Accessing arguments in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Java program?

Consider the following Java program that accesses command-line arguments. What will it print when run with arguments: java Main Hello World?

Java

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

A HelloWorld

B args[0] args[1]

C Hello World

D ArrayIndexOutOfBoundsException

💡 Hint

Remember that args is an array of strings passed from the command line.

The program prints the first and second command-line arguments separated by a space. Since args[0] is "Hello" and args[1] is "World", the output is "Hello World".

❓ Predict Output

intermediate

What happens if no arguments are passed?

What will be the output or result of running this Java program with no command-line arguments?

Java

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

A 0 followed by null

B Compilation error

C 0 followed by 0

D 0 followed by ArrayIndexOutOfBoundsException

💡 Hint

Check what args.length is and what happens when accessing args[0] if no arguments exist.

args.length is 0 because no arguments are passed. Accessing args[0] causes ArrayIndexOutOfBoundsException because the array is empty.

🔧 Debug

advanced

Identify the error in accessing arguments

This Java program tries to print all command-line arguments but throws an error. What is the cause?

Java

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

A ArrayIndexOutOfBoundsException because loop index goes out of bounds

B NullPointerException because args is null

C Compilation error due to wrong loop syntax

D No error, prints all arguments

💡 Hint

Remember array indices start at 0 and go up to length-1.

The loop uses i from 1 to args.length inclusive, but valid indices are 0 to args.length-1. Accessing args[args.length] causes ArrayIndexOutOfBoundsException.

📝 Syntax

advanced

Which option correctly accesses the last argument?

Given a Java program with command-line arguments, which code snippet correctly prints the last argument?

A System.out.println(args[-1]);

B System.out.println(args[args.length - 1]);

C System.out.println(args[args.length + 1]);

D System.out.println(args[args.length]);

💡 Hint

Array indices start at 0 and end at length-1.

args.length is the number of arguments, but the last index is args.length - 1. Accessing args[args.length] or args[args.length + 1] causes errors. Negative indices are invalid in Java.

🚀 Application

expert

What is the output of this program with arguments: java Main 5 10 15?

This Java program sums integer command-line arguments. What does it print when run with arguments 5 10 15?

Java

public class Main {
    public static void main(String[] args) {
        int sum = 0;
        for (String arg : args) {
            sum += Integer.parseInt(arg);
        }
        System.out.println(sum);
    }
}

A 30

B NumberFormatException

C Compilation error

D 51015

💡 Hint

Integer.parseInt converts strings to numbers before adding.

The program converts each argument to an integer and sums them: 5 + 10 + 15 = 30. It then prints 30.

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