来自Boost的无等待队列。原子示例崩溃 我在寻找用原子实现等待队列的C++实现，发现Boost。Atomic例子： template<typename T> class waitfree_queue { public: struct node { T data; node * next; }; void push(const T &data) { node * n = new node; n->data = data; node * stale_head = head_.load(boost::memory_order_relaxed); do { n->next = stale_head; } while (!head_.compare_exchange_weak(stale_head, n, boost::memory_order_release)); } node * pop_all(void) { T * last = pop_all_reverse(), * first = 0; while(last) { T * tmp = last; last = last->next; tmp->next = first; first = tmp; } return first; } waitfree_queue() : head_(0) {} // alternative interface if ordering is of no importance node * pop_all_reverse(void) { return head_.exchange(0, boost::memory_order_consume); } private: boost::atomic<node *> head_; }; int main() { // pop elements waitfree_queue<int>::node * x = q.pop_all() while(x) { X * tmp = x; x = x->next; // process tmp->data, probably delete it afterwards delete tmp; } }

当我编译原始boost时，它运行时会发生崩溃

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它是Boost.Atomic中的bug还是Atomic的MSVC实现中的bug？

它看起来像MSVC实现中的bug。断言失败，因为

Order2

是

memory\u order\u release

。但是，正如所指出的（与C++标准一样）（强调矿山）：

3参数过载相当于4参数过载 成功_order==order，失败_order==order，除非 顺序是std:：memory\u order\u acq\u rel，则故障顺序是 std:：memory\u order\u acquire，如果顺序为std:：memory\u order\u release，则为 然后故障顺序是std:：内存顺序\u松弛

换句话说，

Order2

必须是

std:：memory\u order\u released

，因为您传递了

memory\u order\u release

作为

order

。MSVC的实现不是这样的，它是一个bug。如果可能，请将其报告为错误

Assertion failed: _Order2 != memory_order_release, file c:\program files (x86)\m
icrosoft visual studio 11.0\vc\include\xxatomic, line 742