Syntax of command line arguments - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this C program with command line arguments?

Consider this C program that prints the number of command line arguments and the first argument:

#include <stdio.h>
int main(int argc, char *argv[]) {
    printf("%d %s\n", argc, argv[1]);
    return 0;
}

If you run this program as ./program Hello, what will be the output?

C

#include <stdio.h>
int main(int argc, char *argv[]) {
    printf("%d %s\n", argc, argv[1]);
    return 0;
}

A 1 Hello

B 2 Hello

C 2 ./program

D 1 ./program

💡 Hint

Remember, argc counts the program name as the first argument.

The argc counts all arguments including the program name, so it is 2. argv[1] is the first argument after the program name, which is "Hello".

❓ Predict Output

intermediate

What does this program print when no arguments are given?

Look at this C program:

#include <stdio.h>
int main(int argc, char *argv[]) {
    if (argc < 2) {
        printf("No arguments\n");
    } else {
        printf("Argument: %s\n", argv[1]);
    }
    return 0;
}

What will it print if you run it as ./program with no extra arguments?

C

#include <stdio.h>
int main(int argc, char *argv[]) {
    if (argc < 2) {
        printf("No arguments\n");
    } else {
        printf("Argument: %s\n", argv[1]);
    }
    return 0;
}

A Segmentation fault

B Argument: ./program

C Argument: (null)

D No arguments

💡 Hint

Check the value of argc when no extra arguments are passed.

When no arguments are passed, argc is 1 (only the program name). So the condition argc < 2 is true and it prints "No arguments".

❓ Predict Output

advanced

What error occurs when accessing argv out of bounds?

Consider this program:

#include <stdio.h>
int main(int argc, char *argv[]) {
    printf("%s\n", argv[2]);
    return 0;
}

If you run this program as ./program one, what happens?

C

#include <stdio.h>
int main(int argc, char *argv[]) {
    printf("%s\n", argv[2]);
    return 0;
}

A Prints garbage or causes segmentation fault

B Prints program name

C Prints "one"

D Compilation error

💡 Hint

Think about how many arguments are passed and what argv[2] points to.

Running with one argument means argc is 2 (program name and "one"). argv[2] is out of bounds and accessing it causes undefined behavior, often a segmentation fault or garbage output.

🧠 Conceptual

advanced

How are command line arguments passed to a C program?

Which statement best describes how command line arguments are passed to a C program?

A Arguments are passed as an integer count and an array of strings to <code>main()</code>.

B Arguments are passed as environment variables only.

C Arguments are passed as a single string to <code>main()</code>.

D Arguments are passed through global variables automatically.

💡 Hint

Recall the signature of main() that receives arguments.

The standard main() function receives two parameters: argc (argument count) and argv (array of argument strings).

❓ Predict Output

expert

What is the output of this program with multiple arguments?

Analyze this program:

#include <stdio.h>
int main(int argc, char *argv[]) {
    for (int i = 0; i < argc; i++) {
        printf("%d:%s ", i, argv[i]);
    }
    printf("\n");
    return 0;
}

What will be printed if you run ./program apple banana?

C

#include <stdio.h>
int main(int argc, char *argv[]) {
    for (int i = 0; i < argc; i++) {
        printf("%d:%s ", i, argv[i]);
    }
    printf("\n");
    return 0;
}

A 0:apple 1:banana 2:./program

B 0:./program 1:banana 2:apple

C 0:./program 1:apple 2:banana

D 1:./program 2:apple 3:banana

💡 Hint

Remember the order of arguments in argv.

The first argument argv[0] is always the program name. Then the arguments follow in order: argv[1] is "apple", argv[2] is "banana".