python function as parameter to c++ exposed class using ::boost::python

Question

Ive been working with Python and C++ together for some time, but never tried to implement what follows:

Id like the python user to be able to write something like:

def foo(a,b):
    return a+b

myclass.myfunc(foo)

where myclass is a c++ class exposed to python with Boost.Python, with one of its methods (myfunc) that takes a function with:

int func(int,int)

signature, and only that.

Is this possible?

Im thinking about declaring:

myclass::myfunc(boost::python::object)

and extracting the typedef'ed function signature, but im just guessing..

maybe there's a better/feasible way to do this, maybe with some 'function' object?

maybe a better title would have been 'python function as parameter to boost::python c++ exposed class' — user815129
– user815129, Commented Nov 20, 2011 at 8:46
This is an interesting question and I'm curious to see the answer. — Omnifarious
– Omnifarious, Commented Nov 20, 2011 at 9:19

Paul Manta · Accepted Answer · 2011-11-20 09:41:10Z

3

You pretty much guessed the answer. Python functions are indeed just boost::python::object instances. You can then just have a boost::function<int (int, int)> and put the Python object inside that.

I just installed my OS and I don't have Boost yet, so I can't test it, but I think it will work if you just do this (without any wrapper functions):

void function(boost::function<int (int, int)> func) {
    // ...
}

// And then you expose the function as you normally would

I expect the above to work; if it doesn't this surely will:

void function_wrap(boost::python::object func)
{
    auto lambda = [func](int a, int b) -> int {
        return boost::python::extract<int>(func(a, b));
    };
    function(boost::function<int (int, int)>(lambda));
}

// And then you expose the wrapper, not the original function

answered Nov 20, 2011 at 9:41

Paul Manta

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11 Comments

user815129 Over a year ago

that is what i was after.. ill test it asap and post results, thank you

Omnifarious Over a year ago

Interesting. What does ::boost::python do if func is not a callable? I presume ::boost::python::object has a templated version of operator () to make this work.

Paul Manta Over a year ago

@Omnifarious Yes, there's a templated operator(). In case the object is not callable there's an exception thrown (I can't remember what it was called) — pretty much the same thing that happens when you try to call an uncallable object in Python.

user815129 Over a year ago

Ok.. so: the first one doesnt work, it (predictably) complains about not having a converter to get a boost::function from a boost::python::object, then I upgraded to g++ 4.5 (in order o use lambdas) and works great. Now, can you tell me exactly how can it be? please forgive my ignorance.. but even if I know about lambdas I cant figure it out WHY it does work.. and whats that function() at the last line? Thanx in advance

Paul Manta Over a year ago

@user815129 function is your original function; function_wrap is a helper function that makes your C++ code compatible with Python. Your original function expects to receive a function object, so you can't pass it an int; instead you pass it a lambda that will call the python::object and convert its return type to int.

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