double

问题 When I use samsung galaxy mini 4.4 I see two progressDialog(picture below). When I use another device - everething is okay if (progressDialog == null) { progressDialog = new ProgressDialog(getActivity(), R.style.AlertDialogThemeBlackColor); progressDialog.setMessage(getResources().getString(R.string.loading_your_data)); progressDialog.setCancelable(false); } progressDialog.show(); 回答1: I didn't found reason, but I found a way to avoid this. Need to create a dialog in the following manner: new

问题 I have a stupid thing, but I really can´t see what am I missing: I have a test: @Test public void testeBerechneRendite() { get("/rendite?jahresnettomiete=8000&kaufpreis=100000&nebenkosten=500") .then().body(is(closeTo(0.079, 0.01))); } Error is: Response body doesn't match expectation. Expected: is a numeric value within <0.01> of <0.079> Actual: 0.07960199004975124 It seams I don`t understand the closeTo(double, double). From my understanding all numbers between 0.069 and 0.089 should be

问题 I want to serialize/deserialize doubles to a readable String. I'm happy to lose any accuracy beyond the 6th dp. Whats the correct approach for doing this? I'm concerned about System.out.println(1.03 - .42) printing 0.6100000000000001 and not 0.61. Thxs. 回答1: The simplest approach is to use PrintWriter and BufferedReader. double[][] doubles = new double[10][10]; for(double[] ds : doubles) for(int i=0;i<ds.length;i++) ds[i] = Math.random() * 100; PrintWriter pw = new PrintWriter("double.tsv");

问题 I've been trying to solve a double integral equation for a third variable in matlab. An example: At first, I tried to solve it symbolically (when k=1) as below: syms x y h F = @(x,y,h) 2*x*y+4*h; f = @(x) x/2; solve(int(int(F(x,y,h)*f(x),x,0,3)*f(y),y,0,1)-3, h) The code gives the right answer, i.e. h=2/3 when k=1. For my real problem, however, the functions F and f are so much more complex. And when I applied the same code above with the complex F and f, the matlab does not solve it in an

问题 I am looking for a function similar to atol (char * to long int) but atofl (char to long double), does anyone know of a library that does this, or a simple way to do this, since using atof on "0.00000005" casts off the 5? Thanks. 回答1: There is a function called strtod that does what you want. 回答2: I think you have use the wrong argument of printf, so you only saw the console giving you "0.000000", you can try printf("%0.10f", atof(str_num)); to show the your number, and you will find what you

问题 I'm currently working on a program and im converting my java code into c# . but i'm having some trouble. public double round(double value){ BigDecimal b = new BigDecimal(value); b = b.setScale(2,BigDecimal.ROUND_UP); return (b.doubleValue()); } i wrote this converting code but i cant convert it to c#.BigDecimal type causes some problem and im totally new to .Net.Definitely need some help. Edit : Ok buds i got it , sorry for the dumb question. 回答1: Couldn't you just do this to round to 2

问题 To begin with, allow me to confess that I'm an experienced programmer, who has over 10 years of programming experience. However, the question I'm asking here is the one, which has bugged me ever since, I first picked up a book on C about a decade back. Below is an excerpt from a book on Python, explaining about Python Floating type. Floating-point numbers are represented using the native double-precision (64-bit) representation of floating-point numbers on the machine. Normally this is IEEE

问题 I'm a newbie to Swift and xCode so apologies if some of my terminology is incorrect. I've just started learning CoreData and am attempting to produce a basic function where users can create a 'location.' I set up the data model with the attributes name (type = string), latitude (type = double) and longitude (type = double). I've set up a TableViewController (which is working fine) with a segue to another Controller which is set up to enable people to enter a name, latitude and longitude. As

问题 I am getting a weird problem while parsing a double value in managed C++. It may be that I am doing something wrong. When I do: double value = 0.006; result = Math::Parse( value) The output of result is 0.006000000000001 . Why it is appending a 1? Also when I go an round the value to 5 decimal places, it fails. I am doing: result2 = Math::Round(result, 5) But result2 is always 0.006000000000001 . What am I doing wrong? 回答1: It is normal. This problem caused by IEEE format of double - in real

问题 I would like to convert an integer [] [] to a Vector<Vector<Double>> . After much reading, it seems no one has left a searchable post on the web for something of this nature. plenty of int vector to double vector, arraylist to vector, etc. Sadly I haven't found what i am looking for. So..Do any of you folks know an appropriate method for this? I was thinking of converting my int[][] to strings then convert that to vector<vector<Double>> . Opinions? Would something like this be useful, ie.