How to get the next IP address?

Question

I'm considering to call net.IP.String(), strings.Split(ip, "."), some code to calculate all the corner cases and finally net.ParseIP(s). Is there a better way to this?. Below is the code of my current implementation (no special case handled).

package main

import (
    "fmt"
    "net"
    "strconv"
    "strings"
)

func main() {
    ip := net.ParseIP("127.1.0.0")
    next, err := NextIP(ip)
    if err != nil {
        panic(err)
    }
    fmt.Println(ip, next)
}

func NextIP(ip net.IP) (net.IP, error) {
    s := ip.String()
    sa := strings.Split(s, ".")
    i, err := strconv.Atoi(sa[2])
    if err != nil {
        return nil, err
    }
    i++
    sa[3] = strconv.Itoa(i)
    s = strings.Join(sa, ".")
    return net.ParseIP(s), nil
}

Answer 1

Just increment the last octet in the IP address

ip := net.ParseIP("127.1.0.0")
// make sure it's only 4 bytes
ip = ip.To4()
// check ip != nil
ip[3]++  // check for rollover
fmt.Println(ip)
//127.1.0.1

That however is technically incorrect, since the first address in the 127.1.0.1/8 subnet is 127.0.0.1. To get the true "first" address, you will also need an IPMask. Since you didn't specify one, you could use DefaultMask for IPv4 addresses (for IPv6 you can't assume a mask, and you must provide it).

http://play.golang.org/p/P_QWwRIBIm

ip := net.IP{192, 168, 1, 10}
ip = ip.To4()
if ip == nil {
    log.Fatal("non ipv4 address")
}

ip = ip.Mask(ip.DefaultMask())
ip[3]++

fmt.Println(ip)
//192.168.1.1