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嗯......标题有点拗口，但我真的不确定这是哪一部分导致问题，我已经经历了很多次，并且无法确定原因......

这个想法是让单个 Choice 实例能够存储传递给它的模板列表的任何类型的任何一个值......它有点像一个联合，除了它跟踪被存储的类型并考虑值每种类型都是不同的，这允许它绕过联合成员中构造函数的 C++ 约束。

它在某些情况下确实有效，但清理代码似乎存在一些问题。当我开始使用这个结构和参数列表中传递的 std::basic_string 或类似类型时，我开始遇到段错误，但我不明白为什么会导致任何问题。

这对我自己来说是一种虽然实验，但我看不出它不应该工作的任何原因（在 g++ 中以 C++0x 模式编译）：

// virtual methods should provide a way of "remembering"
// the type stored within the choice at any given time
struct ChoiceValue
{
   virtual void del(void* value) = 0;
   virtual bool is(int choice) = 0;
};

// Choices are initialized with an instance
// of this structure in their choice buffer
// which should handle the uninitialized case
struct DefaultChoiceValue : public IChoiceValue
{
   virtual void del(void* value) {}
   virtual bool is(int choice) { return false; }
};

// When a choice is actually initialized with a value
// an instance of this structure (with the appropriate value
// for T and TChoice) is created and stored in the choice
// buffer, allowing it to be cleaned up later (using del())
template<int TChoice, typename T>
struct ChoiceValue
{
    virtual void del(void* value) { ((T*)value)->~T(); }
    virtual bool is(int choice) { return choice == TChoice; }
};

template<typename ... TAll>
struct Choice
{
};

template<typename T1, typename ... TRest>
struct Choice<T1, TRest...>
{
  // these two constants should compute the buffer size needed to store
  // the largest possible value for the choice and the actual value
  static const int CSize = sizeof(ChoiceValue<0, T1>) > Choice<TRest...>::CSize
         ? sizeof(ChoiceValue<0, T1>) : Choice<TRest...>::CSize;
  static const int VSize = sizeof(T1) > Choice<TRest...>::VSize
         ? sizeof(T1) : Choice<TRest...>::VSize;

   IChoiceValue* _choice;
   char* _choiceBuffer;
   char* _valueBuffer;

   Choice()
   {
      _choiceBuffer = new char[CSize];
      _valueBuffer = new char[VSize];
      _choice = new (_choiceBuffer) DefaultChoiceValue();
   }
   ~Choice()
   {
      _choice->del(_valueBuffer);
      delete[] _choiceBuffer;
      delete[] _valueBuffer;
   }
   template<int TChoice, typename T>
   T& get()
   {
      if(_choice->is(TChoice))
        return *(T*)_valueBuffer;
      else
      {
         _choice->del(_valueBuffer);
         new (_valueBuffer) T();
         _choice = new (_choiceBuffer) ChoiceValue<TChoice, T>();
         return *(T*)_valueBuffer;
      }
   }
};

template<typename T1>
struct Choice<T1>
{
  // required for the base case of a template
  // with one type argument
  static const int CSize = sizeof(ChoiceValue<0, T1>) > sizeof(DefaultChoiceValue)
              ? sizeof(ChoiceValue<0, T1>) : sizeof(DefaultChoiceValue);
  static const int VSize = sizeof(T1);

  // I have an implementation here as well in my code
  // but it is pretty much just a copy of the above code
  // used in the multiple types case
};

非常感谢，如果有人能找出我做错了什么:)

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1 回答 1

4

您没有发布任何与实际崩溃相关的代码，但我猜您要么Choice<...>按值返回一个实例，要么通过其他方式调用复制构造函数。由于您没有定义复制构造函数，因此您可能会双重释放您分配的内存Choice<...>::Choice。

于 2010-10-04T21:25:49.917 回答