In the following program you can see that each value slightly less than .5 is rounded down, except for 0.5. for (int i = 10; i >= 0; i–) { long l = Double.doubleToLongBits(i + 0.5); double x; do { x = Double.longBitsToDouble(l); System.out.println(x + ” rounded is ” + Math.round(x)); l–; } while (Math.round(x) … Read more
I want to use a Track-Bar to change a Form‘s opacity. This is my code: decimal trans = trackBar1.Value / 5000; this.Opacity = trans; When I build the application, it gives the following error: Cannot implicitly convert type decimal to double I have tried using trans and double, but then the Control doesn’t work. This … Read more
This question already has answers here: When should I use double instead of decimal? (12 answers) Closed 8 years ago. I keep seeing people using doubles in C#. I know I read somewhere that doubles sometimes lose precision. My question is when should a use a double and when should I use a decimal type? … Read more
What is the difference between decimal, float and double in .NET? When would someone use one of these? 1 18 float and double are floating binary point types. In other words, they represent a number like this: 10001.10010110011 The binary number and the location of the binary point are both encoded within the value. decimal … Read more
You can use Double.parseDouble() to convert a String to a double: String text = “12.34”; // example String double value = Double.parseDouble(text); For your case it looks like you want: double total = Double.parseDouble(jlbTotal.getText()); double price = Double.parseDouble(jlbPrice.getText());
is there a possibility that casting a double created via Math.round() will still result in a truncated down number No, round() will always round your double to the correct value, and then, it will be cast to an long which will truncate any decimal places. But after rounding, there will not be any fractional parts remaining. Here are the docs from Math.round(double): … Read more
Round a double to 2 decimal places [duplicate]
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 public static double round(double value, int places) { if (places < 0) throw new IllegalArgumentException(); long factor = (long) Math.pow(10, places); value = value * factor; long tmp = Math.round(value); return (double) tmp / factor; … Read more
Use a DecimalFormatter: double number = 0.9999999999999; DecimalFormat numberFormat = new DecimalFormat(“#.00”); System.out.println(numberFormat.format(number)); Will give you “0.99”. You can add or subtract 0 on the right side to get more or less decimals. Or use ‘#’ on the right to make the additional digits optional, as in with #.## (0.30) would drop the trailing 0 to … Read more
I was reading a book on Java and came across an example in which an array of type double was initialized in a way that I haven’t seen before. What type of initialization is it and where else can it be used? double m[][]={ {0*0,1*0,2*0,3*0}, {0*1,1*1,2*1,3*1}, {0*2,1*2,2*2,3*2}, {0*3,1*3,2*3,3*3} };
