This one takes microseconds into account! import math def ceil_dt(dt): # how many secs have passed this hour nsecs = dt.minute*60 + dt.second + dt.microsecond*1e-6 # number of seconds to next quarter hour mark # Non-analytic (brute force is fun) way: # delta = next(x for x in xrange(0,3601,900) if x>=nsecs) – nsecs # analytic … Read more
Why use external script languages? You get floor by default. To get ceil, do $ divide=8; by=3; (( result=(divide+by-1)/by )); echo $result 3 $ divide=9; by=3; (( result=(divide+by-1)/by )); echo $result 3 $ divide=10; by=3; (( result=(divide+by-1)/by )); echo $result 4 $ divide=11; by=3; (( result=(divide+by-1)/by )); echo $result 4 $ divide=12; by=3; (( result=(divide+by-1)/by … Read more
std::ceil rounds up to the nearest integer std::floor rounds down to the nearest integer std::round performs the behavior you expect please give a use case with numbers if this does not provide you with what you need!
After a night lost trying to solve this problem I believe I’ve found a rather simple solution, here it is: function bcceil($number) { if (strpos($number, ‘.’) !== false) { if (preg_match(“~\.[0]+$~”, $number)) { return bcround($number, 0); } if ($number[0] != ‘-‘) { return bcadd($number, 1, 0); } return bcsub($number, 0, 0); } return $number; } … Read more
Since you’re looking for fourths (.00, .25, .50, .75), multiply your number by 4, round to nearest whole number as desired (floor if down, ceil if up), then divide by 4. 1.32, down to nearest fourth: 1.32 * 4 = 5.28 floor(5.28) = 5.00 5.00 / 4 = 1.25 Same principle applies for any other … Read more