c++ - using c++11 auto as return type for const function object -

i have const function object , timebeing, returnes void. can return int or double. writing code in c++11 style , trying use auto return type. although code compiles, not sure 100% correct or not. here code.

template <typename graph_t>  struct my_func {     public:     my_func() {  }      my_func (graph_t&  _g) : g(_g) {  }      template <typename edge_t>    auto operator()(edge_t   edge) -> void const {       //do edge.    } //operator     private:     graph_t& g;    };     //call functor: (pass graph g template parameter)    std::for_each(beginedge, endedge, my_func<graph>(g)); 

this code compiles , works in serial mode perfectly. try parallelize above for_each using intel tbb parallel_for_each(). requires function object const. meaning threads should not allowed modify or change private variables of function object.

   //tbb::parallel_for_each    tbb::paralle_for_each(beginedge, endedge, my_func<graph>(g));     now, compiler error comes:     passing const my_func< ... > ..  discards qualifiers 

so had change operator()() following:

   template <typename edge_t>    void operator()(edge_t  edge) const {      //     } //operator 

my question is: how use "auto operator()() ->void" , make operator "const" becomes valid ?

my question is: how use "auto operator()() ->void" , make operator "const" becomes valid ?

   template <typename edge_t>    auto operator()(edge_t   edge) const -> void    {       //do edge.    } 

remember, declarator cv-qualifier has following form:

( parameter-declaration-clause ) cv-qualifier-seq [ref-qualifier] [exception-specification] [trailing-return-type]

(omitted attributes)