Chapter 5. Frequently Asked Questions (FAQs)

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

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

void foo(funcptr fp)


And then:

>>> 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.

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.