IP 地址从 maxAddress 开始

Question

我需要在给定 CIDR 和一些缓存地址的情况下开始生成 IP 地址。我在这里有一些代码，我正在做字节。与存储的地址进行比较，只选择那些更大的。https://play.golang.org/p/yT_Mj4fR_jK这里出了什么问题？基本上我需要“62.76.47.12/28”中“62.76.47.9”的所有地址。在给定的 CIDR 范围内生成 IP 是众所周知的。谢谢。

Answer 1

此示例将打印从第一个地址 62.76.47.12/28 到 62.76.47.9 的地址。

游乐场： https: //play.golang.org/p/MUtbiKaQ_3-

package main

import (

    "fmt"

    "net"

)

func main() {

    cidr := "62.76.47.12/28"

    _, ipnet, _ := net.ParseCIDR(cidr)

    ipFirst := ipnet.IP

    ipFirstValue := toHost(ipFirst)

    ipLast := net.ParseIP("62.76.47.9")

    ipLastValue := toHost(ipLast)

    fmt.Println("cidr:  ", cidr)

    fmt.Println("first: ", ipFirst)

    fmt.Println("last:  ", ipLast)

    if ipLastValue < ipFirstValue {

        fmt.Println("ugh")

        return

    }

    for i := ipFirstValue; i < ipLastValue; i++ {

        addr := toIP(i)

        fmt.Println(addr)

    }

}

func toHost(ip net.IP) uint32 {

    i := ip.To4()

    return uint32(i[0])<<24 + uint32(i[1])<<16 + uint32(i[2])<<8 + uint32(i[3])

}

func toIP(v uint32) net.IP {

    v3 := byte(v & 0xFF)

    v2 := byte((v >> 8) & 0xFF)

    v1 := byte((v >> 16) & 0xFF)

    v0 := byte((v >> 24) & 0xFF)

    return net.IPv4(v0, v1, v2, v3)

}

Answer 2

如果你打印ìpMax，你会看到它的底层表示使用了 16 个字节。（另请参阅文档

fmt.Printf("'%#v'\n",ipMax)

'net.IP{0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0xff, 0xff, 0x3e, 0x4c, 0x2f, 0x9}'

您可以转换ipMax为 IPv4 表示以获得所需的结果：

ipMax := net.ParseIP("62.76.47.9").To4()

Answer 3

使用IPAddress Go library，我可以提供两种方法，一种方法类似于您通过迭代整个子网的建议，另一种方法是从给定地址开始并递增直到达到最大地址。请注意，这些解决方案与 IPv6 的工作方式相同。免责声明：我是项目经理。

62.76.47.12/28请注意，和 的封闭 CIDR 子网62.76.47.9/28是62.76.47.0/28，因此您实际上不需要同时提供62.76.47.12/28和62.76.47.9，您可以简单地提供单个字符串62.76.47.9/28。但是为了坚持您提出的示例，我将使用这两个字符串。

import (

    "fmt"

    "github.com/seancfoley/ipaddress-go/ipaddr"

)

func main() {

    cidrString, addrString := "62.76.47.12/28", "62.76.47.9"

    addr := ipaddr.NewIPAddressString(addrString).GetAddress()

    cidrAddr := ipaddr.NewIPAddressString(cidrString).GetAddress()

    iterateThroughSubnet(addr, cidrAddr) // solution 1

    incrementAddressToMax(addr, cidrAddr) // solution 2

}

func iterateThroughSubnet(addr, cidrAddr *ipaddr.IPAddress) {

    block := cidrAddr.ToPrefixBlock().WithoutPrefixLen()

    fmt.Printf("\nstarting with block %s, equivalent to %v,\nand address %s\n",

        cidrAddr.ToPrefixBlock(), block, addr)

    iterator := block.Iterator()

    for next := iterator.Next(); next != nil; next = iterator.Next() {

        fmt.Printf("compare %s with %s: %d\n", next, addr, next.Compare(addr))

    }

}

func incrementAddressToMax(addr, cidrAddr *ipaddr.IPAddress) {

    block := cidrAddr.ToPrefixBlock()

    max := block.GetUpper()

    fmt.Printf("\nstarting with block %s and address %s,\n"+

        "incrementing to block max %s\n", block, addr, max)

    for ; addr.Compare(max) <= 0; addr = addr.Increment(1) {

        fmt.Println(addr)

    }

}

输出：

starting with block 62.76.47.0/28, equivalent to 62.76.47.0-15,

and address 62.76.47.9

compare 62.76.47.1 with 62.76.47.9: -1

compare 62.76.47.2 with 62.76.47.9: -1

compare 62.76.47.3 with 62.76.47.9: -1

compare 62.76.47.4 with 62.76.47.9: -1

compare 62.76.47.5 with 62.76.47.9: -1

compare 62.76.47.6 with 62.76.47.9: -1

compare 62.76.47.7 with 62.76.47.9: -1

compare 62.76.47.8 with 62.76.47.9: -1

compare 62.76.47.9 with 62.76.47.9: 0

compare 62.76.47.10 with 62.76.47.9: 1

compare 62.76.47.11 with 62.76.47.9: 1

compare 62.76.47.12 with 62.76.47.9: 1

compare 62.76.47.13 with 62.76.47.9: 1

compare 62.76.47.14 with 62.76.47.9: 1

compare 62.76.47.15 with 62.76.47.9: 1

starting with block 62.76.47.0/28 and address 62.76.47.9,

incrementing to block max 62.76.47.15/28

62.76.47.9

62.76.47.10

62.76.47.11

62.76.47.12

62.76.47.13

62.76.47.14

62.76.47.15

对于小子网，这两种方法可能同样有效，但是随着子网变大，您希望避免第一个递增遍历整个子网的解决方案。如果需要，使用GetNetIp的方法ipaddr.IPAddress切换到net.IP。