c++ 获取用户选择的日期和实际日期之间的年数(计算天数、月数、年数)

Question

我试过这样做:

struct Den_t
{
    int day, month, year;
}; 

int main()
{
        struct Den_t* Datum = new struct Den_t;
        struct Den_t* Dnes = new struct Den_t;

    time_t theTime = time(NULL);
    struct tm aTime;
    localtime_s(&aTime, &theTime);

    Dnes->day = aTime.tm_mday;
    Dnes->month = aTime.tm_mon + 1;
    Dnes->year = aTime.tm_yday + 1900;

    cin >> Datum->day >> Datum->month >> Datum->year;
    if (Dnes->year - Datum->year >= 18 )
        cout << "full aged " << endl;
    else
        cout << "not full aged " << endl;
    system("PAUSE");
    return 0;
}

但我不知何故无法理解我什至应该比较和递减什么，有人能解释一下吗

what else i need to do to tell people's date for example in float by comparing year,month and day of actual time and date user inputs in the program?

Answer 1

您的代码逻辑有问题。例如:

Datum is 31/12/1982
Dnes is 01/01/2000

年差是 18，但年龄是 17 和 2 天。

考虑使用标准库函数而不是重新发明轮子。 difftime可能有用，例如

这是一个非常肮脏的例子，但它可以完成工作:

   time_t dnes;
   time(&dnes);

   // Set datum here ... 
   cin >> Datum->day >> Datum->month >> Datum->year;
   datum.tm_mday = Datum->day;
   datum.tm_mon =  Datum->month - 1;
   datum.tm_yday = Datum->year - 1900;

   datum->tm_yday+=18;

   if (difftime(dnes, mktime(&datum)) <0 )
        cout << "not full aged " << endl;
    else
        cout << "full aged " << endl;