Use strconv.FormatFloat like such:

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

Outputs

3.1415

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.

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 package doc of fmt explains it: The %v verb is the default format, which for floating numbers means / reverts to %g which is

%e for large exponents, %f otherwise. Precision is discussed below.

If you always want "decimal point but no exponent, e.g. 123.456", use %f explicitly.

But you can only use that for floating numbers, so you have to check the type of the value you print. For that you may use a type switch or type assertion.

Example:

switch v.(type) {
case float64, float32:
    fmt.Printf("%f\n", v)
default:
    fmt.Printf("%v\n", v)
}

Output (try it on the Go Playground):

mydata
1234567890.123000

I think you are just using the wrong format specifier:

package main

import (
    "fmt"
    "strconv"
)

func main() {
    fmt.Println(strconv.FormatFloat(-3.6739403974420544e-15, 'e', -1, 64))
}

Output:

-3.6739403974420544e-15

The docs show the other formats as well.