You want to cast the numbers: double num3 = (double)num1/(double)num2; Note: If any of the arguments in C# is a double, a double divide is used which results in a double. So, the following would work too: double num3 = (double)num1/num2; For more information see: Dot Net Perls
A BigDecimal is an exact way of representing numbers. A Double has a certain precision. Working with doubles of various magnitudes (say d1=1000.0 and d2=0.001) could result in the 0.001 being dropped alltogether when summing as the difference in magnitude is so large. With BigDecimal this would not happen. The disadvantage of BigDecimal is that … Read more
According to the IEEE standard, NaN values have the odd property that comparisons involving them are always false. That is, for a float f, f != f will be true only if f is NaN. Note that, as some comments below have pointed out, not all compilers respect this when optimizing code. For any compiler … Read more
You can use Swift’s round function to accomplish this. To round a Double with 3 digits precision, first multiply it by 1000, round it and divide the rounded result by 1000: let x = 1.23556789 let y = Double(round(1000 * x) / 1000) print(y) /// 1.236 Unlike any kind of printf(…) or String(format: …) solutions, … Read more
“%f” is the (or at least one) correct format for a double. There is no format for a float, because if you attempt to pass a float to printf, it’ll be promoted to double before printf receives it1. “%lf” is also acceptable under the current standard — the l is specified as having no effect … Read more
Here’s an utility that rounds (instead of truncating) a double to specified number of decimal places. For example: round(200.3456, 2); // returns 200.35 Original version; watch out with this public static double round(double value, int places) { if (places < 0) throw new IllegalArgumentException(); long factor = (long) Math.pow(10, places); value = value * factor; … Read more
If the idea is to print integers stored as doubles as if they are integers, and otherwise print the doubles with the minimum necessary precision: public static String fmt(double d) { if(d == (long) d) return String.format(“%d”,(long)d); else return String.format(“%s”,d); } Produces: 232 0.18 1237875192 4.58 0 1.2345 And does not rely on string manipulation.
Summary In Java 6 (and presumably earlier), round(x) is implemented as floor(x+0.5).1 This is a specification bug, for precisely this one pathological case.2 Java 7 no longer mandates this broken implementation.3 The problem 0.5+0.49999999999999994 is exactly 1 in double precision: static void print(double d) { System.out.printf(“%016x\n”, Double.doubleToLongBits(d)); } public static void main(String args[]) { double … Read more
An explicit cast to double like this isn’t necessary: double trans = (double) trackBar1.Value / 5000.0; Identifying the constant as 5000.0 (or as 5000d) is sufficient: double trans = trackBar1.Value / 5000.0; double trans = trackBar1.Value / 5000d;
For money, always decimal. It’s why it was created. If numbers must add up correctly or balance, use decimal. This includes any financial storage or calculations, scores, or other numbers that people might do by hand. If the exact value of numbers is not important, use double for speed. This includes graphics, physics or other … Read more