C program days between two dates - c

Question

I have written a program that should find the days between two dates, but it has some hiccups. The logic makes perfect sense in my head when I read through it, so I'm assuming I have some syntax errors that I keep glancing over or something.
Firstly, when entering two dates in different years, the output is always off by about one month (31 in most cases, but 32 in one case...go figure). Second, two dates exactly one month apart will return the number of days in the second month (i.e. 1/1/1 to 2/1/1 yields 28). There are inevitably some other weird things that this program does, but I am hoping that is enough information to help you guys figure out what I'm doing wrong. For the life of me I can't figure this one out on my own. I am relatively new to C, so please be gentle =)
Thanks
// Calculates the number of calendar days between any two dates in history (beginning with 1/1/1).
#include <stdio.h>
#include <stdlib.h>
void leap(int year1, int year2, int *leap1, int *leap2);
void date(int *month1, int *day1, int *year1, int *month2, int *day2, int *year2, int *leap1, int *leap2);
int main(void)
{
int month1, day1, year1, month2, day2, year2, leap1, leap2;
int daysPerMonth[] = {31,28,31,30,31,30,31,31,30,31,30,31};
int daysPerMonthLeap[] = {31,29,31,30,31,30,31,31,30,31,30,31};
leap(year1, year2, &leap1, &leap2);
date(&month1, &day1, &year1, &month2, &day2, &year2, &leap1, &leap2);
if(year1 == year2)
{
int i, total;
if(month1 == month2) // Total days if month1 == month2
{
total = day2 - day1;
printf("There are %d days between the two dates.", total);
}
else
{
if(leap1 == 1)
total = daysPerMonthLeap[month1] - day1;
else
total = daysPerMonth[month1] - day1;
for(i = month1 + 1; i < month2; i++) // Days remaining between dates (excluding last month)
{
if(leap1 == 1)
total += daysPerMonthLeap[i];
else
total += daysPerMonth[i];
}
total += day2; // Final sum of days between dates (including last month)
printf("There are %d days between the two dates.", total);
}
}
else // If year1 != year2 ...
{
int i, total, century1 = ((year1 / 100) + 1) * 100, falseleap = 0;
if(leap1 == 1)
total = daysPerMonthLeap[month1] - day1;
else
total = daysPerMonth[month1] - day1;
for(i = month1 + 1; i <= 12; i++) // Day remaining in first year
{
if(leap1 == 1)
total += daysPerMonthLeap[i];
else
total += daysPerMonth[i];
}
for(i = 1; i < month2; i++) // Days remaining in final year (excluding last month)
{
if(leap2 == 1)
total += daysPerMonthLeap[i];
else
total += daysPerMonth[i];
}
int leapcount1 = year1 / 4; // Leap years prior to and including first year
int leapcount2 = year2 / 4; // Leap years prior to and NOT including final year
if(year2 % 4 == 0)
leapcount2 -= 1;
int leaptotal = leapcount2 - leapcount1; // Leap years between dates
for(i = century1; i < year2; i += 100) // "False" leap years (divisible by 100 but not 400)
{
if((i % 400) != 0)
falseleap += 1;
}
total += 365 * (year2 - year1 - 1) + day2 + leaptotal - falseleap; // Final calculation
printf("There are %d days between the two dates.", total);
}
return 0;
}
void leap(int year1, int year2, int *leap1, int *leap2) // Determines if first and final years are leap years
{
if(year1 % 4 == 0)
{
if(year1 % 100 == 0)
{
if(year1 % 400 == 0)
*leap1 = 1;
else
*leap1 = 0;
}
else
*leap1 = 1;
}
else
*leap1 = 0;
if(year2 % 4 == 0)
{
if(year2 % 100 == 0)
{
if(year2 % 400 == 0)
*leap2 = 1;
else
*leap2 = 0;
}
else
*leap2 = 1;
}
else
*leap2 = 0;
}
void date(int *month1, int *day1, int *year1, int *month2, int *day2, int *year2, int *leap1, int *leap2)
{
for(;;) // Infinite loop (exited upon valid input)
{
int fail = 0;
printf("\nEnter first date: ");
scanf("%d/%d/%d", month1, day1, year1);
if(*month1 < 1 || *month1 > 12)
{
printf("Invalid entry for month.\n");
fail += 1;
}
if(*day1 < 1 || *day1 > 31)
{
printf("Invalid entry for day.\n");
fail += 1;
}
if(*year1 < 1)
{
printf("Invalid entry for year.\n");
fail += 1;
}
if(daysPerMonth[month1] == 30 && *day1 > 30)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
if(*month1 == 2)
{
if(*leap1 == 1 && *day1 > 29)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
else if(*day1 > 28)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
}
if(fail > 0)
continue;
else
break;
}
for(;;)
{
int fail = 0;
printf("\nEnter second date: ");
scanf("%d/%d/%d", month2, day2, year2);
if(*year1 == *year2)
{
if(*month1 > *month2)
{
printf("Invalid entry.\n");
fail += 1;
}
if(*month1 == *month2 && *day1 > *day2)
{
printf("Invalid entry.\n");
fail += 1;
}
}
if(*month2 < 1 || *month2 > 12)
{
printf("Invalid entry for month.\n");
fail += 1;
}
if(*day2 < 1 || *day2 > 31)
{
printf("Invalid entry for day.\n");
fail += 1;
}
if(*year2 < 1)
{
printf("Invalid entry for year.\n");
fail += 1;
}
if(daysPerMonth[month2] == 30 && *day2 > 30)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
if(*month2 == 2)
{
if(*leap2 == 1 && *day2 > 29)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
else if(*day2 > 28)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
}
if(fail > 0)
continue;
else
break;
}
}

Answer 1

First, that leap function feels overly complicated; you don't need to do both dates in one function call, and I'm sure that can be written more succinctly so that it is more obviously correct. Here's a version I've got laying around that isn't succinct but I'm confident it is easy to check the logic:
int is_leap_year(int year) {
if (year % 400 == 0) {
return 1;
} else if (year % 100 == 0) {
return 0;
} else if (year % 4 == 0) {
return 1;
} else {
return 0;
}
}
You could call it like this:
int year1, year2, leap1, leap2;
year1 = get_input();
year2 = get_input();
leap1 = is_leap_year(year1);
leap2 = is_leap_year(year2);
No pointers and significantly less code duplication. Yes, I know that is_leap_year() can be reduced to a single if(...) statement, but this is easy for me to read.
Second, I think you're got a mismatch between 0-indexed arrays and 1-indexed human months:
if(*month1 < 1 || *month1 > 12)
vs
int daysPerMonth[] = {31,28,31,30,31,30,31,31,30,31,30,31};
Third, I think that days per month can be calculated slightly nicer:
int days_in_month(int month, int year) {
int leap = is_leap_year(year);
/* J F M A M J J A S O N D */
int days[2][12] = {{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
{31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};
if (month < 0 || month > 11 || year < 1753)
return -1;
return days[leap][month];
}
Here, I assume January is 0; you would need to force the rest of the code to match. (I learned this double-array trick from The Elements of Programming Style (page 54).) The best part of using a routine like this is that it removes the leap condition from the difference calculation.
Fourth, you're indexing arrays outside their bounds:
for(i = month1 + 1; i <= 12; i++)
{
if(leap1 == 1)
total += daysPerMonthLeap[i];
This is just another instance of the problem with 0-indexed arrays and 1-indexed months -- but be sure that you fix this, too, when you fix the months.
I have a fear that I haven't yet found all the issues -- you may find it easier to sort the first and the second date after input and remove all that validation code -- and then use names before and after or something to give names that are easier to think through in the complicated core of the calculation.

Answer 2

This is not a complete answer. I just wanted to mention a better way to calculate leap year (this is taken from The C Programming Language - Page #41)
if ((year % 4 == 0 && year % 100 != 0) || year % 400 ==0)
printf("%d is a leap year \n", year);
else
printf("%d is not a leap year \n", year);

Answer 3

Reduce all month indexes by 1.
What I mean to say is January will correspond to daysPerMonth[0] or daysPerMonthLeap[0] and not daysPerMonth[1] or daysPerMonthLeap[1].
The reason for this being array indexes start from 0.
So, wherever you are using month1, month2 insidedaysPerMonth[] or daysPerMonthLeap[], use month1-1 and month2-1 instead.
I hope this is clear enough.
Otherwise, feel free to comment.

Answer 4

Change
int daysPerMonth[] = {31,28,31,30,31,30,31,31,30,31,30,31};
int daysPerMonthLeap[] = {31,29,31,30,31,30,31,31,30,31,30,31};
to
int daysPerMonth[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
int daysPerMonthLeap[] = {0,31,29,31,30,31,30,31,31,30,31,30,31};
i.e. pad the arrays at the beginning since all the code relies on the array values to start at element 1 rather than element 0.
That will get rid of the error you complained of.
The other problem is an off-by-one error when you add day2 to the total. In both cases you should add day2 - 1 rather than day2. This is also due to the date indexes starting at 1 instead of 0.
After I made these changes (plus a couple just to get the code to compile), it works properly.

Answer 5

There are multiple problems in your code snippet.. but I must say it is a very good attempt. There are many short cuts to what you're try to achieve.
I have written the following program which finds the number of days between two given dates. You may use this as a reference.
#include <stdio.h>
#include <stdlib.h>
char *month[13] = {"None", "Jan", "Feb", "Mar",
"Apr", "May", "June", "July",
"Aug", "Sept", "Oct",
"Nov", "Dec"};
/*
daysPerMonth[0] = non leap year
daysPerMonth[1] = leap year
*/
int daysPerMonth[2][13] = {{-1, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
{-1, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};
typedef struct _d {
int day; /* 1 to 31 */
int month; /* 1 to 12 */
int year; /* any */
}dt;
void print_dt(dt d)
{
printf("%d %s %d \n", d.day, month[d.month], d.year);
return;
}
int leap(int year)
{
return ((year % 4 == 0 && year % 100 != 0) || year % 400 ==0) ? 1 : 0;
}
int minus(dt d1, dt d2)
{
int d1_l = leap(d1.year), d2_l = leap(d2.year);
int y, m;
int total_days = 0;
for (y = d1.year; y >= d2.year ; y--) {
if (y == d1.year) {
for (m = d1.month ; m >= 1 ; m--) {
if (m == d1.month) total_days += d1.day;
else total_days += daysPerMonth[leap(y)][m];
// printf("%d - %5s - %d - %d \n", y, month[m], daysPerMonth[leap(y)][m], total_days);
}
} else if (y == d2.year) {
for (m = 12 ; m >= d2.month ; m--) {
if (m == d2.month) total_days += daysPerMonth[leap(y)][m] - d2.day;
else total_days += daysPerMonth[leap(y)][m];
// printf("%d - %5s - %d - %d \n", y, month[m], daysPerMonth[leap(y)][m], total_days);
}
} else {
for (m = 12 ; m >= 1 ; m--) {
total_days += daysPerMonth[leap(y)][m];
// printf("%d - %5s - %d - %d \n", y, month[m], daysPerMonth[leap(y)][m], total_days);
}
}
}
return total_days;
}
int main(void)
{
/* 28 Oct 2018 */
dt d2 = {28, 10, 2018};
/* 30 June 2006 */
dt d1 = {30, 6, 2006};
int days;
int d1_pt = 0, d2_pt = 0;
if (d1.year > d2.year) d1_pt += 100;
else d2_pt += 100;
if (d1.month > d2.month) d1_pt += 10;
else d2_pt += 10;
if (d1.day > d2.day) d1_pt += 1;
else d2_pt += 1;
days = (d1_pt > d2_pt) ? minus(d1, d2) : minus(d2, d1);
print_dt(d1);
print_dt(d2);
printf("number of days: %d \n", days);
return 0;
}
The output is as follows:
$ gcc dates.c
$ ./a.out
30 June 2006
28 Oct 2018
number of days: 4503
$
Note: this is not a complete program. It lacks input validation.
Hope it helps!

Answer 6

//Difference/Duration between two dates
//No need to calculate leap year offset or anything
// Author: Vinay Kaple
# include <iostream>
using namespace std;
int main(int argc, char const *argv[])
{
int days_add, days_sub, c_date, c_month, b_date, b_month, c_year, b_year;
cout<<"Current Date(dd mm yyyy): ";
cin>>c_date>>c_month>>c_year;
cout<<"Birth Date(dd mm yyyy): ";
cin>>b_date>>b_month>>b_year;
int offset_month[12] = {0,31,59,90,120,151,181,212,243,273,304,334};
days_add = c_date + offset_month[c_month-1];
days_sub = b_date + offset_month[b_month-1];
int total_days = (c_year-b_year)*365.2422 + days_add - days_sub+1;
cout<<"Total days: "<<total_days<<"\n";
int total_seconds = total_days*24*60*60;
cout<<"Total seconds: "<<total_seconds<<"\n";
return 0;
}