Round currency closest to five

Question

I\'d like to round my values to the closest of 5 cent for example:

5.31 -> 5.30
5.35 -> 5.35
5.33 -> 5.35
5.38 -> 5.40

Currentl

Answer 1

Turns out..it's really simple

let x: Float = 1.03 //or whatever value, you can also use the Double type
let y = round(x * 20) / 20

Answer 2

Depending on how you are storing your currency data, I would recommend using a dictionary or an array to look up the original cents value, and return a pre-computed result. There's no reason to do the calculations at all, since you know that 0 <= cents < 100.

If your currency is a string input, just chop off the last couple of digits and do a dictionary lookup.

round_cents = [ ... "12":"10", "13":"15", ... ]

If your currency is a floating point value, well, you have already discovered the joys of trying to do that. You should change it.

If your currency is a data type, or a fixed point integer, just get the cents part out and do an array lookup.

...
round_cents[12] = 10
round_cents[13] = 15
...

In either case, you would then do:

new_cents = round_cents[old_cents]

and be done with it.

Answer 3

I will use round function and NSNumberFormatter also but slightly different algorithm

I was thinking about using % but I changed it to /

let formatter = NSNumberFormatter()
formatter.minimumFractionDigits = 2
formatter.maximumFractionDigits = 2

//5.30
formatter.stringFromNumber(round(5.31/0.05)*0.05)
//5.35
formatter.stringFromNumber(round(5.35/0.05)*0.05)
//5.35
formatter.stringFromNumber(round(5.33/0.05)*0.05)
//5.40
formatter.stringFromNumber(round(5.38/0.05)*0.05)
//12.15
formatter.stringFromNumber(round(12.13/0.05)*0.05)

Answer 4

It's really better to stay away from floating-point for this kind of thing, but you can probably improve the accuracy a little with this:

import Foundation

func roundToFive(n: Double) -> Double {
  let f = floor(n)
  return f + round((n-f) * 20) / 20
}

roundToFive(12.12) // 12.1