rounding

问题 How do I round a double to 5 decimal places, without using DecimalFormat ? 回答1: You can round to the fifth decimal place by making it the first decimal place by multiplying your number. Then do normal rounding, and make it the fifth decimal place again. Let's say the value to round is a double named x : double factor = 1e5; // = 1 * 10^5 = 100000. double result = Math.round(x * factor) / factor; If you want to round to 6 decimal places, let factor be 1e6 , and so on. 回答2: Whatever you do, if

问题 How do I round a double to 5 decimal places, without using DecimalFormat ? 回答1: You can round to the fifth decimal place by making it the first decimal place by multiplying your number. Then do normal rounding, and make it the fifth decimal place again. Let's say the value to round is a double named x : double factor = 1e5; // = 1 * 10^5 = 100000. double result = Math.round(x * factor) / factor; If you want to round to 6 decimal places, let factor be 1e6 , and so on. 回答2: Whatever you do, if

问题 How do I round a double to 5 decimal places, without using DecimalFormat ? 回答1: You can round to the fifth decimal place by making it the first decimal place by multiplying your number. Then do normal rounding, and make it the fifth decimal place again. Let's say the value to round is a double named x : double factor = 1e5; // = 1 * 10^5 = 100000. double result = Math.round(x * factor) / factor; If you want to round to 6 decimal places, let factor be 1e6 , and so on. 回答2: Whatever you do, if

问题 How do I round a double to 5 decimal places, without using DecimalFormat ? 回答1: You can round to the fifth decimal place by making it the first decimal place by multiplying your number. Then do normal rounding, and make it the fifth decimal place again. Let's say the value to round is a double named x : double factor = 1e5; // = 1 * 10^5 = 100000. double result = Math.round(x * factor) / factor; If you want to round to 6 decimal places, let factor be 1e6 , and so on. 回答2: Whatever you do, if

问题 How to get the rounding number of 0.5, previously I found in this link: Click here But this is not what I want: I want like this example: 10.24 = 10.25 17.90 = 17.90 3.89 = 3.90 7.63 = 7.65 jQuery will be fine with me. 回答1: Use Math.round(num * 20) / 20 , if you are looking to round a number to nearest 0.05 . 回答2: This will do what you need... function roundoff(value) { return (Math.round(value / 0.05) * 0.05).toFixed(2); } Using Math.round will round the value to the nearest 0.05, and then

问题 Hi i want to round double numbers like this (away from zero) in C++: 4.2 ----> 5 5.7 ----> 6 -7.8 ----> -8 -34.2 ----> -35 What is the efficient way to do this? 回答1: inline double myround(double x) { return x < 0 ? floor(x) : ceil(x); } As mentioned in the article Huppie cites, this is best expressed as a template that works across all float types See http://en.cppreference.com/w/cpp/numeric/math/floor and http://en.cppreference.com/w/cpp/numeric/math/floor or, thanks to Pax, a non-function

问题 I want to round a number and I need a proper integer because I want to use it as an array key. The first "solution" that comes to mind is: $key = (int)round($number) However, I am unsure if this will always work. As far as I know (int) just truncates any decimals and since round($number) returns a float with theoretically limited precision, is it possible that round($number) returns something like 7.999999... and then $key is 7 instead of 8? If this problem actually exists (I don't know how

问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: How to iterate between 0.1f and 1.0f with 0.1f increments in Java? Part of my program needs to use values inside a while loop as: 0.1 0.2 0.3 ... 0.9 so I need to provide them inside that loop. Here is the code: double x = 0.0; while ( x<=1 ) { // increment x by 0.1 for each iteration x += 0.1; } I need the output to be EXACTLY: 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 But it actually gives me something like: 0.1 0.2

问题 What is your advice on: compensation of accumulated error in bulk math operations on collections of Money objects. How is this implemented in your production code for your locale? theory behind rounding in accountancy. any literature on topic. I currently read Fowler. He mentions Money type, it's typcal structure (int, long, BigDecimal), but says nothing on strategies. Older posts on money-rounding (here, and here) do not provide a details and formality I need. Thoughts I found in the inet

问题 I'm trying to get Delphi to Round like Excel but I can't. Here is the code: procedure TForm1.Button1Click(Sender: TObject); var s : string; c : currency; begin c := 54321.245; s := ''; s := s + Format('Variable: %m',[c]); s := s + chr(13); s := s + Format(' Literal: %m',[54321.245]); ShowMessage(s); end; I'm using a currency variable that is set to 54321.245 and when I format this variable it rounds using Bankers Rounding. However, when I format the same value as a literal it rounds the way