Accessing arguments - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this C program?

Consider the following C program that prints command line arguments:

#include <stdio.h>

int main(int argc, char *argv[]) {
    printf("%d\n", argc);
    return 0;
}

If you run this program as ./program arg1 arg2, what will it print?

C

#include <stdio.h>

int main(int argc, char *argv[]) {
    printf("%d\n", argc);
    return 0;
}

A 1

B 2

C 3

D 0

💡 Hint

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

The argc variable counts all arguments including the program name. So ./program arg1 arg2 has 3 arguments total.

❓ Predict Output

intermediate

What does this program print?

Look at this C program that prints the first command line argument:

#include <stdio.h>

int main(int argc, char *argv[]) {
    if (argc > 1) {
        printf("%s\n", argv[1]);
    } else {
        printf("No argument\n");
    }
    return 0;
}

What will it print if run as ./program Hello?

C

#include <stdio.h>

int main(int argc, char *argv[]) {
    if (argc > 1) {
        printf("%s\n", argv[1]);
    } else {
        printf("No argument\n");
    }
    return 0;
}

A Hello

B No argument

C ./program

D argv[1]

💡 Hint

Remember argv[0] is the program name, argv[1] is the first argument.

The program checks if there is at least one argument after the program name. Since Hello is passed, it prints Hello.

❓ Predict Output

advanced

What is the output of this program?

Analyze this C program that prints all command line arguments:

#include <stdio.h>

int main(int argc, char *argv[]) {
    for (int i = 0; i < argc; i++) {
        printf("%s ", argv[i]);
    }
    printf("\n");
    return 0;
}

What will it print if run as ./prog one two?

C

#include <stdio.h>

int main(int argc, char *argv[]) {
    for (int i = 0; i < argc; i++) {
        printf("%s ", argv[i]);
    }
    printf("\n");
    return 0;
}

A one two

B ./prog one two

C ./prog

D prog one two

💡 Hint

Remember argv[0] is the program name including the path used to run it.

The loop prints all arguments including the program name ./prog followed by one and two, each separated by space.

❓ Predict Output

advanced

What error does this program cause?

Consider this C program:

#include <stdio.h>

int main(int argc, char *argv[]) {
    printf("%s\n", argv[argc]);
    return 0;
}

What happens when you run it?

C

#include <stdio.h>

int main(int argc, char *argv[]) {
    printf("%s\n", argv[argc]);
    return 0;
}

A Segmentation fault (accessing invalid memory)

B Prints the last argument

C Prints the program name

D Compilation error

💡 Hint

Remember that argv array indices go from 0 to argc - 1.

The argv[argc] is out of bounds because valid indices are 0 to argc - 1. Accessing it causes a segmentation fault.

🧠 Conceptual

expert

How to access the third command line argument safely?

You want to print the third command line argument in a C program. Which code snippet safely does this?

A printf("%s\n", argv[2]);

B if (argc > 2) printf("%s\n", argv[2]); else printf("No third argument\n");

C if (argc > 3) printf("%s\n", argv[3]); else printf("No third argument\n");

D if (argc >= 4) printf("%s\n", argv[3]); else printf("No third argument\n");

💡 Hint

Remember that argv[0] is the program name, so the first argument is argv[1].

The third command line argument (excluding the program name) is argv[3]. To access it safely, check argc >= 4 since argc includes the program name as the first. Option D does this correctly. Option D accesses the second argument (argv[2]), and D has no bounds check.