Firstly, this wouldn’t be a C# bug anyway – it would be a .NET bug. C# is the language – it doesn’t decide how Math.Round is implemented. And secondly, no – if you read the docs, you’ll see that the default rounding is “round to even” (banker’s rounding): Return ValueType: System.DoubleThe integer nearest a. If … Read more
You can use the round function, which takes as its first argument the number and the second argument is the precision after the decimal point. In your case, it would be: answer = str(round(answer, 2))
Math.round(num * 10) / 10 works, here is an example… var number = 12.3456789 var rounded = Math.round(number * 10) / 10 // rounded is 12.3 if you want it to have one decimal place, even when that would be a 0, then add… var fixed = rounded.toFixed(1) // fixed is always to 1 d.p. … 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
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
To format a number using fixed-point notation, you can simply use the toFixed method: (10.8).toFixed(2); // “10.80” var num = 2.4; alert(num.toFixed(2)); // “2.40” Note that toFixed() returns a string. IMPORTANT: Note that toFixed does not round 90% of the time, it will return the rounded value, but for many cases, it doesn’t work. For … Read more
The math.ceil (ceiling) function returns the smallest integer higher or equal to x. For Python 3: import math print(math.ceil(4.2)) For Python 2: import math print(int(math.ceil(4.2)))
You can use number_format(): return number_format((float)$number, 2, ‘.’, ”); Example: $foo = “105”; echo number_format((float)$foo, 2, ‘.’, ”); // Outputs -> 105.00 This function returns a string.
Use setRoundingMode, set the RoundingMode explicitly to handle your issue with the half-even round, then use the format pattern for your required output. Example: DecimalFormat df = new DecimalFormat(“#.####”); df.setRoundingMode(RoundingMode.CEILING); for (Number n : Arrays.asList(12, 123.12345, 0.23, 0.1, 2341234.212431324)) { Double d = n.doubleValue(); System.out.println(df.format(d)); } gives the output: 12 123.1235 0.23 0.1 2341234.2125 EDIT: … Read more
You are running into the old problem with floating point numbers that not all numbers can be represented exactly. The command line is just showing you the full floating point form from memory. With floating point representation, your rounded version is the same number. Since computers are binary, they store floating point numbers as an … Read more