Editor’s Note: The following answer provides a simplistic solution that contains several implementation flaws (see Shafik Yaghmour’s answer for a full explanation). Note that C++11 includes std::round, std::lround, and std::llround as builtins already. There’s no round() in the C++98 standard library. You can write one yourself though. The following is an implementation of round-half-up: double … Read more
There are many ways to do just this, provided you are not concerned about reliance on the original decimal data. The first and perhaps most popular method would be the Largest Remainder Method Which is basically: Rounding everything down Getting the difference in sum and 100 Distributing the difference by adding 1 to items in … Read more
Using Math.floor() is one way of doing this. More information: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/floor
You could cast your result as numeric(x,2). Where x <= 38. select round(630/60.0,2), cast(round(630/60.0,2) as numeric(36,2)) Returns 10.500000 10.50
The other answers with reasons why the Banker’s algorithm (aka round half to even) is a good choice are quite correct. It does not suffer from negative or positive bias as much as the round half away from zero method over most reasonable distributions. But the question was why .NET use Banker’s actual rounding as … Read more
This works: inputValue = Math.Round(inputValue, 2);
Use 1, not 2. js> 2.3 % 1 0.2999999999999998
Background: Some answers suggested on this page (e.g., signif, options(digits=…)) do not guarantee that a certain number of decimals are displayed for an arbitrary number. I presume this is a design feature in R whereby good scientific practice involves showing a certain number of digits based on principles of “significant figures”. However, in many domains … Read more
Number((6.688689).toFixed(1)); // 6.7
Here’s some examples: decimal a = 1.994444M; Math.Round(a, 2); //returns 1.99 decimal b = 1.995555M; Math.Round(b, 2); //returns 2.00 You might also want to look at bankers rounding / round-to-even with the following overload: Math.Round(a, 2, MidpointRounding.ToEven); There’s more information on it here.