Base-10 floating point in Swift

Question

I read about swift floating point types which are :

The Swift standard library provides three signed floating-point number >types: Float for 32-bit floating-point numbers, Double for 64-bit floating >point numbers, and Float80 for extended-precision 80-bit floating-point > numbers.

But I didn't see a "base 10" decimal type. and I think it is represented in base 2 because :

let g1 = (Double(0.1)+Double(0.2) );
println(g1==Double(0.3)) //false

Question:

How can I make this to evaluate to true ?

Answer 1

Swift has the same base-10 numerical type as ObjC has had for years: NSDecimalNumber.

One nice change in Swift is that you can, if you choose, add operator overloads to simplify its usage.

import Foundation

public func +(lhs: NSDecimalNumber, rhs: NSDecimalNumber) -> NSDecimalNumber {
    return lhs.decimalNumberByAdding(rhs)
}

let g1 = (NSDecimalNumber(double:0.1) + NSDecimalNumber(double:0.2) );
println(g1==NSDecimalNumber(double:0.3)) // true

If you want to have even more certainty of avoiding rounding errors (there's a possible rounding error when you take a value through the implicit double-convertion to get it to NSDecimalNumber(double:)), you can use the more explicit, integer-based interface that cannot suffer rounding:

let point1 = NSDecimalNumber(mantissa: 1, exponent: -1, isNegative: false);
let point2 = NSDecimalNumber(mantissa: 2, exponent: -1, isNegative: false);

You may be tempted to use NSDecimalNumber(string:), and it's useful, but you have to be very careful about locales. You probably want to use systemLocale() (which is really a "fallback" locale with fixed settings) to avoid comma-vs-dot confusion:

let g2 = NSDecimalNumber(string: "0.1", locale:NSLocale.systemLocale())

But I'd avoid strings here and use the other techniques that are safer.

Answer 2

Honestly? There's just no clean solution.

Others have written a fuzzyEquals method that compares only to the nth decimal place.
Here's a similar question.