Golang : Check if IP address is version 4 or 6
Problem :
Your Golang program needs to determine if a given IP address is version 4 or 6.
Solution :
Not really a fool proof solution, but it should be sufficient to detect if a given string has a decimal or colon in it. If the string has decimal as separator, then it is IP version 4 and if colon, then it is IP version 6.
Here you go :
package main
import (
"fmt"
"net"
)
func ip4or6(s string) string {
for i := 0; i < len(s); i++ {
switch s[i] {
case '.':
return "version 4"
case ':':
return "version 6"
}
}
return "unknown"
}
func main() {
var ipAddress string = "192.168.0.1"
//sanity check
testInput := net.ParseIP(ipAddress)
if testInput.To4() == nil {
fmt.Printf("%v is not a valid IPv4 address\n", testInput)
}
fmt.Printf("%s is IP address of : %s \n", ipAddress, ip4or6(ipAddress))
var ip6Address string = "FE80::0202:B3FF:FE1E:8329"
//sanity check
testInputIP6 := net.ParseIP(ip6Address)
if testInputIP6.To16() == nil {
fmt.Printf("%v is not a valid IP address\n", testInputIP6)
}
fmt.Printf("%s is IP address of : %s \n", ip6Address, ip4or6(ip6Address))
var ip6AddressURLPort string = "http://[2001:db8:0:1]:80"
//sanity check
testInputIP6URLPort := net.ParseIP(ip6AddressURLPort)
if testInputIP6URLPort.To16() == nil {
fmt.Printf("%v is not a valid IP address\n", ip6AddressURLPort)
}
fmt.Printf("%s is IP address of : %s \n", ip6AddressURLPort, ip4or6(ip6AddressURLPort))
}
Output :
192.168.0.1 is IP address of : version 4
FE80::0202:B3FF:FE1E:8329 is IP address of : version 6
http://[2001:db8:0:1]:80 is not a valid IP address
http://[2001:db8:0:1]:80 is IP address of : version 6
References :
See also : Golang : How to check if IP address is in range
By Adam Ng
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