std::bind to std::function with return value

Question

I am facincg some difficulties when using transform a member function to a std::function, by means of std::bind. I am using g++ 7.3.0, and -std=c++11, and the compiler reports:

../untitled2/main.cpp:25:42: error: no match for call to ‘(a) (std::_Bindres_helper<short int, short int (b::*)(), b>::type, int16_t*)’
   a0(std::bind<int16_t>(&b::run, b()), &r);
                                          ^
../untitled2/main.cpp:7:8: note: candidate: template<class t_result> void a::operator()(std::function<t_result()>&&, t_result*)
   void operator()(std::function<t_result()>&& p_function, t_result* p_result)
        ^~~~~~~~
../untitled2/main.cpp:7:8: note:   template argument deduction/substitution failed:
../untitled2/main.cpp:25:42: note:   ‘std::_Bindres_helper<short int, short int (b::*)(), b>::type {aka std::_Bind_result<short int, short int (b::*(b))()>}’ is not derived from ‘std::function<t_result()>’
   a0(std::bind<int16_t>(&b::run, b()), &r);

when it tries to compile this code:

     1  #include <chrono>
     2  
     3  
     4  struct a
     5  {
     6    template<typename t_result>
     7    void operator()(std::function<t_result()>&& p_function, t_result* p_result)
     8    {
     9      *p_result = p_function();
    10    }
    11  };
    12  
    13  struct b
    14  {
    15    int16_t run() { return -5; }
    16  };
    17  
    18  int
    19  main()
    20  {
    21    a a0;
    22  
    23    int16_t r;
    24  
    25    a0(std::bind<int16_t>(&b::run, b()), &r);
    26  
    27    return 1;
    28  }

But, the compiler compiles this code:

     1  #include <chrono>
     2  
     3  
     4  struct a
     5  {
     6    template<typename t_result>
     7    void operator()(std::function<t_result()>&& p_function, t_result* p_result)
     8    {
     9      *p_result = p_function();
    10    }
    11  };
    12  
    13  struct b
    14  {
    15    int16_t run() { return -5; }
    16  };
    17  
    18  int
    19  main()
    20  {
    21    a a0;
    22  
    23    int16_t r;
    24  
    25    std::function<int16_t()> f = std::bind<int16_t>(&b::run, b());
    26  
    27    a0(std::move(f), &r);
    28  
    29    return 1;
    30  }

So, why the second code compiles, and the firt does not, if, AFAIK, I am passing to a::operator() the same object in both cases.

Thanks a lot for your time!

Hint: in the second example, change `std::function f =` into `auto f =`, see whether it compiles, and whether you can infer something from that. — Angew is no longer proud of SO, Jan 25 '19 at 14:55
there are only few cases where you need to use `std::bind`, lambdas are much nicer — 463035818_is_not_an_ai, Jan 25 '19 at 15:00
The dupe I used is about a lambda but `std::bind` has the exact same issue. Everywhere you see lambda just pretend it says `std::bind` — NathanOliver, Jan 25 '19 at 15:04
I appreciate all the comments, but I still do not understand why doing `std::function f = std::bind(&b::run, b());` and then `a0(std::move(f), &r);` works, and `a0(std::bind(&b::run, b()), &r);` does not, as they are, to my knowledge, the same code. — canellas, Jan 25 '19 at 15:12

std::bind to std::function with return value

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