String representation of time_t?

Question

time_t seconds;
time(&seconds);

cout << seconds << endl;

This gives me a timestamp. How can I get that epoch date into a string?<

Answer 1

The top answer here does not work for me.

See the following examples demonstrating both the stringstream and lexical_cast answers as suggested:

#include 
#include 

int main(int argc, char** argv){
 const char *time_details = "2017-01-27 06:35:12";
  struct tm tm;
  strptime(time_details, "%Y-%m-%d %H:%M:%S", &tm);
  time_t t = mktime(&tm); 
  std::stringstream stream;
  stream << t;
  std::cout << t << "/" << stream.str() << std::endl;
}

Output: 1485498912/1485498912 Found here

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#include 
#include 

int main(){
    const char *time_details = "2017-01-27 06:35:12";
    struct tm tm;
    strptime(time_details, "%Y-%m-%d %H:%M:%S", &tm);
    time_t t = mktime(&tm); 
    std::string ts = boost::lexical_cast(t);
    std::cout << t << "/" << ts << std::endl;
    return 0;
}

Output: 1485498912/1485498912 Found: here

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The 2nd highest rated solution works locally:

#include 
#include 
#include 

int main(){
  const char *time_details = "2017-01-27 06:35:12";
  struct tm tm;
  strptime(time_details, "%Y-%m-%d %H:%M:%S", &tm);
  time_t t = mktime(&tm); 

  std::tm * ptm = std::localtime(&t);
  char buffer[32];
  std::strftime(buffer, 32, "%Y-%m-%d %H:%M:%S", ptm);
  std::cout << t << "/" << buffer;
}

Output: 1485498912/2017-01-27 06:35:12 Found: here

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