计算明天日期的 C++ 函数给出今天的日期

Question

这是我编写的一个函数，它首先检查日期是否有效，然后计算出一天后的日期。为简单起见，闰年不适用于此程序。

#include <iostream>
using namespace std;

// Declaration of Date class
class Date {

    public:
    int year;
    int month;
    int day;
};

bool valid_date(Date &today);
bool getTomorrow(Date today, Date &tomorrow);

int main() {
    Date today, tomorrow;
    today.year=2015;
    today.month=2;
    today.day=28;

    bool getTomorrow(today, tomorrow);
}

bool valid_date(Date &today) {
    bool valid = true;
    const int days[12] = {
        31,28,31,30,31,30,31,31,30,31,30,31
    };
    if (today.month<1 || today.month > 12) {
        valid = false;
    }
    else if (today.day <1 || today.day > days[today.month-1]) {
        valid = false;
    }
    return valid;
}


bool getTomorrow(Date today, Date &tomorrow) {
    bool valid = true;
    const int days[12] = {
        31,28,31,30,31,30,31,31,30,31,30,31
    };

    if (valid_date(today)==false) {
        valid = false;
    }
    else if (today.day==31 && today.month==12) {
        tomorrow.day = 1;
        tomorrow.month = 1;
        tomorrow.year= today.year +1;

    }
    else if (today.day == days[today.month-1]) {
        tomorrow.day = 1;
        tomorrow.month = today.month +1;
        tomorrow.year = today.year;
    }
    else {
        tomorrow.day = today.day + 1;
        tomorrow.month = today.month;
        tomorrow.year= today.year;
    }

    return valid;
}

它无法运行，Xcode 向我发出此行的“标量初始化程序中的多余元素”警告:

bool getTomorrow(today, tomorrow);

非常感谢任何帮助!

Answer 1

这样的语句:

int xyzzy(42);

是一种将变量初始化为给定值的方法。这就是您的代码所发生的情况:

bool getTomorrow(today, tomorrow);

除了编译器提示您为初始化程序提供了两个值。

调用它的正确方法是:

bool myBoolVar = getTomorrow(today, tomorrow);

---

而且，为了提供建议，我不是“只有一个返回点”准则的忠实拥护者，尤其是当它使您的代码更长并且更容易出错时。从这个意义上讲，valid_date() 可以写得更简洁，包括分解 days[] 数组，因为它在多个地方使用并且永远不会改变:

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

bool valid_date (Date today) {
    // Check month first.
    if (today.month < 1 || today.month > 12)
        return false;

    // Allow Feb 29 in leap year if needed.
    if (today.month == 2 && today.day == 29) {
        if (today.year % 400 == 0)
            return true;
        if ((today.year % 4 == 0) && (today.year % 100 != 0))
            return true;
    }

    // Then check day.
    if (today.day < 1 || today.day > days[today.month-1])
        return false;

    return true;
}

出于同样的原因，获取明天的代码似乎也有点折磨人，所以我会看类似的东西:

bool getTomorrow (Date today, Date &tomorrow) {
    // Don't do anything for bad dates.
    if (!valid_date (today)) return false;

    // Just blindly add a day with no checks.
    tomorrow.year = today.year;
    tomorrow.month = today.month;
    tomorrow.day = today.day + 1;

    // Allow Feb 29 in leap year if needed.
    if (tomorrow.month == 2 && tomorrow.day == 29) {
        if (tomorrow.year % 400 == 0)
            return true;
        if ((tomorrow.year % 4 == 0) && (tomorrow.year % 100 != 0))
            return true;
    }

    // Catch rolling into new month.
    if (tomorrow.day > days[tomorrow.month-1]) {
        tomorrow.day = 1;
        tomorrow.month++;

        // Catch rolling into new year.
        if (tomorrow.month == 13) {
            tomorrow.month = 1;
            tomorrow.year++;
        }
    }

    return true;
}

您会注意到我还放入了代码(在两个函数中)以实际允许闰年的 2 月 29 日，基于 400 的倍数和 4 的倍数不是 100 的倍数的规则, 是闰年。如果不需要，只需将其删除 - 我只是坚持完整性:-)