英文:
Remove multiple items from a slice
问题
我正在尝试使用这里的删除示例从一个切片中删除多个项：http://code.google.com/p/go-wiki/wiki/SliceTricks
这是我写的代码：
package main
import "fmt"
import "net"
func main() {
a := []string{"72.14.191.202", "69.164.200.202", "72.14.180.202", "2600:3c00::22", "2600:3c00::32", "2600:3c00::12"}
fmt.Println(a)
for index, element := range a {
if net.ParseIP(element).To4() == nil {
//a = append(a[:index], a[index+1:]...)
a = a[:index+copy(a[index:], a[index+1:])]
}
}
fmt.Println(a)
}
当切片中有多个IPv6地址时，这段代码无法正常工作。它会出现"panic: runtime error: slice bounds out of range"的错误。我应该如何修改这段代码，以便能够删除所有IPv6地址？
英文:
I am trying to remove multiple items from a slice by using the Delete examples from here: http://code.google.com/p/go-wiki/wiki/SliceTricks
Here is the code I have:
package main
import "fmt"
import "net"
func main() {
a := []string{"72.14.191.202", "69.164.200.202", "72.14.180.202", "2600:3c00::22", "2600:3c00::32", "2600:3c00::12"}
fmt.Println(a)
for index, element := range a {
if net.ParseIP(element).To4() == nil {
//a = append(a[:index], a[index+1:]...)
a = a[:index+copy(a[index:], a[index+1:])]
}
}
fmt.Println(a)
}
While the code works fine if I have only one IPv6 address in the slice, it fails if there are more than one IPv6 address. It fails with the error "panic: runtime error: slice bounds out of range". What should I do to fix this code so it's able to delete all IPv6 addresses?
答案1
得分: 18
你的问题是你正在修改你正在迭代的切片。下面是稍作修改的代码：
package main
import (
"fmt"
"net"
)
func main() {
a := []string{"72.14.191.202", "69.164.200.202", "72.14.180.202", "2600:3c00::22", "2600:3c00::32", "2600:3c00::12"}
fmt.Println(a)
for i := 0; i < len(a); i++ {
if net.ParseIP(a[i]).To4() == nil {
a = append(a[:i], a[i+1:]...)
//a = a[:i+copy(a[i:], a[i+1:])]
i-- // 由于我们刚刚删除了a[i]，我们必须重新处理该索引
}
}
fmt.Println(a)
}
英文:
Your problem is that you are modifying the slice that you are iterating over. Below is your code a bit modified:
package main
import (
"fmt"
"net"
)
func main() {
a := []string{"72.14.191.202", "69.164.200.202", "72.14.180.202", "2600:3c00::22", "2600:3c00::32", "2600:3c00::12"}
fmt.Println(a)
for i := 0; i < len(a); i++ {
if net.ParseIP(a[i]).To4() == nil {
a = append(a[:i], a[i+1:]...)
//a = a[:i+copy(a[i:], a[i+1:])]
i-- // Since we just deleted a[i], we must redo that index
}
}
fmt.Println(a)
}
答案2
得分: 4
只是提出一点：在迭代的结构上进行修改总是很棘手的。
避免这种情况的常见方法是在一个新变量中构建最终结果：
package main
import (
"fmt"
"net"
)
func main() {
a := []string{"72.14.191.202", "69.164.200.202", "72.14.180.202", "2600:3c00::22", "2600:3c00::32", "2600:3c00::12"}
fmt.Println(a)
var b []string
for _, ip := range a {
if net.ParseIP(ip).To4() != nil {
b = append(b, ip)
}
}
fmt.Println(b)
}
英文:
Just to raise the point : it is always tricky to alter the structure on which you are iterating.
A common way to avoid this is to build the end result in a new variable :
package main
import (
"fmt"
"net"
)
func main() {
a := []string{"72.14.191.202", "69.164.200.202", "72.14.180.202", "2600:3c00::22", "2600:3c00::32", "2600:3c00::12"}
fmt.Println(a)
var b []string
for _, ip := range a {
if net.ParseIP(ip).To4() != nil {
b = append(b, ip)
}
}
fmt.Println(b)
}
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