Hi Jochen,
I think it's possible with a trick.
My first naive thought on how to solve this is to simply shift the decimal
point over, do the rounding on the whole number, and then shift it back.
import "math"
// Round performs rounding by shifting the decimal, rounding, and shifting
back.
func Round(x float64, digits int) float64 {
scale := math.Pow(10, float64(digits))
return math.Round(x*scale) / scale
}
https://go.dev/play/p/_DVsD45FAKb
Am Mi., 13. Aug. 2025 um 14:51 Uhr schrieb robert engels <
reng...@ix.netcom.com>:
> Read up on numerical analysis - what you are asking for is impossible :)
>
> You need to convert to a string, or use BCD/fixed place values - like
> github.com/robaho/fixed
>
> On Aug 13, 2025, at 7:42 AM, Jochen Voss <jochen.v...@gmail.com> wrote:
>
> Dear all,
>
> I would like to define a function "func Round(x float64, digits int)
> float64" which rounds to the given number of digits, in the sense that I
> want "strconv.FormatFloat(x, 'f', -1, 64)" to show at most the given
> number of digits after the decimal point.
>
> The following naive approach does not work:
>
> func Round(x float64, digits int) float64 {
> eps := math.Pow10(-digits)
> return math.Round(x/eps) * eps
> }
>
> For example for rounding math.Pi to five digits I get "3.1415900000000003"
> instead of "3.14159". https://go.dev/play/p/gRtHG6ZgTjj .
>
> Anthropic's Claude suggested the following:
>
> func Round(x float64, digits int) float64 {
> if digits <= 0 {
> pow := math.Pow10(-digits)
> return math.Round(x/pow) * pow
> }
>
> format := "%." + strconv.Itoa(digits) + "f"
> s := fmt.Sprintf(format, x)
> result, _ := strconv.ParseFloat(s, 64)
>
> return result
> }
>
> This seems to work, but also seems quite inefficient.
>
> Is there a better way?
>
> All the best,
> Jochen
>
> PS.: Here is some testing code for experimenting
> https://go.dev/play/p/Xcd6fTvYend
>
>
>
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