英文:
Convert rune to int?
问题
在下面的代码中，我逐个字符（rune）遍历一个字符串（string），但实际上我需要一个整数（int）来执行一些校验和计算。我真的需要将rune编码为[]byte，然后将其转换为字符串，然后使用Atoi从rune中获取一个整数吗？这是惯用的做法吗？
// 字符串s只包含数字。
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
}
在playground上查看：http://play.golang.org/p/noWDYjn5rJ
英文:
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 intfor 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
答案1
得分: 129
问题比看起来的要简单。你可以使用int(r)将一个rune值转换为一个int值。但是你的代码暗示你想要从数字的ASCII（或UTF-8）表示中获取整数值，你可以通过r - '0'作为一个rune或int(r - '0')作为一个int来轻松地实现。请注意，超出范围的rune将破坏这个逻辑。
英文:
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.
答案2
得分: 14
例如，sum += (int(c) - '0') * factor，
package mainimport ("fmt""strconv""unicode/utf8")func main() {s := "9780486653556"var factor, sum1, sum2 intfor 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 * factorsum2 += (int(c) - '0') * factor}fmt.Println(sum1, sum2)}
输出：
124 124
英文:
For example, sum += (int(c) - '0') * factor,
package mainimport ("fmt""strconv""unicode/utf8")func main() {s := "9780486653556"var factor, sum1, sum2 intfor 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 * factorsum2 += (int(c) - '0') * factor}fmt.Println(sum1, sum2)}
Output:
124 124
答案3
得分: 2
为什么你不只使用 "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 是 rune 转换为字符串
xint, _ := strconv.Atoi(xstr)
sum += xint * factor
}
fmt.Println(sum)
英文:
why don't you do only "string(rune)".
s:="12345678910"var factor,sum intfor i,x:=range s{if i%2==0{factor=1}else{factor=3}xstr:=string(x) //x is rune converted to stringxint,_:=strconv.Atoi(xstr)sum+=xint*factor}fmt.Println(sum)
答案4
得分: -1
val, _ := strconv.Atoi(string(v))
其中v是一个rune
更简洁但与上述代码思路相同
英文:
val, _ := strconv.Atoi(string(v))
Where v is a rune
More concise but same idea as above
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