Golang human readable byte sizes

Here is a function that prints out byte sizes, like the return from len() function, into human readable units.

Function and example

import (
	"fmt"
	"strconv"
)
const (
	TB = 1000000000000
	GB = 1000000000
	MB = 1000000
	KB = 1000
)
func lenReadable(length int, decimals int) (out string) {
	var unit string
	var i int
	var remainder int
	// Get whole number, and the remainder for decimals
	if length > TB {
		unit = "TB"
		i = length / TB
		remainder = length - (i * TB)
	} else if length > GB {
		unit = "GB"
		i = length / GB
		remainder = length - (i * GB)
	} else if length > MB {
		unit = "MB"
		i = length / MB
		remainder = length - (i * MB)
	} else if length > KB {
		unit = "KB"
		i = length / KB
		remainder = length - (i * KB)
	} else {
		return strconv.Itoa(length) + " B"
	}
	if decimals == 0 {
		return strconv.Itoa(i) + " " + unit
	}
	// This is to calculate missing leading zeroes
	width := 0
	if remainder > GB {
		width = 12
	} else if remainder > MB {
		width = 9
	} else if remainder > KB {
		width = 6
	} else {
		width = 3
	}
	// Insert missing leading zeroes
	remainderString := strconv.Itoa(remainder)
	for iter := len(remainderString); iter < width; iter++ {
		remainderString = "0" + remainderString
	}
	if decimals > len(remainderString) {
		decimals = len(remainderString)
	}
	return fmt.Sprintf("%d.%s %s", i, remainderString[:decimals], unit)
}
func main() {
	test := []int{425005, 8741208, 114448910, 557891, 557,
		114710, 8933578, 5849684981, 12033687, 742289, 678007439}
	for _, v := range test {
		fmt.Println(v, "\t = ", lenReadable(v, 2))
	}
}

Output

425005          =  425.00 KB
8741208         =  8.74 MB
114448910       =  114.44 MB
557891          =  557.89 KB
557             =  557 B
114710          =  114.71 KB
8933578         =  8.93 MB
5849684981      =  5.84 GB
12033687        =  12.03 MB
742289          =  742.28 KB
678007439       =  678.00 M