c++ - 嵌套类上的 RTTI，VS Bug？

Question

以下代码输出:

struct Property<int>::IValue

但我希望它输出:

struct Property<int>::Value<int>

代码:

struct IProperty
{
    virtual ~IProperty() = 0;
};

IProperty::~IProperty(){}

template<typename T>
struct Property : IProperty
{
    struct IValue
    {
        virtual ~IValue() = 0;
    };

    template<typename Q>
    struct Value : IValue
    {
        Q thing;
    };

    Property() : m_pValue(new Value<T>()){}
    std::shared_ptr<IValue> m_pValue;
};

template<typename T>
Property<T>::IValue::~IValue(){}

int main()
{
    std::unique_ptr<IProperty> p(new Property<int>);

    std::cout << typeid(*static_cast<Property<int>&>(*p).m_pValue).name() << '\n';
}

如果IValue和 Value被移出Property这样它们就不再是嵌套类了，我得到了我期望的结果。这是 VS 错误还是预期行为？

Answer 1

这绝对是 Visual C++ 编译器错误。由于某些奇怪的原因，编译器无法确定 Property<T>::IValue从 shared_ptr 取消引用是多态基类，并为 typeid 使用静态类型信息表达式而不是运行时类型信息。

重现错误的最少代码:

template <class T> struct Property
{
   struct IValue
   {
      void f() {}
      virtual ~IValue() = 0 {}
   };

   struct Value: IValue {};

   Property(): m_pValue(new Value())
   {
      // typeid inside class works as expected
      std::cout << "Inside class: " << typeid(*m_pValue).name() << std::endl;
   }

   // If changed to unique_ptr typeid will work as expected
   // But if changed to raw pointer the bug remains!
   std::shared_ptr<IValue> m_pValue;
};

int main()
{
   Property<int> p;
   // If next line uncommented typeid will work as expected
   //p.m_pValue->f();
   std::cout << "Outside class: " << typeid(*p.m_pValue).name() << std::endl;
}

我试了一下这个样本，发现 typeid如果在 typeid 之前调用了指向对象的某些函数，则开始按预期工作并显示运行时类型名称或者如果 shared_ptr替换为 unique_ptr .有趣，但要替换 shared_ptr<IValue>与 IValue*仍然重现错误!

实际输出:

Inside class: struct Property<int>::ValueOutside class: struct Property<int>::IValue

Expected (correct) output:

Inside class: struct Property<int>::ValueOutside class: struct Property<int>::Value

From assembler listing it's clear that it's really compiler problem:

typeid called from Property ctor generates honest call to __RTtypeid:

; 225 : std::cout << typeid(*m_pValue).name() << std::endl;
  ... irrelevant code skipped ...
; __type_info_root_node
  push  OFFSET ?__type_info_root_node@@3U__type_info_node@@A
  mov ecx, DWORD PTR _this$[ebp]
; std::tr1::shared_ptr<Property<int>::IValue>::operator*
  call ??D?$shared_ptr@UIValue@?...<skipped>
  push eax
  call ___RTtypeid

typeid外部调用生成静态 IValue输入信息:

; 237  :    std::cout << typeid(*p.m_pValue).name() << std::endl;
  ... irrelevant code skipped ...
; __type_info_root_node
  push  OFFSET ?__type_info_root_node@@3U__type_info_node@@A
  mov ecx, OFFSET ??_R0?AUIValue@?$Property@H@@@8

另请注意，VS2008 中存在相同的错误，因此这是过去的行为。