C ++原子CAS(比较和交换)操作不会更改值 [英] C++ atomic CAS(compare-and-swap) operation does not change value

问题描述

在下面的示例中，实际发生了什么?成功交换后，为什么价值不会改变?

In the following example, what actually happens? Why value does not changes after successful exchange?

实时: https://wandbox.org/permlink/f5VYSKfQ9UJqa8FQ

std::atomic<bool> flag{false};

int main()
{
    std::thread thread([](){
        while(true){        
            // wait until flag not becomes true
            {
              bool expect = true;
              while(!flag.compare_exchange_strong(expect, false, std::memory_order_acq_rel)){
                  std::cout << "wait" << std::endl;
              }
            }
            
            std::cout << "work" << std::endl;
        }
    });
    flag.store(true, std::memory_order_release);
    thread.join();
}

输出:

work
wait
work
wait
...

推荐答案

考虑会发生什么:

          bool expect = true;
          while(!flag.compare_exchange_strong(expect, false, std::memory_order_acq_rel)){
              std::cout << "wait" << std::endl;
          }

当标志为假时.在while循环中第一次运行测试时， expect 将为true，因此与该标志不匹配.因此，将 expect 更新为false，并且该函数返回false.因此，将打印 wait ，然后重复循环.循环中的第二个测试 expect 现在将为false，与 flag 匹配，因此 flag 将设置为false(因为已经是)，循环将退出.

when the flag is false. The first time the test in the while loop runs, expect will be true, so does not match the flag. So expect is updated to the false, and the function returns false. So wait is printed and the loop repeats. The second test in the loop, expect will now be false, which matches flag, so flag will be set to false (a noop as it already is), and the loop will exit.

最终效果是始终将 flag 设置为false，并在 wait 已经存在的情况下打印 wait .因此，您看到的输出.

The net effect will be to always set flag to false, and print wait if it was already false. Thus, the output you see.

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