double

问题 This question already has answers here : How do I print a double value with full precision using cout? (10 answers) Closed 5 years ago . I'm trying to convert a string into a double but my double gets cut off at the 3rd decimal point. My string looks like this: "-122.39381636393" After it gets converted it looks like this: -122.394 void setLongitude(string longitude){ this->longitude = (double)atof(longitude.c_str()); cout << "got longitude: " << longitude << endl; cout << "setting longitude:

问题 I tried many ways to convert a Double to it's exact decimal String with String.format() and DecimalFormater and I found this so far : String s = "-.0000000000000017763568394002505"; Double d = Double.parseDouble(s) * 100; System.out.println((new java.text.DecimalFormat("###########################.################################################################################")).format(d)); (https://ideone.com/WG6zeC) I'm not really a fan of my solution because it's ugly to me to make a huge

问题 I need to know how to do this procedure. calculation1: 1/4 = 0,25 calculation2: 1/8 = 0,125 calculation3: 47/183 = 0,25683060109289617486338797814207...... calculation4: 58/889 = 0,06524184476940382452193475815523...... calculation5: 1/5 = 0,2 The results of calculations 1, 2 and 5 will give a short result, no periods or and endless string of digits. The results of calculations 3 and 4 are very long and complicated. How can I check which calculation is an "easy one" and gives a "short" result

问题 I need to know how to do this procedure. calculation1: 1/4 = 0,25 calculation2: 1/8 = 0,125 calculation3: 47/183 = 0,25683060109289617486338797814207...... calculation4: 58/889 = 0,06524184476940382452193475815523...... calculation5: 1/5 = 0,2 The results of calculations 1, 2 and 5 will give a short result, no periods or and endless string of digits. The results of calculations 3 and 4 are very long and complicated. How can I check which calculation is an "easy one" and gives a "short" result

问题 I would like to allow both "comma" and "dot" as separator in double. I could use replace method in string to get only one separator, but problem is that double value is value of JSpinner and I was not able to find any method to allow both separators. If I set locale for example to French only one separator is allowed. 回答1: Just use a custom formatter for the JFormattedTextField of the DefaultEditor of the JSpinner , like the code below: import java.text.ParseException; import javax.swing

问题 I would like to convert double into String . I want it to have as few digits as possible and maximally 6. So I found String.format("%.6f", d) which converts my 100.0 into 100.000000 . Max precision works correctly, but I would like it to be converted to 100 (minimum precision). Have you got any idea what method is working like that? 回答1: Use DecimalFormat : new DecimalFormat("#.0#####").format(d) . This will produce numbers with 1 to 6 decimal digits. Since DecimalFormat will use the symbols

问题 I have a simple example application here where I am multiplying and adding double variables and then comparing them against an expected result. In both cases the result is equal to the expected result yet when I do the comparison it fails. static void Main(string[] args) { double a = 98.1; double b = 107.7; double c = 92.5; double d = 96.5; double expectedResult = 88.5; double result1 = (1*2*a) + (-1*1*b); double result2 = (1*2*c) + (-1*1*d); Console.WriteLine(String.Format("2x{0} - {1} = {2}

问题 I have the following line of code: float top = shape.Y + (shape.Height / 2.0) - 4.5; whic is failing with the error' Can't cast double to float. Shape.Y and shape.height are both type float. What is causing this error and what is the best was to make sure top is a float (as i need to pass it into another function tha expects a float. 回答1: Try to wrap your numbers with f , as normally 2.0 would represent a double number. You can read more about float here. float top = shape.Y + (shape.Height /

问题 i have a list of double values... 1.23, 1.24, 1.78, 1,74... so i want to calculate the differences between the successor -> only adding (negative values should be firstly positive)... above 4 values would be 0,01 +0,53 (-)-0,04 (-) -> to make it positive... with an for-loop, it is easy... any idea how to solve it with linq? 回答1: I'm not sure what you mean about the negative bits, but this might do what you want. It's horrible because it uses side effects, but... double prev = 0d; var

问题 I'm using the following function to approximate the derivative of a function at a point: def prime_x(f, x, h): if not f(x+h) == f(x) and not h == 0.0: return (f(x+h) - f(x)) / h else: raise PrecisionError As a test I'm passing f as fx and x as 3.0. Where fx is: def fx(x): import math return math.exp(x)*math.sin(x) Which has exp(x)*(sin(x)+cos(x)) as derivative. Now, according to Google and to my calculator exp(3)*(sin(3)+cos(3)) = -17.050059 . So far so good. But when I decided to test the