Accessing arguments in Java

Introduction

Arguments let you give information to a program when you start it. Accessing arguments means reading that information inside the program.

You want to tell a program which file to open when it starts.

You want to give a program a number to use in calculations.

You want to change how a program works without changing its code.

You want to pass a username or password to a program at start.

You want to run the same program with different settings easily.

Syntax

Java

public static void main(String[] args) {
    // args is an array of Strings
    String firstArgument = args[0];
}

args is an array holding all arguments as strings.

You access each argument by its position, starting at 0.

Examples

This prints the first argument given to the program.

Java

public class Example {
    public static void main(String[] args) {
        System.out.println("First argument: " + args[0]);
    }
}

This prints all arguments one by one.

Java

public class Example {
    public static void main(String[] args) {
        for (int i = 0; i < args.length; i++) {
            System.out.println("Argument " + i + ": " + args[i]);
        }
    }
}

Sample Program

This program checks if any arguments were given. If yes, it prints how many and lists each one.

Java

public class AccessArguments {
    public static void main(String[] args) {
        if (args.length == 0) {
            System.out.println("No arguments provided.");
        } else {
            System.out.println("You gave " + args.length + " arguments.");
            for (int i = 0; i < args.length; i++) {
                System.out.println("Argument " + i + ": " + args[i]);
            }
        }
    }
}

Important Notes

If you try to access an argument that does not exist, the program will crash with an error.

Arguments are always strings. You need to convert them if you want numbers.

Use args.length to know how many arguments were given.

Summary

Arguments are extra information given when starting a program.

You access arguments inside main using the args array.

Always check how many arguments you have before using them.

Practice

1. What is the correct way to access the first command-line argument in a Java program inside the main method?

easy

A. args.first()

B. args[0]

C. args(0)

D. args.get(0)

Solution

1. Step 1: Understand the type of args

   The args parameter in main is an array of strings.

2. Step 2: Access array elements in Java

   Array elements are accessed using square brackets with the index inside, starting at 0.

3. Final Answer:

   args[0] -> Option B

4. Quick Check:

   Array access uses square brackets [OK]

Hint: Use square brackets to access array elements in Java [OK]

Common Mistakes:

• Using parentheses instead of square brackets
• Trying to call methods like first() on array
• Using get() which is for lists, not arrays

2. Which of the following is the correct signature of the main method to access command-line arguments in Java?

easy

A. public void main(String[] args)

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

C. static public void main(String args)

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

Solution

1. Step 1: Recall the standard 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. public void main(String[] args) is missing static. static public void main(String args) has wrong parameter type. public static int main(String[] args) returns int which is invalid.

3. Final Answer:

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

4. Quick Check:

   Main method must be public static void with String array [OK]

Hint: Main method must be public static void with String array [OK]

Common Mistakes:

• Omitting static keyword
• Wrong return type
• Using single String instead of String array

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

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

medium

A. 3
World

B. 2
Hello

C. 2
World

D. 3
Hello

Solution

1. Step 1: Count the command-line arguments

   The program is run with two arguments: "Hello" and "World", so args.length is 2.

2. Step 2: Access the second argument

   args[1] is the second argument, which is "World".

3. Final Answer:

   2 World -> Option C

4. Quick Check:

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

Hint: args.length counts arguments; args[1] is second argument [OK]

Common Mistakes:

• Counting program name as argument
• Using args[0] as second argument
• Confusing args.length with number of arguments plus one

4. Identify the error in the following Java program that tries to print the first command-line argument:

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

medium

A. Using parentheses instead of square brackets to access array element

B. Missing static keyword in main method

C. Incorrect class name

D. Missing return statement in main

Solution

1. Step 1: Check how array elements are accessed

   In Java, array elements are accessed using square brackets, not parentheses.

2. Step 2: Identify the error in the code

   The code uses args(0) which is invalid syntax and causes a compile error.

3. Final Answer:

   Using parentheses instead of square brackets to access array element -> Option A

4. Quick Check:

   Array access uses square brackets [OK]

Hint: Use square brackets [] for arrays, not parentheses () [OK]

Common Mistakes:

• Using parentheses for array access
• Confusing method call syntax with array access
• Ignoring compile errors due to wrong syntax

5. You want to write a Java program that prints all command-line arguments separated by commas. Which code snippet correctly does this inside main?

hard

A. for(int i=0; i

B. for(String arg : args) {
System.out.print(arg + ", ");
}

C. int i = 0;
while(i <= args.length) {
System.out.print(args[i] + ", ");
i++;
}

D. for(int i=1; i<=args.length; i++) {
System.out.print(args[i] + ", ");
}

Solution

1. Step 1: Understand the goal

   We want to print all arguments separated by commas without trailing comma.

2. Step 2: Analyze each option

   for(int i=0; i

3. Final Answer:

   for(int i=0; i<args.length; i++) { System.out.print(args[i] + (i < args.length - 1 ? ", " : "")); } -> Option A

4. Quick Check:

   Loop with index and conditional comma [OK]

Hint: Use index loop and add comma only if not last element [OK]

Common Mistakes:

• Adding comma after last argument
• Off-by-one errors in loops
• Accessing array out of bounds