How can we truncate float64 type to a particular precision?

Question

package main

import (
    \"fmt\"
    \"strconv\"
    )

func main() {
    k := 10/3.0
    i := fmt.Sprintf(\"%.2f\", k)
    f,_ := strconv.ParseFloat(i, 2)
    fmt.Pri         


        

Answer 1

The following code should work for a lot of simple use cases with relatively small numbers and small precision inputs. However, it may not work for some uses cases because of numbers overflowing out of the range of float64 numbers, as well as IEEE-754 rounding errors (other languages have this issue as well).

If you care about using larger numbers or need more precision, the following code may not work for your needs, and you should use a helper library (e.g. https://github.com/shopspring/decimal).

I picked up a one-liner round function from elsewhere, and also made toFixed() which depends on round():

func round(num float64) int {
    return int(num + math.Copysign(0.5, num))
}

func toFixed(num float64, precision int) float64 {
    output := math.Pow(10, float64(precision))
    return float64(round(num * output)) / output
}

Usage:

fmt.Println(toFixed(1.2345678, 0))  // 1
fmt.Println(toFixed(1.2345678, 1))  // 1.2
fmt.Println(toFixed(1.2345678, 2))  // 1.23
fmt.Println(toFixed(1.2345678, 3))  // 1.235 (rounded up)