Format a date in C++

Question

Hi,

I have a series of date strings in the format: "30-05-2001"

string date1 = "30-05-2001";

I would like to parse the date into Day, Month, Year. Now an easy way of doing this would be just to call the function sscanf. But I'd like to explore other possiblilties and from searching the web the following function from time.h was recommended:

char *strptime(const char *buf, const char *format, struct tm *tm); 

Does any anyone have experience using this function? Some short example would be very helpful.

Thanks!

Answer 1

Source:

#include <time.h>
...


struct tm tm;
time_t t;


if (strptime("6 Dec 2001 12:33:45", "%d %b %Y %H:%M:%S", &tm) == NULL)
    /* Handle error */;


printf("year: %d; month: %d; day: %d;\n",
        tm.tm_year, tm.tm_mon, tm.tm_mday);
printf("hour: %d; minute: %d; second: %d\n",
        tm.tm_hour, tm.tm_min, tm.tm_sec);
printf("week day: %d; year day: %d\n", tm.tm_wday, tm.tm_yday);


tm.tm_isdst = -1;      /* Not set by strptime(); tells mktime()
                          to determine whether daylight saving time
                          is in effect */
t = mktime(&tm);
if (t == -1)
    /* Handle error */;
printf("seconds since the Epoch: %ld\n", (long) t);"

Answer 2

See here for an example. Also good explanation how to use the conversion specifiers etc.

Edit: KMan was faster :)

Edit2: Link leads to a desctiption of

char *strptime(const char *restrict buf, const char *restrict format, struct tm *restrict tm);

and also a short example on how to use. The Site is provided by

The Open Group Base Specifications Issue 6 IEEE Std 1003.1, 2004 Edition

Answer 3

Hello Skaffman,

Just in case you want to understand how it works (no blackbox) or if you need the stuff to fly (no function calls) then here are a few tips for you:

//             0123456789
char date[] = "30-05-2001"; // DD-MM-YYYY
int day, month, year;

day = (date[0] - '0') * 10 + (date[1] - '0');
month = (date[3] - '0') * 10 + (date[4] - '0');
year = (date[6] - '0') * 1000 + (date[7] - '0') * 100 + (date[8] - '0') * 10 + (date[9] - '0');

This (untested) code is clother to the right way to do what you wanted to do as it only does what you need to do (not more, not less). Safe, small, simple, FAST.

If you are unsure of the date (user input) then you will have to check for its validity (or accept the idea to live with invalid dates in your application).

Answer 4

std::istringstream buffer("30-05-2001");
int day = 0;
int month = 0;
int year = 0;
char dummy = 0;

buffer >> day >> dummy >> month >> dummy >> year;
// TODO: validate values here

You can perform different reads on the data if you want to also validate that the dummy equals '-'.