如何获得C++模版实参推断结果

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在《C++ Primer》第五版中，关于模版实参推断，讲述了如下内容

1.类型转换与模版类型参数

2.函数模版的显式实参

3.尾置返回类型

4.类型转换

5.函数指针和实参推断

6.模版参数推断和引用

书中留给大家一些习题，要求大家给出模版参数的推断类型，参见16.2.5节练习。问题是如何验证我们的答案是否正确，我在这里给出一段代码，抛砖引玉！

<span style="font-size:14px;">template <typename T>  void g(T&& val)
{

  if(is_lvalue_reference<T>::value)
  {
    if(is_const<typename remove_reference<T>::type>::value){cout<<"const left reference of "<<typeid(T).name()<<endl;}
    else{cout<<"left reference of "<<typeid(T).name()<<endl;}
  }
  else if(is_rvalue_reference<T>::value)
  {
    if(is_const<typename remove_reference<T>::type>::value){cout<<"const right reference of "<<typeid(T).name()<<endl;}
    else{cout<<"right reference of "<<typeid(T).name()<<endl;}
  }
  else cout<<typeid(T).name()<<endl;

}
//------------------------------------------------------------
</span><pre name="code" class="cpp"><span style="font-size:14px;">template <typename T>  void g(const T& val)
{

  if(is_lvalue_reference<T>::value)
  {
    if(is_const<typename remove_reference<T>::type>::value){cout<<"const left reference of "<<typeid(T).name()<<endl;}
    else{cout<<"left reference of "<<typeid(T).name()<<endl;}
  }
  else if(is_rvalue_reference<T>::value)
  {
    if(is_const<typename remove_reference<T>::type>::value){cout<<"const right reference of "<<typeid(T).name()<<endl;}
    else{cout<<"right reference of "<<typeid(T).name()<<endl;}
  }
  else cout<<typeid(T).name()<<endl;

}
//---------------------------------------------------------------
</span><pre name="code" class="cpp"><pre name="code" class="cpp"><span style="font-size:14px;">template <typename T>  void g(T val)
{

  if(is_lvalue_reference<T>::value)
  {
    if(is_const<typename remove_reference<T>::type>::value){cout<<"const left reference of "<<typeid(T).name()<<endl;}
    else{cout<<"left reference of "<<typeid(T).name()<<endl;}
  }
  else if(is_rvalue_reference<T>::value)
  {
    if(is_const<typename remove_reference<T>::type>::value){cout<<"const right reference of "<<typeid(T).name()<<endl;}
    else{cout<<"right reference of "<<typeid(T).name()<<endl;}
  }
  else cout<<typeid(T).name()<<endl;

}</span>

<span style="font-size:14px;">int main()
{
  double i=0;const double ci=i;
  g(i);
  g(ci);
  g(i*ci);
  g(i=ci);
  return 0;
}
</span>

如何获得C++模版实参推断结果

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原文地址：http://blog.csdn.net/zjq2008wd/article/details/41354569