帮助进行序数时钟转换

Question

目前，我的任务是编写一段代码，在将其转换为序数时钟之前询问用户的输入，以及闰年的例外情况。唯一的问题是，在将数组放入函数的情况下，我不知道自己在做什么。

/*The purpose of this program is to make the user enter a date and return the info as ordinal*/


#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <math.h>
#define MAXDAYSINMONTH 12 /*Define maximums for dates*/
#define MAXDAYS 365

int main()
{
    void findDate(int[], int *, int *); /*Prototype*/
int daysInMonth[MAXDAYSINMONTH]; /*Variables for function*/

int days, months, year, dayTest; /*Variables for user*/

printf("This Program will ask you for the date\n");
printf("Please enter the Day: "); /*Get User input*/
scanf("%d", &days);
printf("\nPlease enter the Month: ");
scanf("%d", &months);
printf("\nPlease enter the Year: ");
scanf("%d", &year);

findDate(days, months, &dayTest);/*call the function*/


/**
* Input:   Actual Date
* Process: Conversion
* Output:  Ordinal Date
*/

printf("The ordinal date for the date entered is: ", dayTest);/*Get Output*/
if (year / 4 == 0) { /*Look for Leap Year*/
    if (dayTest > 59) {
        dayTest = dayTest + 1;
    }
    printf("This date is a leap year");
}
else
    printf("This date is not a leap year");

return 0;
}
/*Convert Regular Date to Ordinal*/
void findDate(int days, int months, int* dayTest, int daysInMonths[]) {
    int daysInMonth[MAXDAYSINMONTH], dayTest; /*Variables for function*/
    daysInMonth[0] = 31;
    daysInMonth[1] = 28;
    daysInMonth[2] = 31;
    daysInMonth[3] = 30;
    daysInMonth[4] = 31;
    daysInMonth[5] = 30;
    daysInMonth[6] = 31;
    daysInMonth[7] = 31;
    daysInMonth[8] = 30;
    daysInMonth[9] = 31;
    daysInMonth[10] = 30;
    daysInMonth[11] = 31;
    if (months > 31) {
        *dayTestaddr = days + daysInMonth;
    }
}

新代码：

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <math.h>
#define MAXDAYSINMONTH 12 /*Define maximums for dates*/
#define MAXDAYS 365

void findDate(int days, int months, int* dayTest, int daysInMonths[]); /*Prototype*/

int main()
{

    int daysInMonth[MAXDAYSINMONTH]; /*Variables for function*/

    int days, months, year, dayTest; /*Variables for user*/

    printf("This Program will ask you for the date\n");
    printf("Please enter the Day: "); /*Get User input*/
    scanf("%d", &days);
    printf("\nPlease enter the Month: ");
    scanf("%d", &months);
    months = months - 1;
    printf("\nPlease enter the Year: ");
    scanf("%d", &year);

    findDate(days, months, &dayTest, daysInMonth); /*call the function*/


    /**
    * Input:   Actual Date
    * Process: Conversion
    * Output:  Ordinal Date
    */

    printf("The ordinal date for the date entered is: %d%d", year, dayTest);/*Get Output*/

    if (year % 4 == 0) { /*Look for Leap Year*/
        if (dayTest > 59) {
            dayTest = dayTest + 1;
        }
        printf("\nThis date is a leap year");
    }
    else
        printf("\nThis date is not a leap year");
    
    return 0;
}
/*Convert Regular Date to Ordinal*/
void findDate(int days, int months, int* dayTest, int daysInMonths[]) {
    daysInMonths[0] = 31;
    daysInMonths[1] = 28;
    daysInMonths[2] = 31;
    daysInMonths[3] = 30;
    daysInMonths[4] = 31;
    daysInMonths[5] = 30;
    daysInMonths[6] = 31;
    daysInMonths[7] = 31;
    daysInMonths[8] = 30;
    daysInMonths[9] = 31;
    daysInMonths[10] = 30;
    daysInMonths[11] = 31;
    if (months > 11 || months < 0)
    {
        // Illegal call
        printf("Your month is over 12 or under 0, try again\n");
    }
    if (months == 0) {
        *dayTest = days;
    }
    else 
        *dayTest = days + (daysInMonths[months] - daysInMonths[1]);
}

我只是保存了这个，以便我的进步更加明显，我确实认为我有一个通用的句柄，但我仍然需要继续前进，以 YYYY-DDD 的格式制作序数日期。我们将不胜感激，谢谢。

Answer 1

findDate的声明不一致，原型应该放在main之前。

你的原型：

void findDate(int[], int *, int *); /*Prototype*/

您的电话：

findDate(days, months, &dayTest);
          ^       ^
          |       Not an int*
          Not an array

你的定义：

void findDate(int days, int months, int* dayTest, int daysInMonths[])

这里有 4 个参数，但之前只有 3 个。此外，单个参数与之前的定义不匹配。

您需要解决此不一致问题。

在main 之前放：

void findDate(int days, int months, int* dayTest, int daysInMonths[]);

在main 中这样称呼它：

findDate(days, months, &dayTest, daysInMonth);

并且不要在函数体中重新定义 daysInMonth 和 dayTest 之类的变量。只需这样做：

void findDate(int days, int months, int* dayTest, int daysInMonths[]) {
    daysInMonth[0] = 31;
    daysInMonth[1] = 28;
    daysInMonth[2] = 31;
    daysInMonth[3] = 30;
    daysInMonth[4] = 31;
    daysInMonth[5] = 30;
    daysInMonth[6] = 31;
    daysInMonth[7] = 31;
    daysInMonth[8] = 30;
    daysInMonth[9] = 31;
    daysInMonth[10] = 30;
    daysInMonth[11] = 31;
    if (months >= 12 || months < 0) 
    {
        // Illegal call
        .. error handling ..
    }
    *dayTest = days + daysInMonth[months];
}

也就是说，为什么每次调用函数时都将daysInMonths 数组作为变量填充？这是对 CPU 周期的浪费。而是将其设为全局常量。

所以之前main做：

const int daysInMonth[] = {31, 28, 31, 30, ....};

那么你根本不需要传递数组并且你的函数变得更加简单：

void findDate(int days, int months, int* dayTest) {
    if (months >= 12 || months < 0) 
    {
        // Illegal call
        .. error handling ..
    }
    *dayTest = days + daysInMonth[months];
}

或者通过使用返回值而不是指针来进一步简化：

int findDate(int days, int months) {
    assert(months >= 0 && months < 12); 
    return days + daysInMonth[months];
}

然后这样称呼它：

dayTest = findDate(days, months);

简单得多...没有指针，没有要传递给函数的数组。

最后注意...数组索引从零开始，但您的用户可能会在 1 月输入 1。所以你有一个必须处理的不一致。