how to efficiently parse date time (boost)

Question

I have a char * with a date string I wish to parse. In this case a very simple format: 2010-10-28T16:23:31.428226 (common ISO format).

I know with Boost I can parse this, but at a horrible cost. I have to create a string-stream, possibly a string, and then copy data back and forth. Is there any way to parse the char * without allocating any additional memory. Stack objects are fine, so is reusing a heap object.

Any easy way would also be great! ;)

Edit: I need the result in microseconds since the epoch.

Answer 1

Sometimes plain old C is simpler. You can almost do it with strptime(...):

struct tm parts = {0};
strptime("2010-10-28T16:23:31", "%Y-%m-%dT%H:%M:%S", &parts);

Unfortunately, you'd have to grab the fractional seconds separately. I suppose you could do it with sscanf(...) too:

unsigned int year, month, day, hour, min;
double sec;
int got = sscanf(
     "2010-10-28T16:23:31.428226", 
     "%u-%u-%uT%u:%u:%lf",
     &year, &month, &day, &hour, &min, &sec
);
assert(got == 6);

Answer 2

If it's a fixed format, why don't you take advantage of the positional information?

for example, you know always that the year is the first four characters, so:

const char* date = "2010-10-28T16:23:31.428226";
int year = (((date[0] ^ 0x30)) * 1000) + ((date[1] ^ 0x30) * 100) + ((date[2] ^ 0x30) * 10) + (date[3] ^ 0x30);
int month = ((date[5] ^ 0x30) * 10) + (date[6] ^ 0x30);
int day = ((date[8] ^ 0x30) * 10) + (date[9] ^ 0x30);

etc.

the microsecond segment is a little trickier, depending on whether it's 0 padded or not... but you know the format, so I should assume it's trivial...

significantly faster than any library routine, of course it's very fragile, but if you can control input, why not?