Accessing arguments - Mini Project: Build & Apply

Accessing Arguments in C

📖 Scenario: Imagine you want to create a simple program that greets people by their names. The names will be given when running the program from the command line.

🎯 Goal: Build a C program that reads names passed as command line arguments and prints a greeting for each name.

📋 What You'll Learn

Create a main function with parameters int argc and char *argv[]

Use a for loop to access each argument starting from index 1

Print a greeting message for each name argument

Handle the case when no names are given

💡 Why This Matters

🌍 Real World

Many programs accept information from the command line to customize their behavior, like tools that process files or scripts that automate tasks.

💼 Career

Understanding how to access and use command line arguments is essential for software developers, especially when creating utilities, scripts, or applications that run in terminal environments.

1

Set up the main function with arguments

Write the main function header with parameters int argc and char *argv[].

C

int main(int argc, char *argv[]) {
    // Your code here
    return 0;
}

Need a hint?

The main function should accept two parameters: argc and argv.

2

Add a variable to count names

Inside main, create an int variable called name_count and set it to argc - 1 to count how many names were passed.

C

int main(int argc, char *argv[]) {
    int name_count = argc - 1;  // Your code here
    return 0;
}

Need a hint?

Subtract 1 from argc because the first argument is the program name.

3

Loop through the names and print greetings

Use a for loop with int i = 1 to i <= name_count to access each name in argv. Inside the loop, print Hello, <name>! using printf and argv[i].

C

int main(int argc, char *argv[]) {
    int name_count = argc - 1;
    for (int i = 1; i <= name_count; i++) {
        // Your code here
    }
    return 0;
}

Need a hint?

Start the loop at 1 because argv[0] is the program name.

4

Print a message if no names are given

Add an if statement to check if name_count is 0. If yes, print No names provided. Otherwise, print greetings as before.

C

int main(int argc, char *argv[]) {
    int name_count = argc - 1;
    if (name_count == 0) {
        // Your code here
    } else {
        for (int i = 1; i <= name_count; i++) {
            printf("Hello, %s!\n", argv[i]);
        }
    }
    return 0;
}

Need a hint?

Use printf to show the message when no names are passed.