Parse date string to remove hours, minutes and seconds from timestamp - c++

Question

Suppose I receive from a web server a string to parse. This string contains a date in the format YYYY-MM-DD.
What I want is to convert it to the timestamp that represents the begin of that day, hence I don't want seconds, minutes and hours.
As a dummy example, I'm trying to extract the timestamp of the current day, once converted to the YYYY-MM-DD format. Here's the code:
#include <chrono>
#include <iomanip>
#include <iostream>
#include <sstream>
int main()
{
// Current time at GMT
std::time_t now = std::time(0);
std::tm *now_tm = std::gmtime(&now);
std::ostringstream oss;
// Extract yyyy-mm-dd = %F
oss << std::put_time(now_tm, "%F");
// Use oss to get a date without seconds from
// current time at gmt
std::tm tm;
std::istringstream ss(oss.str());
ss >> std::get_time(&tm, "%F");
std::time_t current_date = std::mktime(&tm);
std::cout << oss.str() << std::endl;
std::cout << "cd: " << current_date << std::endl;
return 0;
}
The output is:
2017-10-19
cd: 1908337984324104
The extracted timestamp is clearly wrong. Where's the problem in the parsing the 2017-10-19 string using the %F format?

Answer 1

You can do this without leaving the safety of the chrono type system by using Howard Hinnant's, free, open-source, header-only chrono-extension library.
#include "date/date.h"
#include <iostream>
int
main()
{
std::istringstream ss{"2017-10-19"};
date::sys_seconds tp;
ss >> date::parse("%F", tp);
std::cout << date::format("%F\n", tp);
using date::operator<<;
std::cout << "cd: " << tp.time_since_epoch() << '\n';
}
date::sys_seconds is a std::chrono::time_point that counts chrono::seconds in Unix Time. You can parse directly into it using %F. You can also format it, using the same format string (%F), and also inspect the underlying count of chrono::seconds. This program outputs:
2017-10-19
cd: 1508371200s

Answer 2

The documentation of std::get_time lists no conversion specifier %F. When checking the stream flag (which you always should do!), it will also tell that the conversion failed, at least with my compiler.
So by replacing it with %Y-%m-%d, the conversion succeeds. Finally, you default-constructed the tm variable without zeroing it (e.g. by value-initialization). When fixing this as well, the code works as expected:
#include <chrono>
#include <iomanip>
#include <iostream>
#include <sstream>
int main()
{
// Current time at GMT
std::time_t now = std::time(0);
std::tm *now_tm = std::gmtime(&now);
std::ostringstream oss;
// Extract yyyy-mm-dd = %F
oss << std::put_time(now_tm, "%Y-%m-%d");
// Use oss to get a date without seconds from
// current time at gmt
std::tm tm{ }; // value-initialize!
std::istringstream ss(oss.str());
ss >> std::get_time(&tm, "%Y-%m-%d");
if(!ss) std::cout << "conversion error\n";
else {
std::time_t current_date = std::mktime(&tm);
std::cout << current_date << '\n';
std::cout << "cd: " << current_date << '\n';
}
return 0;
}
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