Is this function designed intentionally? boost::date_time::c_time::gmtime doesn't touch the result pointer passed.

[fragrans]

inline static std::tm* gmtime(const std::time_t* t, std::tm* result);

I assume this function will change the result pointer by filling the struct; but it doesn't;

Windows 10; MSVC 16.11.7

date_time/include/boost/date_time/c_time.hpp

Lines 109 to 120 in 7156392

	//! requires a pointer to a user created std::tm struct
	inline
	static std::tm* gmtime(const std::time_t* t, std::tm* result)
	{
	result = std::gmtime(t);
	if (!result)
	boost::throw_exception(std::runtime_error("could not convert calendar time to UTC time"));
	return result;
	}
	#if defined(__clang__) // Clang has to be checked before MSVC
	#pragma clang diagnostic pop
	#elif (defined(_MSC_VER) && (_MSC_VER >= 1400))

[fragrans]

Any one?

[JeffGarland]

I'm not sure what the confusion is. The function calls gmtime to assign the result pointer and then returns it? Also this is really an implementation detail of the library, so not really part of the documented api.

[fragrans]

the content of the 2nd argument (pointer result) passed to the function will not change;
didn't you see that?

[JeffGarland]

result = std::gmtime(t);

That changes it.

[fragrans]

I am sorry, and confused. if you want to change this pointer, you need to

auto temp = std::gmtime(t); 
*result = *temp;

[jeking3]

@fragrans This is a C++03-compatible library so auto would not be appropriate, but I agree with you otherwise.

@JeffGarland if you look in date_clock_device.hpp, the implementation is passed a std::tm& but that is never filled in, and the returned std::tm* is going to be different than what was passed in for result. It does seem a bit confusing and likely a bug as there could be other bad assumptions made by consumers about get_local_time and get_universal_time based on the inputs.