0x03-debugging

0x03. C - Debugging

• By Carrie Ybay

Resources

Read or watch:

• Debugging
• Rubber Duck Debugging

Debugging is the process of finding and fixing errors in software that prevents it from running correctly. As you become a more advanced programmer and an industry engineer, you will learn how to use debugging tools such as gdb or built-in tools that IDEs have. However, it's important to understand the concepts and processes of debugging manually.

Learning Objectives

At the end of this project, you are expected to be able to explain to anyone, without the help of Google:

General

• What is debugging
• What are some methods of debugging manually
• How to read the error messages

Requirements

General

• Allowed editors: vi, vim, emacs
• All your files will be compiled on Ubuntu 20.04 LTS using gcc, using the options -Wall -Werror -Wextra -pedantic -std=gnu89
• All your files should end with a new line
• Your code should use the Betty style. It will be checked using betty-style.pl and betty-doc.pl
• A README.md file at the root of the repo, containing a description of the repository
• A README.md file, at the root of the folder of this project (i.e. 0x03-debugging), describing what this project is about

Tasks

0. Multiple mains

mandatory

In most projects, we often give you only one main file to test with. For example, this main file is a test for a postitive_or_negative() function similar to the one you worked with in an earlier C project:

carrie@ubuntu:/debugging$ cat main.c
#include "main.h"

/**
* main - tests function that prints if integer is positive or negative
* Return: 0
*/

int main(void)
{
        int i;

        i = 98;
        positive_or_negative(i);

        return (0);
}
carrie@ubuntu:/debugging$

carrie@ubuntu:/debugging$ cat main.h
#ifndef MAIN_H
#define MAIN_H

#include <stdio.h>

void positive_or_negative(int i);

#endif /* MAIN_H */
carrie@ubuntu:/debugging$

carrie@ubuntu:/debugging$ gcc -Wall -pedantic -Werror -Wextra -std=gnu89 positive_or_negative.c main.c
carrie@ubuntu:/debugging$ ./a.out
98 is positive
carrie@ubuntu:/debugging$

Based on the main.c file above, create a file named 0-main.c. This file must test that the function positive_or_negative() gives the correct output when given a case of 0.

You are not coding the solution / function, you're just testing it! However, you can adapt your function from 0x01. C - Variables, if, else, while - Task #0 to compile with this main file to test locally.

• You only need to upload 0-main.c and main.h for this task. We will provide our own positive_or_negative() function.
• You are not allowed to add or remove lines of code, you may change only one line in this task.

carrie@ubuntu:/debugging$ gcc -Wall -pedantic -Werror -Wextra -std=gnu89 positive_or_negative.c 0-main.c -o 0-main
carrie@ubuntu:/debugging$ ./0-main
0 is zero
carrie@ubuntu:/debugging$ wc -l 0-main.c
16 1-main.c
carrie@ubuntu:/debugging$

Repo:

• GitHub repository: alx-low_level_programming
• Directory: 0x03-debugging
• File: 0-main.c, main.h

1. Like, comment, subscribe

mandatory

Copy this main file. Comment out (don't delete it!) the part of the code that is causing the output to go into an infinite loop.

• Don't add or remove any lines of code, as we will be checking your line count. You are only allowed to comment out existing code.
• You do not have to compile with -Wall -Werror -Wextra -pedantic for this task.

carrie@ubuntu:/debugging$ cat 1-main.c
#include <stdio.h>

/**
* main - causes an infinite loop
* Return: 0
*/

int main(void)
{
        int i;

        printf("Infinite loop incoming :(\n");

        i = 0;

        while (i < 10)
        {
                putchar(i);
        }

        printf("Infinite loop avoided! \\o/\n");

        return (0);
}
carrie@ubuntu:/debugging$

Your output should look like this:

carrie@ubuntu:/debugging$ gcc -std=gnu89 1-main.c -o 1-main
carrie@ubuntu:/debugging$ ./1-main
Infinite loop incoming :(
Infinite loop avoided! \o/
carrie@ubuntu:/debugging$ wc -l 1-main.c
24 1-main.c
carrie@ubuntu:/debugging$

Repo:

• GitHub repository: alx-low_level_programming
• Directory: 0x03-debugging
• File: 1-main.c

2. 0 > 972?

mandatory

This program prints the largest of three integers.

carrie@ubuntu:/debugging$ cat 2-main.c
#include <stdio.h>
#include "main.h"

/**
* main - prints the largest of 3 integers
* Return: 0
*/

int main(void)
{
        int a, b, c;
        int largest;

        a = 972;
        b = -98;
        c = 0;

        largest = largest_number(a, b, c);

        printf("%d is the largest number\n", largest);

        return (0);
}
carrie@ubuntu:/debugging$

carrie@ubuntu:/debugging$ cat 2-largest_number.c
#include "main.h"

/**
 * largest_number - returns the largest of 3 numbers
 * @a: first integer
 * @b: second integer
 * @c: third integer
 * Return: largest number
 */

int largest_number(int a, int b, int c)
{
    int largest;

    if (a > b && b > c)
    {
        largest = a;
    }
    else if (b > a && a > c)
    {
        largest = b;
    }
    else
    {
        largest = c;
    }

    return (largest);
}

carrie@ubuntu:/debugging$

carrie@ubuntu:/debugging$ gcc -Wall -Werror -Wextra -pedantic -std=gnu89 2-largest_number.c 2-main.c -o 2-main
carrie@ubuntu:/debugging$ ./2-main
0 is the largest number
carrie@ubuntu:/debugging$

? That's definitely not right.

Fix the code in 2-largest_number.c so that it correctly prints out the largest of three numbers, no matter the case.

• Line count will not be checked for this task.

Repo:

• GitHub repository: alx-low_level_programming
• Directory: 0x03-debugging
• File: 2-largest_number.c, main.h

3. Leap year

mandatory

This program converts a date to the day of year and determines how many days are left in the year, taking leap year into consideration.

carrie@ubuntu:/debugging$ cat 3-main_a.c
#include <stdio.h>
#include "main.h"

/**
* main - takes a date and prints how many days are left in the year, taking
* leap years into account
* Return: 0
*/

int main(void)
{
    int month;
    int day;
    int year;

    month = 4;
    day = 01;
    year = 1997;

    printf("Date: %02d/%02d/%04d\n", month, day, year);

    day = convert_day(month, day);

    print_remaining_days(month, day, year);

    return (0);
}

carrie@ubuntu:/debugging$

carrie@ubuntu:/debugging$ cat 3-convert_day.c
#include "main.h"

/**
* convert_day - converts day of month to day of year, without accounting
* for leap year
* @month: month in number format
* @day: day of month
* Return: day of year
*/

int convert_day(int month, int day)
{
    switch (month)
    {
        case 2:
            day = 31 + day;
            break;
        case 3:
            day = 59 + day;
            break;
        case 4:
            day = 90 + day;
            break;
        case 5:
            day = 120 + day;
            break;
        case 6:
            day = 151 + day;
            break;
        case 7:
            day = 181 + day;
            break;
        case 8:
            day = 212 + day;
            break;
        case 9:
            day = 243 + day;
            break;
        case 10:
            day = 273 + day;
            break;
        case 11:
            day = 304 + day;
            break;
        case 12:
            day = 334 + day;
            break;
        default:
            break;
    }
    return (day);
}

carrie@ubuntu:/debugging$

carrie@ubuntu:/debugging$ cat 3-print_remaining_days.c
#include <stdio.h>
#include "main.h"

/**
* print_remaining_days - takes a date and prints how many days are
* left in the year, taking leap years into account
* @month: month in number format
* @day: day of month
* @year: year
* Return: void
*/

void print_remaining_days(int month, int day, int year)
{
    if ((year % 4 == 0 || year % 400 == 0) && !(year % 100 == 0))
    {
        if (month >= 2 && day >= 60)
        {
            day++;
        }

        printf("Day of the year: %d\n", day);
        printf("Remaining days: %d\n", 366 - day);
    }
    else
    {
        if (month == 2 && day == 60)
        {
            printf("Invalid date: %02d/%02d/%04d\n", month, day - 31, year);
        }
        else
        {
            printf("Day of the year: %d\n", day);
            printf("Remaining days: %d\n", 365 - day);
        }
    }
}

carrie@ubuntu:/debugging$

carrie@ubuntu:/debugging$ gcc -Wall -Werror -Wextra -pedantic -std=gnu89 3-convert_day.c 3-print_remaining_days.c 3-main_a.c -o 3-main_a
carrie@ubuntu:/debugging$ ./3-main_a
Date: 04/01/1997
Day of the year: 91
Remaining days: 274
carrie@ubuntu:/debugging$

Output looks good for 04/01/1997! Let's make a new main file 3-main_b.c to try a case that is a leap year: 02/29/2000.

carrie@ubuntu:/debugging$ gcc -Wall -Werror -Wextra -pedantic -std=gnu89 3-convert_day.c 3-print_remaining_days.c 3-main_b.c -o 3-main_b
carrie@ubuntu:/debugging$ ./3-main_b
Date: 02/29/2000
Invalid date: 02/29/2000
carrie@ubuntu:/debugging$

? That doesn't seem right.

Fix the print_remaining_days() function so that the output works correctly for all dates and all leap years.

• Line count will not be checked for this task.
• You can assume that all test cases have valid months (i.e. the value of month will never be less than 1 or greater than 12) and valid days (i.e. the value of day will never be less than 1 or greater than 31).
• You can assume that all test cases have valid month/day combinations (i.e. there will never be a June 31st or November 31st, etc.), but not all month/day/year combinations are valid (i.e. February 29, 1991 or February 29, 2427).

Repo:

• GitHub repository: alx-low_level_programming
• Directory: 0x03-debugging
• File: 3-print_remaining_days.c, main.h