将double转换为struct tm

Question

I have a double containing seconds. I would like to convert this into a struct tm .

I can't find a standard function which accomplishes this. Do I have to fill out the struct tm by hand?

I just accidentally asked this about converting to a time_t and http://www.StackOverflow.com will not let me post unless I link it.

Answer 1

Well, you accidentally asked the right question before. Convert double to time_t , and then convert that to a struct tm . There's no subsecond field in struct tm anyway.

Answer 2

For grins, using this chrono -based header-only library :

#include "date.h"
#include <iostream>

int
main()
{
    using namespace std::chrono;
    using namespace date;
    auto recovery_time = 320.023s;  // Requires C++14 for the literal 's'
    std::cout << make_time(duration_cast<milliseconds>(recovery_time)) << '\n';
}

outputs:

00:05:20.023

The object returned by make_time has getters if you want to query each field:

constexpr std::chrono::hours hours() const noexcept {return h_;}
constexpr std::chrono::minutes minutes() const noexcept {return m_;}
constexpr std::chrono::seconds seconds() const noexcept {return s_;}
constexpr precision subseconds() const noexcept {return sub_s_;}

You don't need to choose milliseconds . You could choose any precision you want from hours to picoseconds (if you also supply a type alias for picoseconds ). For example:

std::cout << make_time(duration_cast<seconds>(recovery_time)) << '\n';

Outputs:

00:05:20

Answer 3

MSalters answer is correct, but I thought I'd add a bit of detail on how you should convert to time_t and how you should convert to tm .

So given a number of seconds in double input you can use the implementation dependent method of casting:

const auto temp = static_cast<time_t>(input);

But since time_t is implementation defined there is no way to know that this is a primitive that can simply be cast to. So the guaranteed method would be to use the chrono library's implementation independent method of converting:

const auto temp = chrono::system_clock::to_time_t(chrono::system_clock::time_point(chrono::duration_cast<chrono::seconds>(chrono::duration<double>(input))));

The conversion options are discussed in more detail here: https://stackoverflow.com/a/50495821/2642059 but once you have obtained your time_t through one of these methods you can simply use localtime to convert temp to a struct tm .

const auto output = *localtime(&temp);

Note the dereference is important. It will use the default copy assignment operator so output is captured by value, which is essential because:

The structure may be shared between std::gmtime , std::localtime , and std::ctime , and may be overwritten on each invocation.