I want to get each hours of a day, local time. For places without daylight saving time it is easy, but when there DST start or ends, current days have 23 or 25 hours!

I wrote this code to properly get it, but I’m wondering if there is a simpler way with chrono to write the function PrintHours()

#include <iostream>
#include <chrono>

using namespace std::chrono;

constexpr const char* zoneName = "Europe/Berlin";

void PrintHours(std::chrono::local_days date)
{
    zoned_time utcTime{"UTC", zoned_time{zoneName, date}};
    auto localTime = zoned_time{zoneName, utcTime};

    auto dp = std::chrono::floor<days>(localTime.get_local_time());
    // keep looping until day change!
    for (auto startingDay = year_month_day{dp}.day(); startingDay == year_month_day{dp}.day(); dp = std::chrono::floor<days>(localTime.get_local_time()))
    {
        std::cout << zoneName << localTime<< " UTC: " <<  utcTime << std::endl;

        // bump time to next hour
        utcTime = zoned_time("UTC", utcTime.get_local_time() + hours(1));
        localTime= zoned_time{zoneName, utcTime};
    }
}
int main(int argc, char **argv)
{
    using namespace std::chrono;
    year_month_day ymd{year(2023), month(10), day(29)};
    PrintHours(local_days(ymd));
    return 1;
}