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Chapter 5. Frequently Asked Questions (FAQs)

How can I wrap a function which takes a function pointer as an argument?

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

typedefboost::function\<void(strings)\>funcptr;voidfoo(funcptrfp){fp("hello,world!");}BOOST\_PYTHON\_MODULE(test){def("foo",foo);}

And then:

\>\>\>defhello(s):...prints...\>\>\>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.

Distributed under the Boost Software License, Version 1.0. (See accompanying file LICENSE_1_0.txt or copy at http://www.boost.org/LICENSE_1_0.txt)