[Bug c++/57464] c++11 create a std::function object with lambda expr

[redi at gcc dot gnu.org]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=57464

Jonathan Wakely <redi at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|UNCONFIRMED                 |RESOLVED
         Resolution|---                         |FIXED
   Target Milestone|---                         |4.7.0
            Summary|c++11 crate a std::function |c++11 create a
                   |object  with lambda expr    |std::function object  with
                   |                            |lambda expr

--- Comment #1 from Jonathan Wakely <redi at gcc dot gnu.org> ---
GCC 4.7 implements
http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2013/n3687.html#2132 so it
rejects the program, because the inner lambda expression has void return type
so is not convertible to function<int()>

Modifying it to not use std::function shows the bug is fixed already in GCC
4.7.0

template<typename Sig>
class function;

template<typename R, typename... A>
class function< R(A...) >
{
    struct interface
    {
        virtual ~interface() { }
        virtual R invoke(A... a) = 0;
    };

    template<typename F>
        struct impl : interface
        {
            impl(F f) : f(f) { }

            F f;

            virtual R invoke(A... a) { return f(a...); }
        };

    interface* i;

public:
    template<typename F>
        function(F f) : i(new impl<F>(f)) { }

    ~function() { delete i; }

    R operator()(A... a) const { return i->invoke(a...); }
};

int main()
{
  function<function<int()>(int)> rclouse =
      [](int i) { return [&i](){return ++i;}; };

  function<int()> seed3 = rclouse(3);
  __builtin_printf("%d\n", seed3());
}