Can I use (boost) bind with a function template?

问题:

Is it possible to bind arguments to a function template with (boost) bind?

// Define a template function (just a silly example)
template<typename ARG1, typename ARG2>
ARG1 FCall2Templ(ARG1 arg1, ARG2 arg2)
{
    return arg1 + arg2;
}

// try to bind this template function (and call it)
...
boost::bind(FCall2Templ<int, int>, 42, 56)(); // This works

boost::bind(FCall2Templ, 42, 56)(); // This emits 5 pages of error messages on VS2005
// beginning with: error C2780: 
//   'boost::_bi::bind_t<_bi::dm_result<MT::* ,A1>::type,boost::_mfi::dm<M,T>,_bi::list_av_1<A1>::type> 
//   boost::bind(M T::* ,A1)' : expects 2 arguments - 3 provided

boost::bind<int>(FCall2Templ, 42, 56)(); // error C2665: 'boost::bind' : none of the 2 overloads could convert all the argument types

Ideas?

回答1:

I don't think so, only because boost::bind in this case is looking for a function pointer, not a function template. When you pass in FCall2Templ<int, int>, the compiler instantiates the function and it is passed as a function pointer.

However, you can do the following using a functor

struct FCall3Templ {

  template<typename ARG1, typename ARG2>
  ARG1 operator()(ARG1 arg1, ARG2 arg2) {
    return arg1+arg2;
  }
};
int main() {
  boost::bind<int>(FCall3Templ(), 45, 56)();
  boost::bind<double>(FCall3Templ(), 45.0, 56.0)();
  return 0;
}

You have to specify the return type, since the return type is tied to the inputs. If the return doesn't vary, then you can just add typedef T result_type to the template, so that bind can determine what the result is

回答2:

It seems to work if you create a function reference:

int (&fun)(int, int) = FCall2Templ;
int res2 = boost::bind(fun, 42, 56)();

Or:

typedef int (&IntFun)(int, int);
int res3 = boost::bind(IntFun(FCall2Templ), 42, 56)();

(Tested on GCC)