In this article, you will learn how to find the number of days between two dates. That means, if the given two dates are dt1 and dt2, you need to find the total number of days between them.

For example,

Input: 
       dt1 = {10, 2, 2022}
       dt2 = {8, 1, 2023}


Output:
     Difference between two dates is 332

Disclaimer: In my code date is only valid if lying in the range of 1800 to 9999. You can change it as per your requirement.

Approach (Using Subtraction):

Follow the steps below to solve the problem: But you should try the problem yourself first.

• First, validate the entered date. Because it is a primary prerequisite.
• Calculate the total number of days in the given first date (dt₁).
• Add extra days for dt₁ if it has leap years.
• Calculate the total number of days in the given first date (dt₂).
• Add extra days for dt₂ if it has leap years.
• Now just subtract dt₁ and dt₂ to get the total number of days between the dates.

C Program to find the number of days between two given dates:

The following is a C program to calculate the difference between two dates in days. You must need to enter the date in the format of mentioned structure (SDate).

#include <stdio.h>

#define MAX_YR  9999
#define MIN_YR  1800
#define INVALID_YEAR -1
#define INVALID_DAYS_CNT -1


//structure for day, month, year
typedef struct
{
    int dd;
    int mm;
    int yyyy;
} SDate;

// Function to check leap year.
//Function returns 1 if leap year
int  isLeapYear(int year)
{
    return (((year % 4 == 0) &&
             (year % 100 != 0)) ||
            (year % 400 == 0));
}

// returns 1 if given date is valid.
int validDate(const SDate *const pValidDate)
{
    //check range of year,month and day
    if (pValidDate->yyyy > MAX_YR ||
            pValidDate->yyyy < MIN_YR)
        return 0;
    if (pValidDate->mm < 1 || pValidDate->mm > 12)
        return 0;
    if (pValidDate->dd < 1 || pValidDate->dd > 31)
        return 0;
    //Handle feb days in leap year
    if (pValidDate->mm == 2)
    {
        if (isLeapYear(pValidDate->yyyy))
            return (pValidDate->dd <= 29);
        else
            return (pValidDate->dd <= 28);
    }
    //handle months which has only 30 days
    if (pValidDate->mm == 4 || pValidDate->mm == 6 ||
            pValidDate->mm == 9 || pValidDate->mm == 11)
        return (pValidDate->dd <= 30);
    return 1;
}


//number of days in month
const int monthDays[12]
    = { 31, 28, 31, 30, 31, 30,
        31, 31, 30, 31, 30, 31
      };


// This function counts number of
// leap years before the given date
int countLeapYears(const SDate * const pDate)
{
    int cntLeapYear = INVALID_YEAR;

    if(pDate != NULL)
    {
        int years = pDate->yyyy;

        //If current is not aplicable for leap
        // year then exclude it.
        if (pDate->mm <= 2)
            years--;

        // A year is a leap year if it
        // is a multiple of 4,
        // multiple of 400 and not a
        // multiple of 100.
        cntLeapYear =  ((years / 4)
                        - (years / 100)
                        + (years / 400));
    }
    return cntLeapYear;
}

// Function returns number of
// days between two given dates
int daysBetweenDates(const SDate *const pDt1, const SDate * const pDt2)
{
    if((pDt1 == NULL) && (pDt2 == NULL))
    {
        return -1;
    }

    int daysCnt = INVALID_DAYS_CNT;
    int isDateValid = validDate(pDt1);
    isDateValid=  isDateValid? validDate(pDt2):0;
    if(isDateValid)
    {

        // COUNT TOTAL NUMBER OF DAYS
        // BEFORE FIRST DATE 'pDt1'

        // initialize count using years and day
        long int dayCnt1 = (pDt1->yyyy * 365) + pDt1->dd;
        int index = 0;

        // Add days for months in given date
        for (index = 0; index < (pDt1->mm - 1); ++index)
        {
            dayCnt1 += monthDays[index];
        }

        // Add a day extra day for every leap year
        // beacause leap year has 366 days,
        dayCnt1 += countLeapYears(pDt1);

        // SIMILARLY, COUNT TOTAL NUMBER OF
        // DAYS BEFORE 'pDt2'

        long int dayCnt2 = (pDt2->yyyy * 365) + pDt2->dd;
        for (index = 0; index < (pDt2->mm - 1); ++index)
        {
            dayCnt2 += monthDays[index];
        }
        //Add extra day of leap years
        dayCnt2 += countLeapYears(pDt2);

        //number of days between two date
        daysCnt = (dayCnt2>=dayCnt1)? (dayCnt2 - dayCnt1):(dayCnt1-dayCnt2);
    }
    else
    {
        printf("Date is invalid.\n");
    }

    return daysCnt;

}

// Driver code
int main()
{
    SDate dt1 = {10, 2, 2022};
    SDate dt2 = {8, 1, 2023};

    //calling function
    const int dayCntBetweenDays = daysBetweenDates(&dt1,&dt2);
    if(dayCntBetweenDays != INVALID_DAYS_CNT)
    {
        printf( "Difference between two dates is %d\n",dayCntBetweenDays);
    }

    return 0;
}

Output:

Difference between two dates is 332

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