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Count not Updating in Double For-Loop

Time:10-11

I have a program that is meant to count a number of days that a Monday occurs in a specific range. I am having trouble updating the count through each iteration. I tried changing up the count methods even putting it in some if loops to no effect.

Current Code

#include <stdio.h>

#define SIZE 1000

#include "printVday.h"

int main() {
    //update date opposite of year so go through all the days first then the year
    //while loop that checks if the date was on a Monday and gets all the Mondays from the month then checks the second last one 
    int d = 15, m = 5, y, day, month, year, yearplus, i = 0;
    char dates[SIZE];

    printf("Enter a year: ");
    scanf("%d", &y);
    
    for (year = y; year <= y   20; year  ) {
        i  ;
        for (day = 15; day <= 24; day  ) {
            int dow = weekday(day, m, year);
            if (dow == 1) {
                printf("\n%d /%d /%d count = %d\n", day, m, year, i);
                i = 0;
            }
        }
    }
    return 0;
}

int weekday(int d, int m, int y) {

    return (d  = m < 3 ? y-- : y - 2, 23 * m / 9   d   4   y / 4 - y / 100   y / 400) % 7; //Michael Keith and Tom Craver expression to minimise number of keystrokes for conversion of a gregorian date into numerical day of week. Algorithm invented by John H. Conway

}

Currently my count is updating but not properly

Current Input & Output

Enter a year: 2015

18 /5 /2015 count = 1

16 /5 /2016 count = 1

23 /5 /2016 count = 0

15 /5 /2017 count = 1

22 /5 /2017 count = 0

21 /5 /2018 count = 1

20 /5 /2019 count = 1

18 /5 /2020 count = 1

17 /5 /2021 count = 1

24 /5 /2021 count = 0

... all the way to 2035

Desired Output

18 /5 /2015 count = 1

16 /5 /2016 count = 2

23 /5 /2016 count = 0

15 /5 /2017 count = 2

22 /5 /2017 count = 0

21 /5 /2018 count = 1

20 /5 /2019 count = 1

18 /5 /2020 count = 1

17 /5 /2021 count = 2

24 /5 /2021 count = 0

If there is just one day in the specified range that happens to land on a Monday then it will count as 1... if there are 2 days that happen to land on a Monday in the specified range the first instance will be counted as 2 while the latter is updated to 0.

CodePudding user response:

Your desired output is a very specific. You want to distinguish the first matching Monday, for which you print the total count of Mondays in the range, from the subsequent ones for which you print a count of 0.

Here is a modified version:

#include <stdio.h>

#define SIZE 1000

#include "printVday.h"

int weekday(int d, int m, int y) {
    // Michael Keith and Tom Craver expression to minimise number 
    // of keystrokes for conversion of a Gregorian date into 
    // numerical day of week. Algorithm invented by John H. Conway
    return (d  = m < 3 ? y-- : y - 2, 23 * m / 9   d   4   y / 4 - y / 100   y / 400) % 7;
}

#define MONTH      5
#define START_DAY 15
#define END_DAY   24

int main() {
    // while loop that checks if the date is a Monday and gets all the Mondays
    // from the month then checks the second last one 
    int y, day, m = MONTH, year, i;

    printf("Enter a year: ");
    if (scanf("%d", &y) != 1)
        return 1;
    
    for (year = y; year <= y   20; year  ) {
        i = 1;
        for (day = START_DAY; day <= END_DAY; day  ) {
            int dow = weekday(day, m, year);
            if (dow == 1) {
                if (i != 0) {
                    /* on first match, add number of other Mondays in range */
                    i  = (END_DAY - day) / 7;
                }
                printf("\n%d /%d /%d count = %d\n", day, m, year, i);
                i = 0;
                day  = 6;  // skip other weekdays
            }
        }
    }
    return 0;
}
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