Take advantage of integer division, which truncates the decimal portion of the quotient. To make it look like it’s rounding up, add 99 first. int rounded = ((num + 99) / 100 ) * 100; Examples: 801: ((801 + 99) / 100) * 100 → 900 / 100 * 100 → 9 * 100 = … Read more
You sort of answered your own question – build your format string dynamically… valid format strings follow the conventions outlined here: http://java.sun.com/j2se/1.5.0/docs/api/java/util/Formatter.html#syntax. If you want a formatted decimal that occupies 8 total characters (including the decimal point) and you wanted 4 digits after the decimal point, your format string should look like “%8.4f”… To my … Read more
Example of using decimal, as Kibbee pointed out double d = 1.275; Math.Round(d, 2); // 1.27 Math.Round((decimal)d, 2); // 1.28
As long as your grid is regular, just find a transformation from integers to this grid. So let’s say your grid is 0.2 0.4 0.6 … Then you round by float round(float f) { return floor(f * 5 + 0.5) / 5; // return std::round(f * 5) / 5; // C++11 }
A common trick is to do it with math: value = std::round(value * 1000.0) / 1000.0; This scales the value, applies ordinary rounding and then scales it back. You may have some precision loss as is typical with floating-point math, and this won’t be effective at extreme ranges of what can be represented by double, … Read more
As an alternative: <fmt:formatNumber var=”expiry” value=”${(expire.time – now.time) / (60 * 1000)}” maxFractionDigits=”0″ /> This way you do not lose localization (commas and dots).
First off, in most languages an undecorated constant like “1.55” is treated as a double precision value. However, 1.55 is not exactly representable as double precision value, because it doesn’t have a terminating representation in binary. This causes many curious behaviors, but one effect is that when you type 1.55, you don’t actually get the … Read more
For a datetime.datetime rounding, see this function: https://stackoverflow.com/a/10854034/1431079 Sample of use: print roundTime(datetime.datetime(2012,12,31,23,44,59,1234),roundTo=60*60) 2013-01-01 00:00:00
You can get the precision before you change it, with std::ios_base::precision and then use that to change it back later. You can see this in action with: #include <ios> #include <iostream> #include <iomanip> int main (void) { double pi = 3.141592653590; std::streamsize ss = std::cout.precision(); std::cout << “Initial precision = ” << ss << ‘\n’; … Read more