leap year and long/shot months at next day date program

Question

there is an easy way that includes all of the "days" conditions that can be in the year? leap year, long/shot months, etc... trying to make it shorter but easy to understand for someone new at code like me

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>

typedef struct date
{
    int day;
    int month;
    int year;
} date;

int main()
{
    date d;
    printf("\nPlease enter the date DD/MM/YYYY format and i will raise it up \n");
    scanf_s("%d/%d/%d", &d.day, &d.month, &d.year);
    printf("you enter: %02d/%02d/%04d", d.day, d.month, d.year);
    d.day++;
    if (d.day == 29 && d.month == 2)
    {
        d.month++;
        d.day = 1;
    }
    if (d.day == 32 && d.month == 3)
    {
        d.month++;
        d.day = 1;
    }
    if (d.day == 32 && d.month == 12)
    {
        d.year++;
        d.month = 1;
        d.day = 1;
    }
    printf("  tomorrow date: %02d/%02d/%04d", d.day, d.month, d.year);
}

Answer 1

An easy way to include all conditions would be to use the standard library <time.h>. It defines a structure to hold the date, struct tm (so you don't need to create your own) and has many useful functions such as mktime, which adjusts the date taking care of everything (change of month, change of year, leap years etc). Here is an example:

#include <stdio.h>
#include <conio.h>
#include <time.h>

int main(void)
{
    struct tm d;

    printf("Please enter the date DD/MM/YYYY format and I will raise it up\n");
    scanf_s("%d/%d/%d", &d.tm_mday, &d.tm_mon, &d.tm_year);
    d.tm_mon -= 1; // struct tm saves months from 0 to 11
    d.tm_year -= 1900; // struct tm requires years passed since 1900

    printf("You entered: %02d/%02d/%04d\n", d.tm_mday, d.tm_mon + 1, d.tm_year + 1900);

    d.tm_mday++;
    mktime(&d); // Make the necessary adjustments after having modified the day

    printf("Tomorrow date: %02d/%02d/%04d\n", d.tm_mday, d.tm_mon + 1, d.tm_year + 1900);
}

Here I used printf so I had to re-add the values I subtracted before, but <time.h> actually includes functions to easily print the date both in standard (asctime, ctime) and custom formats (strftime). There are many resources online (including SO!) to understand in depth everything it offers, have a look!

Sidenote: you don't use anything from <stdlib.h>, you should't include unnecessary libraries

Answer 2

To answer your original question: "What am I doing wrong?"

if (d.day == 29 && d.month == 2)

Should be

if (d.day == 28 && d.month == 2)

And in the same way 32 should be 31

To answer the question that you posted as an answer: "there is an easy way that includes all of the "days" conditions that can be in the year?"

const int dInM[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

if( d.day == dInM[d.month-1] ) {
    
    ++d.month;
    d.day = 1;
}

Answer 3

#include <stdio.h>

typedef struct date
{
    int day, month,year;
}date;

int main()
{
  date d;
  printf("\nPlease enter the date DD/MM/YYYY format and i will raise it up \n");
  scanf("%d%d%d",&d.day, &d.month,&d.year);
  printf("you enter: %02d/%02d/%04d", d.day, d.month, d.year); //Keep in mind that the %02d means two characters minimum width,The 0 means to pad the field using zeros and the 2 means that the field is two characters wide
  d.day++;

  switch(d.month)
  {
      case 1:
      case 3:
      case 5:
      case 7:
      case 8:
      case 10:
      case 12:
      {
           if(d.day>=32)
           {
                 d.month++;
                 d.day=d.day-31;
           }
      }break;
      case 4:
      case 6:
      case 9:
      case 11:
      {
           if(d.day>=31)
           {
                 d.month++;
                 d.day=d.day-30;
           }break;
      }
      case 2:
      {
             if ((d.year%400==0)||(d.year%4==0&&d.year%100!=0)&&d.day>=30)
             {
                   d.month++;
                   d.day=d.day-29;
             }
             else
             {
                   d.month++;
                   d.day=d.day-28;
             }
      }break;

  }

   printf("\ntomorrow date: %02d/%02d/%04d\n", d.day, d.month, d.year);
}