c – 将不同线程中同一位置的两次轻松写入总是以其他线程以相同

发布时间：2020-12-16 03:01:33 所属栏目：百科来源：网络整理

导读：在x86体系结构中,存储到同一内存位置的总顺序,例如参见 this video.C11内存模型有哪些保证？更准确地说, -- Initially --std::atomicint x{0};-- Thread 1 --x.store(1,std::memory_order_release);-- Thread 2 --x.store(2,std::memory_order_release);--

在x86体系结构中,存储到同一内存位置的总顺序,例如参见 this video.C11内存模型有哪些保证？

更准确地说,

-- Initially --
std::atomic<int> x{0};

-- Thread 1 --
x.store(1,std::memory_order_release);

-- Thread 2 --
x.store(2,std::memory_order_release);

-- Thread 3 --
int r1 = x.load(std::memory_order_acquire);
int r2 = x.load(std::memory_order_acquire);

-- Thread 4 --
int r3 = x.load(std::memory_order_acquire);
int r4 = x.load(std::memory_order_acquire);

将结果r1 == 1,r2 == 2,r3 == 2,r4 == 1被允许(在某些架构以外的x86)？如果我要用std :: memory_order_relaxed替换所有的memory_order怎么办？

解决方法

不,这样的结果是不允许的. §1.10[intro.multithread] / p8,18(引用N3936 / C 14; N3337 / C 11第6和16段中的相同文字)：

8 All modifications to a particular atomic object M occur in some
particular total order,called the modification order of M.

18 If a value computation A of an atomic object M happens before a
value computation B of M,and A takes its value from a side effect X
on M,then the value computed by B shall either be the value stored by
X or the value stored by a side effect Y on M,where Y follows X in
the modification order of M. [ Note: This requirement is known as
read-read coherence. —end note ]

在你的代码中有两个副作用,而p8它们以一些特定的总顺序发生.在线程3中,用于计算存储在r1中的值的值计算发生在r2之前,因此给定r1 == 1和r2 == 2,我们知道由线程1执行的存储先于线程2执行的存储x的修改顺序.在这种情况下,线程4无法观察到r3 == 2,r4 == 1,而不会碰到p18.这与使用的memory_order无关.

p21中的一个注释(N3337中的p19)是相关的：

[ Note: The four preceding coherence requirements effectively
disallow compiler reordering of atomic operations to a single object,
even if both operations are relaxed loads. This effectively makes the
cache coherence guarantee provided by most hardware available to C++
atomic operations. —end note ]

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!