c++ - 编译器优化打破了惰性迭代器

Question

我用自定义迭代器编写了一个自定义容器。由于容器的特殊特性，必须延迟计算迭代器。为了这个问题，代码的相关部分是以这种方式实现的迭代器的取消引用运算符

template<typename T>
struct Container
{
  vector<T> m_Inner;

  // This should calculate the appropriate value.
  // In this example is taken from a vec but in 
  //the real use-case is calculated on request
  T Value(int N)
  { m_Inner.at(N); }
}

template<typename T>
struct Lazy_Iterator
{
  mutable pair<int, T> m_Current;
  int Index
  Container<T>* C

  Lazy_Iterator(const Container& Cont, int N):
    m_Current{Index, T{}}, Index{N}, C{&Cont}
  {      }

  pair<int, T>&
  operator*() const // __attribute__((noinline)) (this cures the symptom)
  {
      m_Current.first = Index; /// Optimized out
      m_Current.second = C->Value(Index); /// Optimized out
      return m_Current;
  }

}

因为迭代器本身就是一个模板，它的函数可以被编译器自由内联。

当我在没有优化的情况下编译代码时，返回值会按预期更新。在某些情况下，当我使用发布编译器优化(GCC 4.9 中的 -O2)时，编译器优化了我标记为 Optimized out 的行，即使 m_Current 成员被标记为可变。因此，返回值与迭代器应指向的值不匹配。

这是预期的行为吗？您是否知道任何可移植的方法来指定该函数的内容即使被标记为 const 也应该被评估？

我希望这个问题足够详尽以便有用。如果更多详细信息对这种情况有帮助，请提出建议。

编辑:

回答一个评论，这是从一个小测试程序中提取的潜在用法:

Container<double> myC;
Lazy_Iterator<double> It{myC, 0}
cout << "Creation: " << it->first << " , " << it->second << endl;

auto it2 = it;
cout << "Copy: "<<  it2->first << " , " << it2->second << endl;

cout << "Pre-increment: " << (it++)->first << " , " << it->second << endl;
cout << "Post-increment: " << (++it)->first << " , " << it->second << endl;
cout << "Pre-decrement: " << (it--)->first << " , " << it->second << endl;
cout << "Post-decrement: " << (--it)->first << " , " << it->second << endl;
cout << "Iterator addition: " << (it+2)->first << " , " << (it+2)->second << endl;
cout << "Iterator subtraction: "<< (it-2)->first << " , " << (it-2)->second << endl;

reverse_iterator<Lazy_Iterator> rit{it};
cout << "Reverse Iterator: " << rit->first << " , " << rit->second << endl;

auto rit2 = rit;
cout << "Reverse Iterator copy: " << rit2->first << " , " << rit2->second << endl;

cout << "Rev Pre-increment: " << (rit++)->first << " , " << rit->second << endl;
cout << "Rev Post-increment: " << (++rit)->first << " , " << rit->second << endl;
cout << "Rev Pre-decrement: " << (rit--)->first << " , " << rit->second << endl;
cout << "Rev Post-decrement: " << (--rit)->first << " , " << rit->second << endl;
cout << "Rev Iterator addition: " << (rit+2)->first << " , " << (rit+2)->second << endl;
cout << "Rev Iterator subtraction: "<< (rit-2)->first << " , " << (rit-2)->second << endl;

除最后两行外，所有测试的测试结果都符合预期

测试的最后两行在打开优化时中断。

该系统实际上运行良好，并不比任何其他迭代器更危险。当然，如果容器在他眼皮子底下被删除，它会失败，通过复制使用返回值可能更安全，而不仅仅是保留引用，但这是题外话

Answer 1

reverse_iterator 持有的物理迭代器(.base() 返回的内容)和它指向的逻辑值之间存在差异:它们是关闭的-一个。 reverse_iterator might do return *(--internal_iterator); on dereference ，这给您留下了对已销毁的局部函数临时内部结构的悬空引用。

再次阅读标准后，我发现它有额外的要求来避免这种情况，请阅读注释。

我还发现 GCC 4.9 标准库不兼容。它使用一个临时的。所以，我认为这是一个 GCC 错误。

编辑:标准报价

24.5.1.3.4 operator*     [reverse.iter.op.star]

reference operator*() const;

1 Effects:

deref_tmp = current;  
--deref_tmp; 
return *deref_tmp;

2 [ Note: This operation must use an auxiliary member variable rather than a temporary variable to avoid returning a reference that persists beyond the lifetime of its associated iterator. (See 24.2.) —end note ]

后续阅读: Library Defect Report 198 .

和it seems它是returned to old behaviour .

后期编辑:P0031 在 C++17 工作草案中投票。它指出 reverse_iterator 使用临时的，而不是成员来保存中间值。