python - Scapy:向 fields_desc 添加更多字段

Question

我正在尝试创建一个新协议(protocol)。问题是，需要将更多字段动态添加到 fields_desc，但我不确定如何实现。在 Scapy 文档中，我读到 Scapy 中的一层实际上只是一个字段列表，可以对其进行操作，但 Scapy 似乎不喜欢以下内容:

>>> class SomePacket(Packet):
...         
...     name = "SomePacket"
...     fields_desc = [ IntField("Number",0) ]
...             
...     def add_IntField(self, name, value):
...         self.fields_desc.append(IntField(name, value))
... 
>>> packet = SomePacket()
>>> ls(packet)
Number     : IntField             = 0               (0)
>>> packet.add_IntField("X",1)
>>> ls(packet)
Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "/usr/lib/python2.7/dist-packages/scapy/packet.py", line 1212, in ls
    print "%-10s : %-20s = %-15s (%s)" % (f.name, f.__class__.__name__, repr(getattr(obj,f.name)), repr(f.default))
  File "/usr/lib/python2.7/dist-packages/scapy/packet.py", line 176, in __getattr__
    fld,v = self.getfield_and_val(attr)
  File "/usr/lib/python2.7/dist-packages/scapy/packet.py", line 172, in getfield_and_val
    return self.payload.getfield_and_val(attr)
  File "/usr/lib/python2.7/dist-packages/scapy/packet.py", line 1057, in getfield_and_val
    raise AttributeError(attr)
AttributeError: X
>>> packet = SomePacket()
>>> ls(packet)
Number     : IntField             = 0               (0)
X          : IntField             = 1               (1)

因此，当我在添加字段后首次尝试显示数据包内容时，它不起作用。但是，如果我再次创建数据包，属性就会突然出现。我做错了什么？

Answer 1

你不应该那样做。当您创建一个新的 SomePacket 对象时，Packet 的 __init__() 函数采用 fields_desc 列表来初始化一个很多其他的东西。向 fields_desc 动态添加新值会导致不一致，从而产生给定的错误。

一个解决方案是将 add_IntField 函数更改为 classmethod 并在创建新对象之前添加所需的字段，如下所示:

>>> class SomePacket(Packet):
...         
...     name = "SomePacket"
...     fields_desc = [ IntField("Number",0) ]
...     
...     @classmethod
...     def add_IntField(cls, name, value):
...         cls.fields_desc.append(IntField(name, value))

>>> SomePacket.add_IntField('X', 1)
>>> packet = SomePacket()
>>> ls(packet)
Number     : IntField             = 0               (0)
X          : IntField             = 1               (1)