问我是否在updateDate函数中正确地使用了指针？

EN

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

// ALL TESTS OMMITTED!

struct date {
  int month;
  int day;
  int year;
};
// add the pointer mark (asterix)
struct date *dateUpdate(struct date *today);
int numberOfDays(struct date *d);
bool isLeapYear(struct date *d);

int main(void)
{
  // again ,just add the pointer marks
  struct date *thisDay, *nextDay;
  // using a pointer means that all you have is a pointer
  // but you need some memory to store the content
  thisDay = malloc(sizeof(struct date));

  printf("Enter today's date (mm dd yyyy) : ");
  // use the "->" notation to get to the respective storages
  scanf("%i%i%i", &thisDay->month, &thisDay->day, &thisDay->year);

  // dateUpdate() has been changed to accept and return a pointer,
  // so no change in notations here 
  nextDay = dateUpdate(thisDay);
  // again: use the "->" notation to get to the respective storages
  printf("Tomorrow's date is %i/%i/%.2i.\n", nextDay->month, nextDay->day,
     nextDay->year % 100);

  // memory allocated by 'alloc() needs to be free'd, too
  free(nextDay);
  free(thisDay);

  return 0;
}

// just added pointer markings
struct date *dateUpdate(struct date *today)
{
  struct date *tomorrow;
  // again, we need to allocated some memory
  // not only to get storage but also to be able to return it
  tomorrow = malloc(sizeof(struct date));
  // again: use the "->" notation to get to the respective storages
  if (today->day != numberOfDays(today)) {
    // the notation of the cast does not change, the target does
    *tomorrow = (struct date) {
    today->month, today->day + 1, today->year};
  } else if (today->month == 12) {
    *tomorrow = (struct date) {
    1, 1, today->year + 1};
  } else {
    *tomorrow = (struct date) {
    today->month + 1, 1, today->year};
  }
  return tomorrow;
}

int numberOfDays(struct date *d)
{
  int days;
  const int daysPerMonth[12] =
      { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

  if (isLeapYear(d) && d->month == 2) {
    days = 29;
  } else {
    days = daysPerMonth[d->month - 1];
  }

  return days;
}

bool isLeapYear(struct date * d)
{
  bool leapYearFlag;
  // you have one "d.year %" too much in your code
  if ((d->year % 4 == 0 && d->year % 100 != 0) || d->year % 400 == 0) {
    leapYearFlag = true;
  } else {
    leapYearFlag = false;
  }

  return leapYearFlag;
}

/* Exercise 10.11

   Given the definition of a date structure as defined in this chapter, write
   a function called dateUpdate() that takes a pointer to a date structure as
   its argument and that updates the structure to the following day.
*/

#include <stdio.h>
#include <stdbool.h>

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

void dateUpdate (struct date *ptr)
{
    int numberOfDays (struct date d);

    if ( ptr->day != numberOfDays (*ptr) )
        ++ptr->day;
    else if ( ptr->month == 12 ) {                  // end of year
        ptr->day = 1;
        ptr->month = 1;
        ++ptr->year;
    }
    else {                                          // end of month
        ptr->day = 1;
        ++ptr->month;
    }
}

int numberOfDays (struct date d)
{
    bool isLeapYear (struct date d);

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

    if ( isLeapYear (d) && d.month == 2 )
        days = 29;
    else
        days = daysPerMonth[d.month - 1];

    return days;        
}

bool isLeapYear (struct date d)
{
    bool leapYearFlag;

    if ( (d.year % 4 == 0 && d.year % 100 != 0) || d.year % 400 == 0 )
        leapYearFlag = true;                        // leap year
    else
        leapYearFlag = false;                       // not a leap year

    return leapYearFlag;        
}

int main (void)
{
    void dateUpdate (struct date *ptr);

    struct date calendar;

    printf ("Enter a day's date (mm dd yyyy): ");
    scanf ("%i%i%i", &calendar.month, &calendar.day, &calendar.year);

    dateUpdate (&calendar);

    printf ("The next day's date is %i/%i/%.2i.\n", calendar.month,
             calendar.day, calendar.year % 100);

    return 0;
}