C两个日期之间的程序日

Question

我已经写了一个程序，应该找出两天之间的日期，但它有一些打嗝。当我阅读它的时候，这个逻辑在我的脑海中是非常有意义的，所以我假设我有一些语法错误，我一直在浏览或者别的什么。

首先，在不同年份输入两个日期时，输出总是会减少大约一个月(多数情况下为31个，但在一个case...go数字中为32个)。第二，相隔一个月的两个日期将返回第二个月的天数(即1/1/1至2/1/1 /1产生28天)。当然，这个程序还会做一些奇怪的事情，但我希望这是足够的信息来帮助你们知道我做错了什么。为了我的生命，我自己想不出这件事。我对C比较陌生，所以请温柔点

谢谢

// 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

首先，这个leap函数感觉过于复杂；您不需要在一个函数调用中同时执行两个日期，我确信可以更简洁地编写它，这样它就更明显地正确了。这里有一个不简洁的版本，但我相信它很容易检查逻辑：

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;
        }
}

你可以这样说：

int year1, year2, leap1, leap2;
year1 = get_input();
year2 = get_input();
leap1 = is_leap_year(year1);
leap2 = is_leap_year(year2);

没有指针，代码复制显著减少。是的，我知道is_leap_year()可以简化为一个if(...)语句，但这对我来说很容易读懂。

其次，我认为0索引数组和1索引人工月之间存在不匹配：

            if(*month1 < 1 || *month1 > 12)

vs

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

第三，我认为每月的日数可以计算得稍微好一点：

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];
}

在这里，我假设一月是0；您需要强制其余的代码匹配。(我从The Elements of Programming Style (page 54)那里学到了这个双数组技巧。)使用像这样的例程最好的部分是，它从差分计算中删除了leap条件。

第四，索引数组超出了它们的界限：

            for(i = month1 + 1; i <= 12; i++)
            {
                if(leap1 == 1)
                    total += daysPerMonthLeap[i];

这只是0-索引数组和1-索引月份问题的另一个实例--但在修复月份时，请确保也修复了这个问题。

我担心我还没有发现所有的问题--在输入和删除所有验证代码之后，您可能会发现更容易对第一次和第二次日期进行排序--然后使用名称before和after或其他名称来给出在计算的复杂核心中更容易考虑的名称。

Answer 2

将所有月份索引减少1。

我想说的是一月将与daysPerMonth[0]或daysPerMonthLeap[0]相对应，而不是daysPerMonth[1]或daysPerMonthLeap[1]。出现这种情况的原因是数组索引从0开始。

因此，无论您在哪里使用month1、month2 and daysPerMonth[]或daysPerMonthLeap[]，请使用month1-1和month2-1。

我希望这是足够清楚的。否则，请随意评论。

Answer 3

这不是一个完整的答案。我只想提一种更好的计算闰年的方法(这取自The C Programming Language -第41页)。

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);