Python CIDR地址块算法
字数
368 字
阅读时间
2 分钟
更新日期
5/21/2018
最近写扫描器有个模块需要把CIDR地址块(类似与192.168.1.1/24)之类的地址解析成IP地址。
写了一会已经可以把地址快的首地址二进制和尾地址二进制读取出来,剩下就是循环二进制解码成IP就行了。
但是解码二进制又要涉及到好多IP转换,看了下寻风的代码,有现成的CIDR转换代码。就贴上来记录一下。
#!/usr/bin/env python
# -*- encoding: utf-8 -*-
def stringxor(str1, str2):
orxstr = ""
for i in range(0, len(str1)):
rst = int(str1[i]) & int(str2[i])
orxstr = orxstr + str(rst)
return orxstr
def bin2dec(string_num):
return str(int(string_num, 2))
def getip(ip, type):
result = ''
for i in range(4):
item = bin2dec(ip[0:8])
if i == 3:
if type == 0:
item = str(int(item) + 1)
else:
item = str(int(item) - 1)
result = result + item + '.'
ip = ip[8:]
return result.strip('.')
def CIDR(input):
try:
ip = input.split('/')[0]
pos = int(input.split('/')[1])
ipstr = ''
for i in ip.split('.'):
ipstr = ipstr + bin(int(i)).replace('0b', '').zfill(8)
pstr = '1' * pos + '0' * (32 - pos)
res = stringxor(ipstr, pstr)
_ip = getip(res, 0), getip(res[0:pos] + '1' * (32 - pos), 1)
return _ip[0] + "-" + _ip[1]
except:
return input
def get_ip_list(ip):
ip_list_tmp = []
iptonum = lambda x: sum([256 ** j * int(i) for j, i in enumerate(x.split('.')[::-1])])
numtoip = lambda x: '.'.join([str(x / (256 ** i) % 256) for i in range(3, -1, -1)])
if '-' in ip:
ip_range = ip.split('-')
ip_start = long(iptonum(ip_range[0]))
ip_end = long(iptonum(ip_range[1]))
ip_count = ip_end - ip_start
if ip_count >= 0 and ip_count <= 655360:
for ip_num in range(ip_start, ip_end + 1):
ip_list_tmp.append(numtoip(ip_num))
else:
print 'IP format error'
else:
ip_split = ip.split('.')
net = len(ip_split)
if net == 2:
for b in range(1, 255):
for c in range(1, 255):
ip = "%s.%s.%d.%d" % (ip_split[0], ip_split[1], b, c)
ip_list_tmp.append(ip)
elif net == 3:
for c in range(1, 255):
ip = "%s.%s.%s.%d" % (ip_split[0], ip_split[1], ip_split[2], c)
ip_list_tmp.append(ip)
elif net == 4:
ip_list_tmp.append(ip)
else:
print "IP format error"
return ip_list_tmp
ip = "255.168.2.1/21"
print get_ip_list(CIDR(ip))