我不这么认为，只是因为 boost::bind 在这种情况下正在寻找一个函数指针，而不是一个函数模板.当您传入 FCall2Templ 时，编译器会实例化该函数并将其作为函数指针传递.

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.

但是，您可以使用函子执行以下操作

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;
}

您必须指定返回类型，因为返回类型与输入相关.如果返回没有变化，那么你可以在模板中添加typedef T result_type，这样bind就可以确定结果是什么

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