Rounding NSDecimalNumber

Question

I\'m having the hardest time figuring out something that seems like it should be very simple. I need to accurately round an NSDecimalNumber to a particular number of decimal

Answer 1

For those that prefer example code...

To round to 2 decimal places (12345.68):

NSDecimalNumber *originalNumber = [NSDecimalNumber decimalNumberWithString:@"12345.6789"];
NSDecimalNumberHandler *behavior = [NSDecimalNumberHandler decimalNumberHandlerWithRoundingMode:NSRoundPlain
                                                                                          scale:2
                                                                               raiseOnExactness:NO
                                                                                raiseOnOverflow:NO
                                                                               raiseOnUnderflow:NO
                                                                            raiseOnDivideByZero:NO];

NSDecimalNumber *roundedNumber = [originalNumber decimalNumberByRoundingAccordingToBehavior:behavior];

To round to the nearest thousand (12000):

NSDecimalNumber *originalNumber = [NSDecimalNumber decimalNumberWithString:@"12345.6789"];
NSDecimalNumberHandler *behavior = [NSDecimalNumberHandler decimalNumberHandlerWithRoundingMode:NSRoundPlain
                                                                                          scale:-3
                                                                               raiseOnExactness:NO
                                                                                raiseOnOverflow:NO
                                                                               raiseOnUnderflow:NO
                                                                            raiseOnDivideByZero:NO];

NSDecimalNumber *roundedNumber = [originalNumber decimalNumberByRoundingAccordingToBehavior:behavior];

Answer 2

I got it working using the below code in Swift 3.

let amount = NSDecimalNumber(string: "123.456789")
let handler = NSDecimalNumberHandler(roundingMode: .plain, scale: 2, raiseOnExactness: false, raiseOnOverflow: false, raiseOnUnderflow: false, raiseOnDivideByZero: false)
let roundedAmount = amount.rounding(accordingToBehavior: handler)

Note the scale parameter, used to define the decimal places you need. Outlined here: https://developer.apple.com/reference/foundation/nsdecimalnumberhandler/1578295-decimalnumberhandlerwithrounding

Answer 3

I'm using this solution:

import Foundation

extension NSDecimalNumber {
    public func round(_ decimals:Int) -> NSDecimalNumber {
        return self.rounding(accordingToBehavior:
            NSDecimalNumberHandler(roundingMode: .plain,
                                   scale: Int16(decimals),
                                   raiseOnExactness: false,
                                   raiseOnOverflow: false,
                                   raiseOnUnderflow: false,
                                   raiseOnDivideByZero: false))
    }
}

let amount = NSDecimalNumber(string: "123.456")

amount.round(2)  --> 123.46
amount.round(1)  --> 123.5
amount.round(0)  --> 123
amount.round(-1) --> 120
amount.round(-2) --> 100

Answer 4

You simply call decimalNumberByRoundingAccordingToBehavior:with the desired NSDecimalNumberBehaviors protocol. See the NSDecimalNumberBehaviors reference in the dev docs.

Update: See http://www.cimgf.com/2008/04/23/cocoa-tutorial-dont-be-lazy-with-nsdecimalnumber-like-me/