how to use localtime() correctly? - c++

Question

In one of my utility programs, localtime() is used to covert unix timestamps to human readable date time.
The following code used to work in VS2010 while it fails to work in VS2019:
std::string sec = "1234123456";
int nsec = atoi(sec.c_str());
tm* t = localtime((time_t*)&nsec); // return null pointer
If I change the code in the following way, it will work also in VS2019:
std::string sec = "1234123456";
int nsec = atoi(sec.c_str());
time_t tt = nsec;
tm* t = localtime(&tt); // works
I have no idea why the additional int to time_t conversion is needed, any suggestion would be appreciated.

Answer 1

On most (if not all) modern compilers time_t is now a 64-bit integer. (time_t*)&nsec is therefore undefined behaviour as you are casting from one pointer type to a different one.
You fixed version is well defined but you will run into the reason that time_t is now 64-bit as 32-bit numbers will only work for times up to 2038 (assuming time_t is using the Unix epoch).
Unfortunately c++ doesn't provide a simple method for converting a string to time_t, to do it properly you'd need something like this:
#include <iostream>
#include <charconv>
time_t str_to_time_t(const std::string& str)
{
auto begin = str.c_str();
auto end = begin + str.size();
time_t time;
auto result = std::from_chars(begin, end, time);
if (result.ec != std::errc())
{
throw std::system_error(std::make_error_code(result.ec));
}
if (result.ptr != end)
{
throw std::invalid_argument("invalid time_t string");
}
return time;
}
int main()
{
std::string sec = "1234123456";
auto nsec = str_to_time_t(sec);
tm* t = localtime((time_t*)&nsec);
if (t)
{
std::cout << "parsed OK\n";
}
}