'How to transform [4]byte{1,2,3,4} to "1.2.3.4" in Golang?
Here is my code which was used to transfer a byte array to string but it was failed:
package main
import (
"bytes"
"fmt"
)
type IPAddr [4]byte
// TODO: Add a "String() string" method to IPAddr.
func (ip IPAddr) String() string {
s := [][]byte{ip[:]}
//fmt.Println(s)
sep := []byte(".")
return string(bytes.Join(s, sep))
}
func main() {
hosts := map[string]IPAddr{
"loopback": {127, 0, 0, 1},
"googleDNS": {8, 8, 8, 8},
}
for name, ip := range hosts {
fmt.Printf("%v: %v\n", name, ip)
}
}
The program does not generate a expected output as
loopback: 127.0.0.1
googleDNS: 8.8.8.8
How to transform [4]byte{1,2,3,4} to "1.2.3.4" in Golang?
Solution 1:[1]
1- Use net.IP, like this working sample code:
package main
import "fmt"
import "net"
func main() {
hosts := map[string]net.IP{
"loopback ": {127, 0, 0, 1},
"googleDNS": {8, 8, 8, 8},
}
for name, ip := range hosts {
fmt.Println(name, ":", ip)
}
}
output:
loopback : 127.0.0.1
googleDNS : 8.8.8.8
2- Use this:
func (ip IPAddr) String() string {
return strconv.Itoa(int(ip[0])) + "." +
strconv.Itoa(int(ip[1])) + "." +
strconv.Itoa(int(ip[2])) + "." +
strconv.Itoa(int(ip[3]))
}
like this working sample code:
package main
import "fmt"
import "strconv"
type IPAddr [4]byte
func (ip IPAddr) String() string {
return strconv.Itoa(int(ip[0])) + "." +
strconv.Itoa(int(ip[1])) + "." +
strconv.Itoa(int(ip[2])) + "." +
strconv.Itoa(int(ip[3]))
}
func main() {
hosts := map[string]IPAddr{
"loopback ": {127, 0, 0, 1},
"googleDNS": {8, 8, 8, 8},
}
for name, ip := range hosts {
fmt.Println(name, ":", ip)
}
}
output:
loopback : 127.0.0.1
googleDNS : 8.8.8.8
3- Use strings.Trim(strings.Join(strings.Fields(fmt.Sprint(ip)), "."), "[]")
like this working sample code:
package main
import "fmt"
import "strings"
type IPAddr [4]byte
func (ip IPAddr) String() string {
return strings.Trim(strings.Join(strings.Fields(fmt.Sprint([4]byte(ip))), "."), "[]")
}
func main() {
hosts := map[string]IPAddr{
"loopback ": {127, 0, 0, 1},
"googleDNS": {8, 8, 8, 8},
}
for name, ip := range hosts {
fmt.Println(name, ":", ip)
}
}
output:
loopback : 127.0.0.1
googleDNS : 8.8.8.8
Solution 2:[2]
You don't need to define your own type as there is already net.IP which implements fmt.Stringer.
Example:
package main
import (
"fmt"
"net"
)
func main() {
ip := net.IP{127, 0, 0, 1}
fmt.Println(ip)
}
net.IP is a byte slice, so you can also do:
b := []byte{127, 0, 0, 1}
ip := net.IP(b)
fmt.Println(ip)
Solution 3:[3]
Note that net.IP is a []byte which implements String() string so you can simply do:
net.IP{127, 0, 0, 1}.String() // => "127.0.0.1"
This strategy is much faster than fmt.Sprintf(...) or strings.Join(...):
// ip_string_test.go
var ip = net.IP{127, 0, 0, 1}
func Benchmark_net_IP_String(b *testing.B) {
for i := 0; i < b.N; i++ {
ip.String()
}
}
func Benchmark_fmt_Sprintf(b *testing.B) {
for i := 0; i < b.N; i++ {
fmt.Sprintf("%d.%d.%d.%d", ip[0], ip[1], ip[2], ip[3])
}
}
func Benchmark_strings_Join(b *testing.B) {
ss := make([]string, len(ip))
for i := 0; i < b.N; i++ {
for i, x := range ip {
ss[i] = strconv.FormatInt(int64(x), 10)
}
strings.Join(ss, ".")
}
}
$ go test -bench=. ./ip_string_test.go
goos: darwin
goarch: amd64
Benchmark_net_IP_String-8 50000000 30.8 ns/op
Benchmark_fmt_Sprintf-8 10000000 194 ns/op
Benchmark_strings_Join-8 20000000 113 ns/op
PASS
ok command-line-arguments 6.123s
Solution 4:[4]
type IPAddr struct {
A byte
B byte
C byte
D byte
}
func (p IPAddr) String() string {
return fmt.Sprintf("%v.%v.%v.%v", p.A, p.B, p.C, p.D)
}
func main() {
hosts := map[string]IPAddr{
"loopback": {127, 0, 0, 1},
"googleDNS": {8, 8, 8, 8},
}
for name, ip := range hosts {
fmt.Printf("%v: %v\n", name, ip)
}
}
Solution 5:[5]
Here's how I did it.
package main
import (
"fmt"
)
type IPAddr [4]byte
func (ip IPAddr) String() string {
return fmt.Sprintf("%v.%v.%v.%v", int(ip[0]), int(ip[1]), int(ip[2]), int(ip[3]))
}
func main() {
hosts := map[string]IPAddr{
"loopback": {127, 0, 0, 1},
"googleDNS": {8, 8, 8, 8},
}
for name, ip := range hosts {
fmt.Printf("%v: %v\n", name, ip)
}
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | |
| Solution 3 | |
| Solution 4 | Marston Lam |
| Solution 5 | chicks |