How do you write a c plus plus program to calculate number of days between given date's?

Answer 1

#include<stdio.h>

#include<conio.h>

#include<time.h>

time_t time(time_t *t);

main()

{

int y2,m2,d2,y,m,d;

time_t t = time(NULL);

struct tm tm = *localtime(&t);

printf("Time now: %d-%d-%d %d:%d:%d\n", tm.tm_year + 1900, tm.tm_mon + 1, tm.tm_mday, tm.tm_hour, tm.tm_min, tm.tm_sec);

y2=tm.tm_year+1900;

m2=tm.tm_mon+1;

d2=tm.tm_mday;

printf("Enter the Reference Date in DD MM YY format\n");

scanf("%d%d%d",&d,&m,&y);

int loop,total_days = 0,days_in_a_year;

int day,month,year;

int month_days[] = {31,28,31,30,31,30,31,31,30,31,30,31};

printf("\n%d\n%d\n%d\n",(y2-y),(m2-m),(d2-d));

printf("%d",month_days[m]);

if(((y2-y)==0&&(m2-m)<1&&(d2<d))(y2<y)(m<1)(m>12)(d<0)(d>month_days[m-1]))

{

printf("\nInvalid Reference Date\n");

}

else

{

total_days=month_days[m-1]-d;

if((m==2)&&((y%4)==0))

total_days+=1;

for(int i=m;i<12;i++)

{

if((y%4==0)&&i==1)

total_days+=1;

total_days+=month_days[i];

}

for(int year=y+1;year<y2;year++)

{

if(year%4==0)

total_days+=366;

else

total_days+=365;

}

for(int i=0;i<m2-1;i++)

{

if((y2%4==0)&&i==1)

total_days+=1;

total_days+=month_days[i];

}

total_days+=d2;

if((y2==y)&&(y%4==0))

total_days-=366;

if((y2==y)&&(y%4!=0))

total_days-=365;

}

printf("\nThe date difference is %d\n",total_days);

getch();

}

Answer 2

To calculate the number of days between given dates, you could calculate the Julian date of each date and then subtract.

http://en.wikipedia.org/wiki/Julian_date

Problem is that this may be inconsistent depending on century is being considered, and what convention is used, because conventions have changed over the centuries.

In the simple case, you could consider that a non leap year is 365 days, and that leap year (366 days) are years evenly divisible by four, except that a century year must be divisible by 400 to be considered a leap year. Then you consider the year, month, and day, adding the number of days in the months preceeding the month in question.

Even that "simple" definition introduces inconsistency, because the year 1600 is leap except in the UK. (There is more to it than that, but the point is made)

Answer 3

Program to compute difference between two dates

#include<stdio.h>

#include<math.h>

void main()

{

int day1,mon1,year1,day2,mon2,year2;

int ref,dd1,dd2,i;

clrscr();

printf("Enter first day, month, year");

scanf("%d%d%d",&day1,&mon1,&year1);

scanf("%d%d%d",&day2,&mon2,&year2);

ref = year1;

if(year2<year1)

ref = year2;

dd1=0;

dd1=func1(mon1);

for(i=ref;i<year1;i++)

{

if(i%4==0)

dd1+=1;

}

dd1=dd1+day1+(year1-ref)*365;

printf("No. of days of first date fronm the Jan 1 %d= %d",year1,dd1);

/* Count for additional days due to leap years*/

dd2=0;

for(i=ref;i<year2;i++)

{

if(i%4==0)

dd2+=1;

}

dd2=func1(mon2)+dd2+day2+((year2-ref)*365);

printf("No. of days from the reference year's first Jan = %d",dd2);

printf("Therefore, diff between the two dates is %d",abs(dd2-dd1));

getch();

}

int func1(x) //x for month y for dd

{ int y=0;

switch(x)

{

case 1: y=0; break;

case 2: y=31; break;

case 3: y=59; break;

case 4: y=90; break;

case 5: y=120;break;

case 6: y=151; break;

case 7: y=181; break;

case 8: y=212; break;

case 9: y=243; break;

case 10:y=273; break;

case 11:y=304; break;

case 12:y=334; break;

default: printf("Error encountered"); exit(1);

}

return(y);

}