std :: bind作为另一个std :: bind和类型推断的参数

Question

请考虑以下代码段：

#include <iostream>
#include <functional>

typedef std::function<double()> doubleFunc;

double add(doubleFunc a, doubleFunc b)
{
  return a() + b();
}

double constValue(double value)
{
  return value;
}

int main(int argc, char *argv[])
{
  doubleFunc c1 = std::bind(constValue, 10);
  doubleFunc c2 = std::bind(constValue, 20);
  doubleFunc c3 = std::bind(constValue, 30);

  // this compiles fine:
  std::cout << add(c1, std::bind(add, c2, c3)) << std::endl;

  // here a cast is required:
  doubleFunc result = std::bind(add, c1, static_cast<doubleFunc>(std::bind(add, c2, c3)));
  std::cout << result() << std::endl;

  // similar line (but without cast to doubleFunc) won't compile:
  // doubleFunc result2 = std::bind(add, c1, std::bind(add, c2, c3));
}

为什么在最后（注释掉）的行中调用doubleFunc时第三个参数需要转换为std::bind？

这是GCC 4.8编译错误：

error: conversion from ‘std::_Bind_helper<false, double (&)(std::function<double()>, std::function<double()>), std::function<double()>&, std::_Bind<double (*(std::function<double()>, std::function<double()>))(std::function<double()>, std::function<double()>)> >::type {aka std::_Bind<double (*(std::function<double()>, std::_Bind<double (*(std::function<double()>, std::function<double()>))(std::function<double()>, std::function<double()>)>))(std::function<double()>, std::function<double()>)>}’ to non-scalar type ‘doubleFunc {aka std::function<double()>}’ requested