如何在使用 chrono 库将数字转换为日期的 cpp 的 main() 中正确调用此模板函数？

Question

如何在使用 chrono 库将数字转换为日期的 cpp 的 main() 中正确调用此模板函数？

 #include <iostream>
 #include <chrono>
 #include <tuple>
 
 //using namespace std;
 // Returns year/month/day triple in civil calendar
 // Preconditions:  z is number of days since 1970-01-01 and is in the range:
 //                   [numeric_limits<Int>::min(), numeric_limits<Int>::max()-719468].
 template <class Int>
 constexpr
 std::tuple<Int, unsigned, unsigned>
 civil_from_days(Int z) noexcept
 {
     static_assert(std::numeric_limits<unsigned>::digits >= 18,
              "This algorithm has not been ported to a 16 bit unsigned integer");
     static_assert(std::numeric_limits<Int>::digits >= 20,
              "This algorithm has not been ported to a 16 bit signed integer");
     z += 719468;
     const Int era = (z >= 0 ? z : z - 146096) / 146097;
     const unsigned doe = static_cast<unsigned>(z - era * 146097);          // [0, 146096]
     const unsigned yoe = (doe - doe/1460 + doe/36524 - doe/146096) / 365;  // [0, 399]
     const Int y = static_cast<Int>(yoe) + era * 400;
     const unsigned doy = doe - (365*yoe + yoe/4 - yoe/100);                // [0, 365]
     const unsigned mp = (5*doy + 2)/153;                                   // [0, 11]
     const unsigned d = doy - (153*mp+2)/5 + 1;                             // [1, 31]
     const unsigned m = mp < 10 ? mp+3 : mp-9;                            // [1, 12]
     return std::tuple<Int, unsigned, unsigned>(y + (m <= 2), m, d);
 }
 
 
 int main(){
    std::cout<< civil_from_days(15432)<<'\n';
 }

给出 this 或 这个（相同）编译器错误。

代码来自 Howard Hinnant

Answer 1

对函数的调用不是问题。您得到的错误（您应该将其包含在问题中）是因为元组没有预定义的输出运算符。不过，您可以打印各个成员：

int main(){
    auto res = civil_from_days(15432);
    std::cout<< std::get<0>(res)<<'\n';
    std::cout<< std::get<1>(res)<<'\n';
    std::cout<< std::get<2>(res)<<'\n';
 }