C++ 0x移动构造函数

Question

编辑:我在这里重新问了同样的问题(解决了这个问题后面的问题):为什么这个C++ 0x程序会产生意外的输出？

基本的想法是,如果你不小心,指向可移动的东西可能会给你带来一些奇怪的结果.

---

C++移动构造函数和移动赋值运算符看起来非常积极.它们可以在复制构造函数没有意义的情况下使用,因为它们不需要重复指向的资源.

但有些情况下,如果你不小心,他们会咬你.这是特别相关的,因为我已经看到允许编译器生成移动构造函数的默认实现的建议.如果有人能给我一个,我会提供一个链接.

所以,这里有一些代码有一些可能不完全明显的缺陷.我测试了代码,以确保它使用-std=gnuc++0x标志以g ++编译.这些缺陷是什么？你会如何修复它们？

#if (__cplusplus <= 199711L) && !defined(__GXX_EXPERIMENTAL_CXX0X__)
   #error This requires c++0x
#endif

#include <unordered_set>
#include <vector>
#include <utility>
#include <algorithm>

class ObserverInterface {
 public:
   virtual ~ObserverInterface() {}

   virtual void observedChanged() = 0;
   virtual void observedGoingAway() = 0;
};

class Observed {
 private:
   typedef ::std::unordered_set<ObserverInterface *> obcontainer_t;

 public:
   Observed() {}
   Observed(const Observed &) = delete;
   const Observed &operator =(const Observed &b) = delete;
   // g++ does not currently support defaulting the move constructor.
   Observed(Observed &&b) : observers_(::std::move(b.observers_)) { }
   // g++ does not currently support defaulting move assignment.
   const Observed &operator =(Observed &&b) {
      observers_ = ::std::move(b.observers_);
      return *this;
   }
   virtual ~Observed() {
      for (auto i(observers_.begin()); i != observers_.end(); ++i) {
         (*i)->observedGoingAway();
      }
   }

   void unObserve(ObserverInterface *v) {
      auto loc(observers_.find(v));
      if (loc != observers_.end()) {
         observers_.erase(loc);
      }
   }

   void changed() {
      if (!observers_.empty()) {
         // Copy observers_ to bector so unObserve works
         ::std::vector<ObserverInterface *> tmp;
         tmp.reserve(observers_.size());
         tmp.assign(observers_.begin(), observers_.end());

         for (auto i(tmp.begin()); i != tmp.end(); ++i) {
            (*i)->observedChanged();
         }
      }
   }

 private:
   obcontainer_t observers_;
};

class Observer : public ObserverInterface {
 public:
   Observer() {}
   Observer(const Observer &) = delete;
   const Observer &operator =(const Observer &b) = delete;
   // g++ does not currently support defaulting the move constructor.
   Observer(Observer &&b) : observed_(b.observed_) {
      b.observed_ = 0;
      return *this;
   }
   // g++ does not currently support defaulting move assignment.
   const Observer &operator =(Observer &&b) {
      observed_ = b.observed_;
      b.observed_ = 0;
      return *this;
   }
   virtual ~Observer() {
      if (observed_) {
         observed_->unObserve(this);
         observed_ = 0;
      }
   }

   virtual void observedChanged() {
      doStuffWith(observed_);
   }
   virtual void observedGoingAway() {
      observed_ = 0;
   }

 private:
   Observed *observed_;

   // Defined elsewhere
   void doStuffWith(Observed *);
};

Answer 1

代码有很多问题.

1. Observer::observed_ 在默认构造函数中保持未初始化,导致在调用析构函数时出现未定义的行为.
2. 没有值,只有0分配Observer::observed_,使变量多余.
3. 即使有一种方法可以将观察者与被观察者联系起来,但在移动观察者时你也不会重新注册.
4. 你试图从观察者的移动构造函数返回一个值.
5. Boost.Signals已经解决了你想要解决的任何问题.
6. 从赋值运算符返回非const引用更加惯用.