public static int main(String[] args)

Solution

1. Step 1: Understand main method signature

   The main method must be public, static, return void, and accept a String array parameter.

2. Step 2: Check each option

   public static void main(String[] args) matches the correct signature exactly. Others either miss static, have wrong return type, or wrong parameter type.

3. Final Answer:

   public static void main(String[] args) -> Option B

4. Quick Check:

   Main method signature = public static void main(String[] args) [OK]

Hint: Main method always: public static void main(String[] args) [OK]

Common Mistakes:

• Omitting static keyword
• Using wrong parameter type
• Changing return type from void

2. Which of the following is the correct way to access the first command line argument inside the main method?

easy

A. args[0]

B. args.first()

C. args.get(0)

D. args(0)

Solution

1. Step 1: Recall how arrays are accessed in Java

   Java arrays use square brackets [] with zero-based indexing to access elements.

2. Step 2: Evaluate options

   Only args[0] uses correct syntax. args(0) and args.get(0) are invalid for arrays, and args.first() is not a method for arrays.

3. Final Answer:

   args[0] -> Option A

4. Quick Check:

   Array access uses square brackets = args[0] [OK]

Hint: Use square brackets to access array elements: args[0] [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Trying to call methods on array
• Confusing with List syntax

3. What will be the output of the following Java program if run with command line arguments: java Program Hello World?

public class Program {
    public static void main(String[] args) {
        System.out.println(args.length);
        System.out.println(args[0]);
        System.out.println(args[1]);
    }
}

medium

A. 2
Hello
World

B. 3
Hello
World

C. 2
World
Hello

D. 3
World
Hello

Solution

1. Step 1: Count command line arguments

   Arguments are "Hello" and "World", so args.length is 2.

2. Step 2: Check printed values

   args[0] is "Hello", args[1] is "World". So output lines are 2, Hello, World.

3. Final Answer:

   2 Hello World -> Option A

4. Quick Check:

   args.length=2, args[0]=Hello, args[1]=World [OK]

Hint: args.length counts arguments; args[0] is first argument [OK]

Common Mistakes:

• Counting program name as argument
• Swapping args[0] and args[1]
• Assuming args.length includes program name

4. Identify the error in this Java program that tries to print all command line arguments:

public class Test {
    public static void main(String[] args) {
        for(int i=1; i<=args.length; i++) {
            System.out.println(args[i]);
        }
    }
}

medium

A. System.out.println should use args(i) instead of args[i]

B. args should be declared as int[] not String[]

C. Loop starts at 1 and uses <= causing ArrayIndexOutOfBoundsException

D. main method must return int, not void

Solution

1. Step 1: Analyze loop indexing

   The loop starts at i=1 and runs while i <= args.length. Array indices go from 0 to args.length-1.

2. Step 2: Identify error cause

   Accessing args[args.length] causes ArrayIndexOutOfBoundsException. Also, skipping args[0] misses first argument.

3. Final Answer:

   Loop starts at 1 and uses <= causing ArrayIndexOutOfBoundsException -> Option C

4. Quick Check:

   Array indices 0 to length-1, loop must use i=0 to i<length [OK]

Hint: Loop from 0 to args.length-1 to avoid index errors [OK]

Common Mistakes:

• Starting loop at 1 instead of 0
• Using <= instead of < in loop condition
• Using parentheses instead of brackets for array access

5. You want to write a Java program that sums two integers passed as command line arguments and prints the result. Which code snippet correctly parses and sums the arguments?

hard

A. int sum = Integer.parseInt(args[0]) + args[1].parseInt(); System.out.println(sum);

B. int sum = args[0] + args[1]; System.out.println(sum);

C. int sum = Integer.valueOf(args[0]) + args[1]; System.out.println(sum);

D. int sum = Integer.parseInt(args[0]) + Integer.parseInt(args[1]); System.out.println(sum);

Solution

1. Step 1: Convert String arguments to integers

   Use Integer.parseInt() to convert each String argument to int.

2. Step 2: Sum the parsed integers and print

   Sum the two parsed ints and print the result using System.out.println.

3. Final Answer:

   int sum = Integer.parseInt(args[0]) + Integer.parseInt(args[1]); System.out.println(sum); -> Option D

4. Quick Check:

   Parse strings to int with Integer.parseInt() before adding [OK]

Hint: Parse args with Integer.parseInt() before arithmetic [OK]

Common Mistakes:

• Adding strings directly without parsing
• Using wrong method to parse integers
• Calling parseInt() on string variable incorrectly