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

Package strconv

import "strconv" > func FormatFloat

func FormatFloat(f float64, fmt byte, prec, bitSize int) string

FormatFloat converts the floating-point number f to a string, according to the format fmt and precision prec. It rounds the result assuming that the original was obtained from a floating-point value of bitSize bits (32 for float32, 64 for float64).

The format fmt is one of 'b' (-ddddp±ddd, a binary exponent), 'e' (-d.dddde±dd, a decimal exponent), 'E' (-d.ddddE±dd, a decimal exponent), 'f' (-ddd.dddd, no exponent), 'g' ('e' for large exponents, 'f' otherwise), or 'G' ('E' for large exponents, 'f' otherwise).

The precision prec controls the number of digits (excluding the exponent) printed by the 'e', 'E', 'f', 'g', and 'G' formats. For 'e', 'E', and 'f' it is the number of digits after the decimal point. For 'g' and 'G' it is the total number of digits. The special precision -1 uses the smallest number of digits necessary such that ParseFloat will return f exactly.

Use a precision of -1, not 1. Use a format of f, not g to avoid exponent form for large exponents (see HectorJ's comment).

startLat := strconv.FormatFloat(o.Coordinate.Longitude, 'f', -1, 64)

For example,

package main

import (
    "fmt"
    "strconv"
)

func main() {
    f := 64.2345
    s := strconv.FormatFloat(f, 'g', 1, 64)
    fmt.Println(s)
    s = strconv.FormatFloat(f, 'f', -1, 64)
    fmt.Println(s)
}

Output:

6e+01
64.2345

Both fmt.Sprintf and strconv.FormatFloat use the same string formatting routine under the covers, so should give the same results.

If the precision that the number should be formatted to is variable, then it is probably easier to use FormatFloat, since it avoids the need to construct a format string as you would with Sprintf. If it never changes, then you could use either.

The last argument to FormatFloat controls how values are rounded. From the documentation:

It rounds the result assuming that the original was obtained from a floating-point value of bitSize bits (32 for float32, 64 for float64)

So if you are working with float32 values as in your sample code, then passing 32 is correct.