Boost.Python

If what you're trying to do is something like this:

ȧ黃名ͼբѹ׶㺍

typedef boost::function<void (string s) > funcptr;

void foo(funcptr fp)
{
    fp("hello,world!");
}

BOOST_PYTHON_MODULE(test)
{
    def("foo",foo) ;
}

And then:

Ȼ곣ꍊ

>>> def hello(s):
...    print s
...
>>> foo(hello)
hello, world!

The short answer is: "you can't". This is not a Boost.Python limitation so much as a limitation of C++. The problem is that a Python function is actually data, and the only way of associating data with a C++ function pointer is to store it in a static variable of the function. The problem with that is that you can only associate one piece of data with every C++ function, and we have no way of compiling a new C++ function on-the-fly for every Python function you decide to pass to foo. In other words, this could work if the C++ function is always going to invoke the same Python function, but you probably don't want that.

첶̵Ļشʇ㺡಻ᱡ㍊ բ⻊ǂoost.Python粲ޖƣ춸ʇC++粲ޖơ㍊ Ί̢ԚӚ㬐ythonꯊ톤ʵNJ���㬍 渽늽ꏵ歸븶C++ꯊ햸ի載蒻罷裬 ʇ챣䦎꺯ʽ齉⌬ᤁ 渃︶C++ꯊ햻Ĝᣴ撻睊���㬍 ⢇Ҏ҃ǎ޷莪y趴뵝踼code>foo儐ythonꯊ��� 촊ᵘ᠒뒻趐μă++ꯊ��� 뻾仰˵㬈繻C++ꯊ헜ʇ巔uem>ͬһ趼/em>Pythonꯊ��� ˼ܹ䗷㬵넇ܲ늇ģϫҪ儡㍊

If you have the luxury of changing the C++ code you're wrapping, pass it an object instead and call that; the overloaded function call operator will invoke the Python function you pass it behind the object.

ȧ黃ㄜ輸Ą㒪碗ൄC++亂룬 Ď괫坒븶object⢵瓃˼㻍 ֘Ԙ儺巔Oˋ㷻롵瓃ģ˹䫵ݵļcode>objectᳺ㵄Pythonꯊ���

That exception is protecting you from causing a nasty crash. It usually happens in response to some code like this:

胒쳣ʇ᣻䄺aӚԬ㉎㏕儱