Use strconv.FormatFloat like such:

s := strconv.FormatFloat(3.1415, 'f', -1, 64)
fmt.Println(s)

Outputs

3.1415

Try this

package main

import "fmt"
import "strconv"

func FloatToString(input_num float64) string {
    // to convert a float number to a string
    return strconv.FormatFloat(input_num, 'f', 6, 64)
}

func main() {
    fmt.Println(FloatToString(21312421.213123))
}

If you just want as many digits precision as possible, then the special precision -1 uses the smallest number of digits necessary such that ParseFloat will return f exactly. Eg

strconv.FormatFloat(input_num, 'f', -1, 64)

Personally I find fmt easier to use. (Playground link)

fmt.Printf("x = %.6f\n", 21312421.213123)

Or if you just want to convert the string

fmt.Sprintf("%.6f", 21312421.213123)

The problem is that you try to parse "x.x\n", e.g: 1.8\n. And this returns an error: strconv.ParseFloat: parsing "1.8\n": invalid syntax. You can do a strings.TrimSpace function or to convert feet[:len(feet)-1] to delete \n character

With strings.TrimSpace() (you need to import strings package):

feetFloat, _ := strconv.ParseFloat(strings.TrimSpace(feet), 64)

Wtih feet[:len(feet)-1]:

feetFloat, _ := strconv.ParseFloat(feet[:len(feet)-1], 64)

Output in both cases:

10.8 feet converted to meters give you 3.2918400000000005 meters

You may use fmt.Sprint

fmt.Sprint returns string format of any variable passed to it

Sample

package main

import (
    "fmt"
)

func main() {
    f := fmt.Sprint(5.03)
    i := fmt.Sprint(5)
    fmt.Println("float:",f,"\nint:",i)
}

play link