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:
4 Answers
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
