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 simplest solution is numeric truncation (assuming i is a float and you want a precision of 2 decimal points):

float64(int(i * 100)) / 100

For example:

i := 123456.789
x := float64(int(i * 100)) / 100
// x = 123456.78

BEWARE!

If you're dealing with large numbers (numbers that can cross the max value boundaries), you should know that the above can lead to serious floating point accuracy issues:

i := float64(1<<63) // 9223372036854775808.0
fmt.Println(i, float64(int64(i * 10)) / 10)

Prints: 9.223372036854776e+18 -9.223372036854776e+17

See also:

1. how 32 bit floating point numbers work
2. how 64 bit floating point numbers work
3. golang math numeric value range constants
4. golang math/big