How to fmt.Printf an integer with thousands comma

Question

Does Go\'s fmt.Printf support outputting a number with the thousands comma?

fmt.Printf(\"%d\", 1000) outputs 1000, what format

Answer 1

Here is a function that takes an integer and grouping separator and returns a string delimited with the specified separator. I have tried to optimize for efficiency, no string concatenation or mod/division in the tight loop. From my profiling it is more than twice as fast as the humanize.Commas implementation (~680ns vs 1642ns) on my Mac. I am new to Go, would love to see faster implementations!

Usage: s := NumberToString(n int, sep rune)

Examples

Illustrates using different separator (',' vs ' '), verified with int value range.

s:= NumberToString(12345678, ',')

=> "12,345,678"

s:= NumberToString(12345678, ' ')

=> "12 345 678"

s: = NumberToString(-9223372036854775807, ',')

=> "-9,223,372,036,854,775,807"

Function Implementation

func NumberToString(n int, sep rune) string {

    s := strconv.Itoa(n)

    startOffset := 0
    var buff bytes.Buffer

    if n < 0 {
        startOffset = 1
        buff.WriteByte('-')
    }


    l := len(s)

    commaIndex := 3 - ((l - startOffset) % 3) 

    if (commaIndex == 3) {
        commaIndex = 0
    }

    for i := startOffset; i < l; i++ {

        if (commaIndex == 3) {
            buff.WriteRune(sep)
            commaIndex = 0
        }
        commaIndex++

        buff.WriteByte(s[i])
    }

    return buff.String()
}