Extracting and formatting today's date
I would like to display today's date in the format dd/mm/yyyy. What is the best function to use to do so? I know that the time() function will give me number of seconds since Jan.1,1970, but do I have to parse the rest of the info. out from that or is there an easier way/function to use??
you can use the function asctime(const struct tm* pointer)
This returns a pointer to a string which contains the information that you need ( and a little more besides). You then have to break that string down a little to do what you want.
cant remember the exact format of the string, its something like
day/date/month/time/year /0 or something similar. look it up in your compiler helpfiles.
Free the weed!! Class B to class C is not good enough!!
And the FAQ is here :- http://faq.cprogramming.com/cgi-bin/smartfaq.cgi
Try this:
int tm_year;
struct tm *systime;
time_t t;
t=time(NULL);
systime=localtime(&t);
if(systime->tm_hour>12)
{systime->tm_hour=systime->tm_hour-12; }
///so output isn't military format///
printf("Today is %.2d/ %.2d/ %.2d !!\n ",
systime->tm_mon+1,
systime->tm_mday,
systime->tm_year+1900);
///"systime" is a structure defined in time.h
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
Thanx Sebastiani, for helping me see the light!
#include <stdio.h>
#include <time.h>
int main() {
time_t t;
time(&t);
printf(ctime(&t));
system("PAUSE");
return 0;
}
Sall
