我的大学课程的这个C程序应该找到两个日期之间的天数。它在很大程度上起作用，但是当month1和month2不同时，输出通常只有几天太高。此外，有31天的月份似乎没有被识别出来，导致当“31”作为日期输入时“无效的月和日组合”错误消息。我不确定为什么会出现这些问题。在此先感谢您的帮助！
// Calculates the number of calendar days between any two dates in history (beginning with 1/1/1).
#include <stdio.h>
#include <stdlib.h>
void date(int *month1, int *day1, int *year1, int *month2, int *day2, int *year2);
void leap(int year1, int year2, int *leap1, int *leap2);
int main(void)
{
int month1, day1, year1, month2, day2, year2, leap1, leap2;
date(&month1, &day1, &year1, &month2, &day2, &year2);
leap(year1, year2, &leap1, &leap2);
if(year1 == year2)
{
int i, total = 0;
if(month1 == month2) // Total days if month1 == month2
{
total = day2 - day1;
printf("There are %d days between the two dates.", total);
}
else
{
if(month1 == 1||3||5||7||8||10) // Days remaining in first month
total = 31 - day1;
else if(month1 == 4||6||9||11)
total = 30 - day1;
else
{
if(leap1 == 1)
total = 29 - day1;
else
total = 28 - day1;
}
for(i = month1 + 1; i < month2; i++) // Days remaining between dates (excluding last month)
{
if(i == 3||5||7||8||10)
total += 31;
else if(i == 4||6||9||11)
total += 30;
else
{
if(leap1 == 1)
total += 29;
else
total += 28;
}
}
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(month1 == 1||3||5||7||8||10||12) // Days remaining in first month
total = 31 - day1;
else if(month1 == 4||6||9||11)
total = 30 - day1;
else
{
if(leap1 == 1)
total = 29 - day1;
else
total = 28 - day1;
}
for(i = month1 + 1; i <= 12; i++) // Day remaining in first year
{
if(i == 3||5||7||8||10||12)
total += 31;
else if(i == 4||6||9||11)
total += 30;
else
{
if(leap1 == 1)
total += 29;
else
total += 28;
}
}
for(i = 1; i < month2; i++) // Days remaining in final year (excluding last month)
{
if(i == 1||3||5||7||8||10)
total += 31;
else if(i == 4||6||9||11)
total += 30;
else
{
if(leap2 == 1)
total += 29;
else
total += 28;
}
}
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 date(int *month1, int *day1, int *year1, int *month2, int *day2, int *year2)
{
for(;;) // Infinite loop (exited upon valid input)
{
int fail = 0;
printf("Enter 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((*month1 == 4||6||9||11) && *day1 > 30)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
if(*month1 == 2)
{
if(*year1 % 4 == 0)
{
if(*year1 % 100 == 0)
{
if(*year1 % 400 == 0 && *day1 > 29)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
if(*year1 % 400 != 0 && *day1 > 28)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
}
if(*year1 % 100 != 0 && *day1 > 29)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
}
if(*year1 % 4 != 0 && *day1 > 28)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
}
if(fail > 0)
continue;
else
break;
}
for(;;)
{
int fail = 0;
printf("Enter second date: ");
scanf("%d/%d/%d", month2, day2, year2);
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((*month2 == 4||6||9||11) && *day2 > 30)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
if(*month2 == 2)
{
if(*year2 % 4 == 0)
{
if(*year2 % 100 == 0)
{
if(*year2 % 400 == 0 && *day2 > 29)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
if(*year2 % 400 != 0 && *day2 > 28)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
}
if(*year2 % 100 != 0 && *day2 > 29)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
}
if(*year2 % 4 != 0 && *day2 > 28)
{
printf("Invalid month and day combination.\n");
fail += 1;
}
}
if(fail > 0)
continue;
else
break;
}
}
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;
}
答案 0 :(得分：1)
if(month1 == 1||3||5||7||8||10)
始终评估为true，因为它已被解析
if ((month == 1) || 3 || 5 || 7 || 8 || 10)
并且任何非零整数的计算结果为true。另外，你忘记了12月。
答案 1 :(得分：0)
你的代码中有很多构造：
value == 1||3||5||7||8||10||12
这不符合你在C中的想法，你必须这样写：
value == 1 || value == 3 || value == 5 || etc...
在您修复此问题并清理完代码后，如果问题仍然存在，请回来（尽管提出新问题）。
答案 2 :(得分：0)
i == 3||5||7||8||10||12
此条件将始终评估为true因此||不能用于区分不同的案例。它是一个逻辑运算符，任何不等于0的值都将计算为true。
我建议您在阵列中存储天数：
int daysPerMonth[] = {31,28,31,30 .. };
这样，在daysPerMonth[i]没有太多复杂条件的情况下，您可以轻松了解每月有多少天。它也会提高可读性。
答案 3 :(得分：0)
此：
if (month1 == 1||3||5||7||8||10||12)
并不像你想的那样做。你的意思是：
if (month1 == 1 || month1 == 3 || month1 == 5 ||
month1 == 7|| month1 == 8|| month1 == 10|| month1 == 12)
您的陈述所做的是评估(month == 1) || 3。为此，它需要将3转换为布尔值，这意味着它变为true，因为它不是零。因此，比较的结果是正确的，这意味着整个if子句也是正确的（事实上，懒惰意味着它在此时实现这一权利（或者如果月份实际上是{{1考虑到剩下的月份数，它甚至都没有计算任何东西。）
因此，整个if子句的计算结果为1。在代码中包含大量这些代码，您的代码必然会做一些奇怪的事情。