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.

Practice

1.

What do argc and argv represent in a C program's main function?

easy

A. argc is the count of arguments; argv is an array of argument strings.

B. argc is the first argument; argv is the last argument.

C. argc is the program name; argv is the argument count.

D. argc and argv are used to return values from main.

Solution

1. Step 1: Understand argc meaning

   argc holds the number of command-line arguments passed, including the program name.

2. Step 2: Understand argv meaning

   argv is an array of strings holding each argument, starting with the program name at index 0.

3. Final Answer:

   argc is the count of arguments; argv is an array of argument strings. -> Option A

4. Quick Check:

   argc = count, argv = array [OK]

Hint: Remember: argc counts, argv holds strings [OK]

Common Mistakes:

• Confusing argc as the first argument value
• Thinking argv is a single string, not an array
• Assuming argc excludes the program name

2.

Which of the following is the correct way to declare the main function to access command-line arguments in C?

easy

A. main(int argc, char argv[])

B. void main(char argc, char argv[])

C. int main(char argc, int argv[])

D. int main(int argc, char *argv[])

Solution

1. Step 1: Check return type and parameters

   The standard way is int main(int argc, char *argv[]) where argc is int and argv is array of char pointers.

2. Step 2: Identify incorrect declarations

   Options A, B, and C have wrong types or missing return type.

3. Final Answer:

   int main(int argc, char *argv[]) -> Option D

4. Quick Check:

   Correct main signature uses int and char* array [OK]

Hint: Main must return int and use char* argv[] [OK]

Common Mistakes:

• Using void as return type for main
• Declaring argv as char instead of char* array
• Omitting return type in main declaration

3.

What will be the output of this program if run as ./prog apple banana?

int main(int argc, char *argv[]) {
    printf("%d %s %s\n", argc, argv[0], argv[1]);
    return 0;
}

medium

A. 2 apple banana

B. 3 ./prog apple

C. 3 apple banana

D. 2 ./prog apple

Solution

1. Step 1: Count arguments with argc

   Running as ./prog apple banana passes 3 arguments: program name, apple, banana. So argc = 3.

2. Step 2: Identify argv values

   argv[0] is program name ./prog, argv[1] is apple.

3. Final Answer:

   3 ./prog apple -> Option B

4. Quick Check:

   argc=3, argv[0]=./prog, argv[1]=apple [OK]

Hint: argv[0] is program name; argc counts all args [OK]

Common Mistakes:

• Assuming argv[0] is first user argument
• Counting argc as number of user arguments only
• Mixing argv indexes and argc count

4.

Identify the error in this program snippet that tries to print the second argument:

int main(int argc, char *argv[]) {
    printf("%s\n", argv[2]);
    return 0;
}

medium

A. No error; prints second argument correctly.

B. Syntax error: argv should be argv[1].

C. Possible runtime error if less than 3 arguments are passed.

D. Missing return statement in main.

Solution

1. Step 1: Check argv index usage

   Accessing argv[2] means the third argument (program name is argv[0]).

2. Step 2: Consider argument count safety

   If fewer than 3 arguments are passed, argv[2] is invalid and causes runtime error.

3. Final Answer:

   Possible runtime error if less than 3 arguments are passed. -> Option C

4. Quick Check:

   Accessing argv beyond argc causes runtime error [OK]

Hint: Check argc before accessing argv index [OK]

Common Mistakes:

• Ignoring argc check before argv access
• Confusing argv index with argument number
• Assuming argv[2] always exists

5.

Write a program snippet that safely prints the first user argument if provided, or prints No argument otherwise.

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

hard

A. if (argc > 1) printf("%s\n", argv[1]); else printf("No argument\n");

B. printf("%s\n", argv[1]);

C. if (argc == 1) printf("%s\n", argv[1]); else printf("No argument\n");

D. if (argc < 1) printf("No argument\n"); else printf("%s\n", argv[0]);

Solution

1. Step 1: Check argument count safely

   To print first user argument, ensure argc > 1 because argv[0] is program name.

2. Step 2: Print argument or fallback message

   If argc > 1, print argv[1]; else print "No argument".

3. Final Answer:

   if (argc > 1) printf("%s\n", argv[1]); else printf("No argument\n"); -> Option A

4. Quick Check:

   Check argc > 1 before argv[1] access [OK]

Hint: Always check argc before using argv index [OK]

Common Mistakes:

• Not checking argc before accessing argv[1]
• Printing argv[0] instead of user argument
• Using wrong condition like argc == 1