Convert rune to int?

In the following code, I iterate over a string rune by rune, but I'll actually need an int to perform some checksum calculation. Do I really need to encode the rune into a []byte, then convert it to a string and then use Atoi to get an int out of the rune? Is this the idiomatic way to do it?

// The string `s` only contains digits.
var factor int
for i, c := range s[:12] {
    if i % 2 == 0 {
        factor = 1
    } else {
        factor = 3
    }
    buf := make([]byte, 1)
    _ = utf8.EncodeRune(buf, c)
    value, _ := strconv.Atoi(string(buf))
    sum += value * factor
}

On the playground: http://play.golang.org/p/noWDYjn5rJ

The problem is simpler than it looks. You convert a rune value to an int value with int(r). But your code implies you want the integer value out of the ASCII (or UTF-8) representation of the digit, which you can trivially get with r - '0' as a rune, or int(r - '0') as an int. Be aware that out-of-range runes will corrupt that logic.

For example, sum += (int(c) - '0') * factor,

package main

import (
	"fmt"
	"strconv"
	"unicode/utf8"
)

func main() {
	s := "9780486653556"
	var factor, sum1, sum2 int
	for i, c := range s[:12] {
		if i%2 == 0 {
			factor = 1
		} else {
			factor = 3
		}
		buf := make([]byte, 1)
		_ = utf8.EncodeRune(buf, c)
		value, _ := strconv.Atoi(string(buf))
		sum1 += value * factor
		sum2 += (int(c) - '0') * factor
	}
	fmt.Println(sum1, sum2)
}

Output:

124 124

why don't you do only "string(rune)".

s:="12345678910"
var factor,sum int
for i,x:=range s{
	if i%2==0{
     		factor=1
     	}else{
		factor=3
	}
     	xstr:=string(x) //x is rune converted to string
     	xint,_:=strconv.Atoi(xstr)
     	sum+=xint*factor
}
fmt.Println(sum)

val, _ := strconv.Atoi(string(v))

Where v is a rune

More concise but same idea as above