从前缀长度python3.x获得十进制网络掩码

Question

我创建这段代码是因为我找不到任何功能来完成我的需求。

如果你能减少它会更好。

只需输入de前缀长度从1到32，您将得到十进制掩码。这段代码帮助我完成了思科的脚本。

import math

#Netmask octets
octet1 = [0,0,0,0,0,0,0,0]
octet2 = [0,0,0,0,0,0,0,0]
octet3 = [0,0,0,0,0,0,0,0]
octet4 = [0,0,0,0,0,0,0,0]

#POW list
pow_list = [7,6,5,4,3,2,1,0]

#Introduce prefix lenght
mask = int(input("Introduce the prefix lenght: "))

#According to the number of bits we will change the array elements from 0 to 1
while mask >= 25 and mask <= 32:
    octet4[mask-25] = 1
    mask -= 1

while mask >= 17 and mask <= 24:
    octet3[mask-17] = 1
    mask -= 1

while mask >= 9 and mask <= 16:
    octet2[mask-9] = 1
    mask -= 1

while mask >= 1 and mask <= 8:
    octet1[mask-1] = 1
    mask -= 1

#Obtain the number of ones
ones1 = octet1.count(1)
ones2 = octet2.count(1)
ones3 = octet3.count(1)
ones4 = octet4.count(1)

#Summary and reuslt of each octet.
sum1 = 0
for i in range(0,ones1):
    sum1 = sum1 + math.pow(2,pow_list[i])
sum1 = int(sum1)

sum2 = 0
for i in range(0,ones2):
    sum2 = sum2 + math.pow(2,pow_list[i])
sum2 = int(sum2)

sum3 = 0
for i in range(0,ones3):
    sum3 = sum3 + math.pow(2,pow_list[i])
sum3 = int(sum3)

sum4 = 0
for i in range(0,ones4):
    sum4 = sum4 + math.pow(2,pow_list[i])
sum4 = int(sum4)

#Join the results with a "."
decimal_netmask = str(sum1) + "." + str(sum2) + "." + str(sum3) + "." + str(sum4)

#Result
print("Decimal netmask is: "+ decimal_netmask)

结果:引入前缀长度:23个十进制网罩为: 255.255.254.0

Answer 1

由于您可能正在做的不仅仅是将CIDR转换为网络掩码，所以我建议您检查内置的库ipaddress。

from ipaddress import ip_network
cidr = input("Introduce the prefix length: ")
decimal_netmask = str(ip_network(f'0.0.0.0/{cidr}').netmask)

Answer 2

您可以通过使用公式将整个掩码值计算为整数来简化代码：

mask = 2**32 - 2**(32-prefix_length)

然后，您可以计算掩码的4个8位部分(通过移动和掩蔽)，将结果附加到列表中，然后最后将列表的每个元素与.连接起来。

def decimal_netmask(prefix_length):
    mask = 2**32 - 2**(32-prefix_length)
    octets = []
    for _ in range(4):
        octets.append(str(mask & 255))
        mask >>= 8
    return '.'.join(reversed(octets))

for pl in range(33):
    print(f'{pl:3d}\t{decimal_netmask(pl)}')

输出：

  0     0.0.0.0
  1     128.0.0.0
  2     192.0.0.0
  3     224.0.0.0
  4     240.0.0.0
  5     248.0.0.0
  6     252.0.0.0
  7     254.0.0.0
  8     255.0.0.0
  9     255.128.0.0
 10     255.192.0.0
 11     255.224.0.0
 12     255.240.0.0
 13     255.248.0.0
 14     255.252.0.0
 15     255.254.0.0
 16     255.255.0.0
 17     255.255.128.0
 18     255.255.192.0
 19     255.255.224.0
 20     255.255.240.0
 21     255.255.248.0
 22     255.255.252.0
 23     255.255.254.0
 24     255.255.255.0
 25     255.255.255.128
 26     255.255.255.192
 27     255.255.255.224
 28     255.255.255.240
 29     255.255.255.248
 30     255.255.255.252
 31     255.255.255.254
 32     255.255.255.255