Obtaining member function/function object's operator() signature

Discussion in 'C++' started by psujkov@gmail.com, Feb 12, 2007.

  1. Guest

    Hi everybody,

    in addition to my previous question, first of all this :

    template<typename F>
    void foo(const F& f)
    {
    std::cout << boost::function_traits<F>::arity << std::endl;

    }

    foo(f);

    was a great solution for a free functions, so thanks to everybody for
    answers :) but, alas, this is not a solution for both member functions
    and function objects : boost::function_traits cannot work with any of
    these types. any ideas ?..

    best regards, Paul Sujkov
     
    , Feb 12, 2007
    #1
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  2. Guest

    Thanks for anyone trying to workaround this (if there were any) ;)
    issue is solved, no answers are needed anymore...

    best regards, Paul Sujkov
     
    , Feb 14, 2007
    #2
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