function pointer to functor

Question

I have a static function foo but the API I want to call only accept pointer to a functor (of similar interface ). Is there a way to pass foo to the API? or I need to re-implement foo in terms of functor.

Example code:

template<typename ReturnType, typename ArgT>
struct Functor: public std::unary_function<ArgT,ReturnType>
{
    virtual ~Functor () {}
    virtual ReturnType operator()( ArgT) = 0;
};


// I have a pre written function
static int foo (int a) {
    return ++a;
}

// I am not allowed to change the signature of this function :(     
static void API ( Functor<int,int> * functor ) {
    cout << (*functor) (5);
}

int main (void) {
    API ( ??? make use of `foo` somehow ??? );
    return 0;
}

My question is for calling API, implementing Functor is only solution or there is a way by which I could use foo to pass it to API?

Will boost::bind help here?
I mean boost::bind(foo, _1) will make function object out of foo and then if there is a way to form desired functor out of function object?

@texasbruce, I think he means http://stackoverflow.com/questions/356950/c-functors-and-their-uses — merlin2011, Nov 05 '13 at 06:23
Will `boost::bind` help with *what*? Is something preventing you from wrapping your static method in a functor, or even a lambda? — WhozCraig, Nov 05 '13 at 06:26
you should add some code, the interface of your function foo, the interface of the API you want to call and the error message. — Verena Haunschmid, Nov 05 '13 at 06:36
@merlin I thought about that but that could be any struct or class. Not sure if he knows what he's talking about. If the function takes anything callable, a pointer to function is just the same as pointer to functor. — SwiftMango, Nov 05 '13 at 06:41
@juanchopanza: No I can't. If I could, I would have change it to accept function pointer instead :( — Vishnu Kanwar, Nov 05 '13 at 06:50
No, you would make it a template that can accept function pointers *and* functors :-) — juanchopanza, Nov 05 '13 at 06:51
I feel the question is valid. It could have straight answer like "No it is not possible as object state is missing in this case" _but_ why negative rank for the question? — Vishnu Kanwar, Nov 05 '13 at 06:52
@juanchopanza: Yes, template is a good option _but_ unfortunately I can't change it. — Vishnu Kanwar, Nov 05 '13 at 06:53
Note that `static int foo (int a) { return ++a; }` is the same thing as `static int foo(int a) { return a + 1; }`. — Kaz, Nov 05 '13 at 06:56
@VishnuKanwar I assume the negative points were because the question was not very clear in the beginning. (You didn't get a negative point from me though). Thanks for adding the code. — Verena Haunschmid, Nov 05 '13 at 07:13

score 2 · Accepted Answer · answered Nov 05 '13 at 06:59

2

It seems like you have no option other than writing your own functor as a derived type of Functor<int, int>. However, you could save yourself some trouble by providing an intermediate class template functor that can be instantiated from a functor or funciton pointer:

template<typename R, typename A>
struct GenericFunctor<R, A> : public Functor<R, A>
{
    template <typename F>
    MyFunctor(F f) : f_(f) {}
    ReturnType operator()(A arg) = { return f_(arg);}
private:
    std::function<R(A)> f_; // or boost::function
};

Then you can say

GenericFunctor<int, int> fun = foo;
API(&fun);  // works. GenericFinctor<int,int> is a Functor<int,int>

This is just a workaround for the fact that the stuff you have been given is so awful.

answered Nov 05 '13 at 06:59

juanchopanza

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Thanks for making it bold that it is not possible and implementing Functor is only solution. I wonder why people rank down a question all of a sudden while it could have negative answer which is also helpful. – Vishnu Kanwar Nov 05 '13 at 07:01
@VishnuKanwar you are implementing `Functor`. Using inheritance. – juanchopanza Nov 05 '13 at 07:02
Thanks a ton `juanchopanza` looks like a nice intermediate solution. – Vishnu Kanwar Nov 05 '13 at 07:07
I would like to request you to give your expert comments on this question sir: http://stackoverflow.com/questions/19269438/is-there-a-way-to-call-member-function-without-operator – Vishnu Kanwar Nov 06 '13 at 05:50

function pointer to functor

1 Answers1