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

No one has mentioned using math/big. The results as pertains to the original question are the same as the accepted answer, but if you are working with floats that require a degree of precision ($money$), then you should use big.Float.

Per the original question:

package main

import (
    "math/big"
    "fmt"
)

func main() {
    // original question
    k := 10 / 3.0
    fmt.Println(big.NewFloat(k).Text('f', 2))
}

Unfortunately, you can see that .Text does not use the defined rounding mode (otherwise this answer might be more useful), but rather always seems to round toward zero:

j := 0.045
fmt.Println(big.NewFloat(j).SetMode(big.AwayFromZero).Text('f', 2)

// out -> 0.04

Nevertheless, there are certain advantages to having your float stored as a big.Float.