Command line execution in Java - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print the first command line argument.

Java

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

Drag options to blanks, or click blank then click option'

Aargs.first()

Bargs(0)

Cargs.get(0)

Dargs[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets to access array elements.

Trying to call methods like get() on the array.

The command line arguments are stored in the args array. To access the first argument, use args[0].

2fill in blank

medium

Complete the code to check if any command line arguments were passed.

Java

public class Main {
    public static void main(String[] args) {
        if ([1] > 0) {
            System.out.println("Arguments received.");
        } else {
            System.out.println("No arguments.");
        }
    }
}

Drag options to blanks, or click blank then click option'

Aargs.length

Bargs.count()

Cargs.size()

Dargs.length()

Attempts:

3 left

💡 Hint

Common Mistakes

Using length() with parentheses which causes a compile error.

Using methods like size() or count() which do not exist for arrays.

The number of command line arguments is given by args.length. It returns the size of the array.

3fill in blank

hard

Fix the error in the code to print all command line arguments separated by spaces.

Java

public class Main {
    public static void main(String[] args) {
        for (int i = 0; i < [1]; i++) {
            System.out.print(args[i] + " ");
        }
    }
}

Drag options to blanks, or click blank then click option'

Aargs.count()

Bargs.size()

Cargs.length

Dargs.length()

Attempts:

3 left

💡 Hint

Common Mistakes

Using length() with parentheses causing errors.

Using size() or count() which are not valid for arrays.

The loop should run while i is less than args.length, which is the number of command line arguments.

4fill in blank

hard

Fill both blanks to print the last command line argument safely.

Java

public class Main {
    public static void main(String[] args) {
        if (args.[1] > 0) {
            System.out.println(args[[2]]);
        } else {
            System.out.println("No arguments.");
        }
    }
}

Drag options to blanks, or click blank then click option'

Alength

Blength()

Csize

Dcount

Elength - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using length() with parentheses.

Using size or count which are invalid for arrays.

To safely print the last command line argument, first check if args.length > 0. Then access args[args.length - 1] for the last element.

5fill in blank

hard

Fill all three blanks to print all arguments in uppercase separated by commas.

Java

public class Main {
    public static void main(String[] args) {
        for (int [1] = 0; [2] < args.[3]; [1]++) {
            System.out.print(args[[1]].toUpperCase());
            if ([1] < args.length - 1) {
                System.out.print(", ");
            }
        }
    }
}

Drag options to blanks, or click blank then click option'

Ai

Bj

Clength

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using size instead of length for arrays.

Using different variable names inconsistently in the loop.

The loop variable is commonly i. The condition uses args.length to get the number of arguments.

Practice

1. What command do you use to compile a Java file named Example.java from the command line?

easy

A. javac Example.java

B. java Example.java

C. compile Example.java

D. run Example.java

Solution

1. Step 1: Understand Java compilation

   Java source files (.java) must be compiled into bytecode (.class) before running.

2. Step 2: Use the correct command

   The javac command compiles Java files, so javac Example.java compiles the file.

3. Final Answer:

   javac Example.java -> Option A

4. Quick Check:

   Compile Java file = javac [OK]

Hint: Use 'javac' to compile Java files before running [OK]

Common Mistakes:

• Using 'java' to compile instead of run
• Typing 'compile' or 'run' commands which don't exist
• Forgetting to compile before running

2. Which command correctly runs a compiled Java program with the main class named App?

easy

A. java App

B. javac App

C. run App

D. java App.class

Solution

1. Step 1: Identify how to run Java bytecode

   Compiled Java classes (.class) are run using the java command followed by the class name without extension.

2. Step 2: Choose the correct syntax

   Running java App executes the App.class file. Adding '.class' or using 'javac' is incorrect here.

3. Final Answer:

   java App -> Option A

4. Quick Check:

   Run Java class = java ClassName [OK]

Hint: Run with 'java' plus class name, no file extension [OK]

Common Mistakes:

• Adding '.class' extension when running
• Using 'javac' to run instead of compile
• Typing 'run' command which is invalid

3. Given the commands executed in order:
javac HelloWorld.java
java HelloWorld
What will be the output if HelloWorld.java contains:

public class HelloWorld {
  public static void main(String[] args) {
    System.out.println("Hello, Java!");
  }
}

medium

A. No output

B. HelloWorld

C. Compilation Error

D. Hello, Java!

Solution

1. Step 1: Compile the Java file

   The javac HelloWorld.java command compiles the source code into HelloWorld.class without errors.

2. Step 2: Run the compiled class

   The java HelloWorld command runs the main method, printing the string inside System.out.println.

3. Final Answer:

   Hello, Java! -> Option D

4. Quick Check:

   Print statement output = Hello, Java! [OK]

Hint: Compile first, then run to see printed output [OK]

Common Mistakes:

• Trying to run without compiling first
• Expecting class name printed instead of message
• Confusing compile errors with runtime output

4. You try to run java MyProgram but get the error:
Error: Could not find or load main class MyProgram
What is the most likely cause?

medium

A. You ran java MyProgram.java instead of java MyProgram

B. You forgot to compile with javac MyProgram.java

C. The class name inside the file is different from the file name

D. You used the wrong Java version

Solution

1. Step 1: Understand the error meaning

   This error means Java cannot find the compiled .class file for MyProgram.

2. Step 2: Check compilation step

   If you did not run javac MyProgram.java first, the .class file does not exist, causing this error.

3. Final Answer:

   You forgot to compile with javac MyProgram.java -> Option B

4. Quick Check:

   Compile before run to avoid class not found [OK]

Hint: Always compile before running to avoid class not found errors [OK]

Common Mistakes:

• Running with .java extension in 'java' command
• Ignoring class name and file name mismatch
• Blaming Java version without checking compilation

5. You have two Java files:

File1.java:
public class File1 {
  public static void main(String[] args) {
    System.out.println("File1 running");
  }
}

File2.java:
public class File2 {
  public static void main(String[] args) {
    System.out.println("File2 running");
  }
}

Which sequence of commands correctly compiles and runs File2 from the command line?

hard

A. java File2javac File2.java

B. javac File2.javajava File2.java

C. javac File2.javajava File2

D. javac File1.java File2.javajava File2.java

Solution

1. Step 1: Compile the correct file

   To run File2, you must compile File2.java first using javac File2.java.

2. Step 2: Run the compiled class properly

   Run the program with java File2 (without .java extension). Running before compiling or adding extensions causes errors.

3. Final Answer:

   javac File2.java java File2 -> Option C

4. Quick Check:

   Compile then run target class without extension [OK]

Hint: Compile target file, then run class name without extension [OK]

Common Mistakes:

• Running with .java extension
• Running before compiling
• Compiling unrelated files unnecessarily