TL;DR Use the following function instead of the currently accepted solution to avoid some undesirable results in certain limit cases, while being potentially more efficient. Know the expected imprecision you have on your numbers and feed them accordingly in the comparison function. bool nearly_equal( float a, float b, float epsilon = 128 * FLT_EPSILON, float … Read more
The Numeric Types section documents this behaviour explicitly: round(x[, n]) x rounded to n digits, rounding half to even. If n is omitted, it defaults to 0. Note the rounding half to even. This is also called bankers rounding; instead of always rounding up or down (compounding rounding errors), by rounding to the nearest even … Read more
It’s not that you’re actually getting extra precision – it’s that the float didn’t accurately represent the number you were aiming for originally. The double is representing the original float accurately; toString is showing the “extra” data which was already present. For example (and these numbers aren’t right, I’m just making things up) suppose you … Read more
Swift 3/4: var distanceFloat1: Float = 5.0 var distanceFloat2: Float = 5.540 var distanceFloat3: Float = 5.03 extension Float { var clean: String { return self.truncatingRemainder(dividingBy: 1) == 0 ? String(format: “%.0f”, self) : String(self) } } print(“Value \(distanceFloat1.clean)”) // 5 print(“Value \(distanceFloat2.clean)”) // 5.54 print(“Value \(distanceFloat3.clean)”) // 5.03 Swift 2 (Original answer) let distanceFloat: … Read more
For example (lesser numbers are faster), 64-bit Intel Xeon X5550 @ 2.67GHz, gcc 4.1.2 -O3 short add/sub: 1.005460 [0] short mul/div: 3.926543 [0] long add/sub: 0.000000 [0] long mul/div: 7.378581 [0] long long add/sub: 0.000000 [0] long long mul/div: 7.378593 [0] float add/sub: 0.993583 [0] float mul/div: 1.821565 [0] double add/sub: 0.993884 [0] double mul/div: … Read more
From this post: How to deal with floating point number precision in JavaScript? You have a few options: Use a special datatype for decimals, like decimal.js Format your result to some fixed number of significant digits, like this: (Math.floor(y/x) * x).toFixed(2) Convert all your numbers to integers
Dogbert’s answer is good, but if your code might have to deal with negative numbers, Math.floor by itself may give unexpected results. E.g. Math.floor(4.3) = 4, but Math.floor(-4.3) = -5 Use a helper function like this one instead to get consistent results: truncateDecimals = function (number) { return Math[number < 0 ? ‘ceil’ : ‘floor’](number); … Read more
Note: This does not work in any recent version of Python. Unfortunately, I believe you have to do this by monkey-patching (which, to my opinion, indicates a design defect in the standard library json package). E.g., this code: import json from json import encoder encoder.FLOAT_REPR = lambda o: format(o, ‘.2f’) print(json.dumps(23.67)) print(json.dumps([23.67, 23.97, 23.87])) emits: … Read more
Your thread’s locale is set to one in which the decimal mark is “,” instead of “.”. Try using this: float.Parse(“41.00027357629127”, CultureInfo.InvariantCulture.NumberFormat); Note, however, that a float cannot hold that many digits of precision. You would have to use double or Decimal to do so.
The IEEE 754-2008 Standard for Floating-Point Arithmetic and the ISO/IEC 10967 Language Independent Arithmetic (LIA) Standard, Part 1 answer why this is so. IEEE 754 ยง 6.3 The sign bit When either an input or result is NaN, this standard does not interpret the sign of a NaN. Note, however, that operations on bit strings … Read more