7

我写了一个程序，应该找到两个日期之间的天数，但它有一些小问题。当我通读它时，这个逻辑在我的脑海中非常有意义，所以我假设我有一些语法错误，我一直在看或者其他什么。

首先，当输入不同年份的两个日期时，输出总是关闭大约一个月（大多数情况下为 31，但在一种情况下为 32……看图）。其次，两个恰好相隔一个月的日期将返回第二个月的天数（即 1/1/1 到 2/1/1 产生 28）。该程序不可避免地会发生其他一些奇怪的事情，但我希望这些信息足以帮助你们找出我做错了什么。对于我的一生，我无法独自解决这个问题。我对C比较陌生，所以请温柔=）

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;
        }
}
4

6 回答 6

9

首先，该leap功能感觉过于复杂；您不需要在一个函数调用中同时执行两个日期，我相信可以更简洁地编写它，以便它更明显正确。这是我所使用的一个不简洁的版本，但我相信很容易检查逻辑：

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;
        }
}

你可以这样称呼它：

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，代码重复也大大减少。是的，我知道这is_leap_year()可以简化为一个if(...)语句，但这对我来说很容易阅读。

其次，我认为您在 0 索引数组和 1 索引人类月份之间存在不匹配：

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

对比

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

第三，我认为每月的天数可以计算得更好一些：

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；您需要强制其余代码匹配。（我从The Elements of Programming Style（第 54 页）中学到了这个双数组技巧。）使用这样的例程最好的部分是它从差异计算中删除了跳跃条件。

第四，您正在索引超出其范围的数组：

            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——但请确保在修复月份时也修复此问题。

我担心我还没有找到所有的问题——你可能会发现在输入后对第一个和第二个日期进行排序并删除所有验证码更容易——然后使用名称before和after/或其他东西来命名在计算的复杂核心中更容易思考。

于 2012-04-25T04:12:06.203 回答
3

这不是一个完整的答案。我只是想提一个更好的方法来计算闰年（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);
于 2012-04-25T04:19:26.497 回答
3

将所有月份索引减 1。

我的意思是一月将对应daysPerMonth[0]ordaysPerMonthLeap[0]而不是daysPerMonth[1]or daysPerMonthLeap[1]。这是数组索引的原因是从 0 开始的。

所以，无论你在哪里使用month1，month2在daysPerMonth[]ordaysPerMonthLeap[]中，使用month1-1andmonth2-1代替。

我希望这足够清楚。否则，请随时发表评论。

于 2012-04-25T04:09:53.963 回答
2

改变

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};

至

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};

即在开头填充数组，因为所有代码都依赖于数组值从元素1而不是元素0开始。

这将摆脱您抱怨的错误。

day2另一个问题是当您添加到总数时出现一个错误。在这两种情况下，您都应该添加day2 - 1而不是day2. 这也是由于日期索引从 1 而不是 0 开始的。

在我进行了这些更改（加上一些只是为了编译代码）之后，它可以正常工作。

于 2012-04-25T04:34:01.670 回答
2

您的代码片段中有多个问题..但我必须说这是一个非常好的尝试。您尝试实现的目标有很多捷径。

我编写了以下程序，该程序查找两个给定日期之间的天数。您可以将此作为参考。

#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;
}

输出如下：

$ gcc dates.c 
$ ./a.out 
30 June 2006 
28 Oct 2018 
number of days: 4503 
$

注意：这不是一个完整的程序。它缺乏输入验证。

希望能帮助到你！

于 2012-04-25T06:25:09.680 回答
0 
//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;
}
于 2016-05-07T04:44:18.230 回答