了解C ++内存模型：不同运行中的不同值 [英] Understanding C++ memory model : Different values on different runs

问题描述

以下代码有什么问题？我期望看到10由consumer1和consumer2产生，但我有时看到-1。

What is wrong with the below code? I expect to see 10 to be produced by consumer1 and consumer2, but I see -1 sometimes.

#include <thread>   
#include <atomic>
#include <cassert>
#include <string>

std::atomic<int> global;
void producer()
{
   global.store(10, std::memory_order_release);
}

void consumer1()
{
   int a = global.load(std::memory_order_acquire);
   printf("a in consumer1 %d\n", a);
}

void consumer2()
{
   int a = global.load(std::memory_order_acquire);
   printf("a in consumer2 %d\n", a);
}

int main()
{
    global.store(-1, std::memory_order_seq_cst);
    std::thread t1(producer);
    std::thread t2(consumer1);
    std::thread t3(consumer2);
    t1.join(); t2.join(); t3.join();
}

我看到

a在consumer2 10
和
code>
a在consumer2 10

如果我理解正确， memory_order_acquire 始终与 memory_order_release 的线程同步。我错了吗？
我在x86-64位机器上运行。我用
编译g ++ file.cpp -pthread -std = c ++ 11

If I understand correctly, the thread which does memory_order_acquire always syncs with the thread which does memory_order_release. Am I wrong? I am running on x86-64 bit machine. I am compiling with g++ file.cpp -pthread -std=c++11

推荐答案

原子变量有非常漂亮的属性，读取的值是之前写过的值。而使用发布/获取语义，它甚至是写入的最后一个值。

Atomic variables have the very nice property that the value read is a value that has been written before. And with release/acquire semantics it's even the last value written.

在这种情况下，您有2次写入和两次读取。只有-1的写入被排序 - 在读取之前，10的写入不被排序。因此，任一值可以是最后写入的。它保证你读取-1或10而不是垃圾。

In this case, you have 2 writes and two reads. Only the write of -1 is sequenced-before the reads, the write of 10 is not sequenced. Therefore either value may be the last written. It's guaranteed that you read -1 or 10 and not garbage.

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