如何原子地否定std :: atomic_bool？ [英] How to atomically negate an std::atomic_bool?

问题描述

天真的布尔否定

  std :: atomic_bool b; 
 b =！b; 
  

似乎不是原子。我怀疑这是因为运算符！触发一个转换为简单 bool 。如何以原子方式执行等价否定？下面的代码说明了天真否定不是原子的：

  #include< thread> 
 #include< vector> 
 #include< atomic> 
 #include< iostream> 
 
 typedef std :: atomic_bool Bool; 
 
 void flipAHundredThousandTimes（Bool& foo）{
 for（size_t i = 0; i< 100000; ++ i）{
 foo =！foo; 
} 
} 
 
 //启动nThreads std :: threads。每个线程调用flipAHundredThousandTimes 
 //在同一个布尔
 void launchThreads（Bool& foo，size_t nThreads）{
 
 std :: vector< std :: thread>线程; 
 for（size_t i = 0; i  threads.emplace_back（flipAHundredThousandTimes，std :: ref（foo））; 
} 
 
 for（auto& thread：threads）thread.join（）; 
 
} 
 
 int main（）{
 
 std :: cout< std :: boolalpha; 
 Bool foo {true}; 
 
 //启动并加入10个线程，20次。 
 for（int i = 0; i <20; ++ i）{
 launchThreads（foo，10）; 
 std :: cout<< Result（should be true）：<< foo< \\\
; 
} 
 
} 
  

代码启动10个线程其中翻转atomic_bool一个larrge，甚至，次数（100000），并打印出布尔值。这重复20次。

编辑：对于那些想要运行这个代码，我使用一个GCC 4.7快照ubuntu 11.10与两个内核。编译选项为：

  -std = c ++ 0x -Wall -pedantic-errors -pthread 
  




解决方案

b =！b

有两个选项可供使用：

1. 而不是 atomic ，使用整数类型（例如<$ c $可以是0或1，并且与1：

     std :: atomic< int>标志（0）; 
    
    flag ^ = 1; //或flag.fetch_xor（1）;不幸的是， fetch_xor 不提供 
    




2. 在循环中执行一个比较/交换操作，直到成功：

     std :: atomic< bool> flag（false）; 
    
    bool oldValue = flag.load（）; 
    while（！flag.compare_exchange_weak（oldValue，！oldValue））{} 
     




The naive boolean negation

std::atomic_bool b;
b = !b;

does not seem to be atomic. I suspect this is because operator! triggers a cast to plain bool. How would one atomically perform the equivalent negation? The following code illustrates that the naive negation isn't atomic:

#include <thread>
#include <vector>
#include <atomic>
#include <iostream>

typedef std::atomic_bool Bool;

void flipAHundredThousandTimes(Bool& foo) {
  for (size_t i = 0; i < 100000; ++i) {
    foo = !foo;
  }
}

// Launch nThreads std::threads. Each thread calls flipAHundredThousandTimes 
// on the same boolean
void launchThreads(Bool& foo, size_t nThreads) {

  std::vector<std::thread> threads;
  for (size_t i = 0; i < nThreads; ++i) {
    threads.emplace_back(flipAHundredThousandTimes, std::ref(foo));
  }

  for (auto& thread : threads) thread.join();

}

int main() {

  std::cout << std::boolalpha;
  Bool foo{true};

  // launch and join 10 threads, 20 times.
  for (int i = 0; i < 20; ++i) {
    launchThreads(foo, 10);
    std::cout << "Result (should be true): " << foo << "\n";
  }

}

The code launches 10 threads, each of which flips the atomic_bool a larrge, even, number of times (100000), and prints out the boolean. This is repeated 20 times.

EDIT: For those who want to run this code, I am using a GCC 4.7 snapshot on ubuntu 11.10 with two cores. The compilation options are:

-std=c++0x -Wall -pedantic-errors -pthread

解决方案

b = !b is not atomic because it is comprised of both a read and a write, each of which is an atomic operation.

There are two options to use:

1. Instead of atomic<bool>, use an integral type (e.g. atomic<int>) which can be 0 or 1, and xor it with 1:

   std::atomic<int> flag(0);
   
   flag ^= 1; //or flag.fetch_xor(1);
   

   Unfortunately, fetch_xor is not provided on atomic<bool>, only on integral types.

2. Perform a compare/exchange operation in a loop, until it succeeds:

   std::atomic<bool> flag(false);
   
   bool oldValue = flag.load();
   while (!flag.compare_exchange_weak(oldValue, !oldValue)) {}
   

这篇关于如何原子地否定std :: atomic_bool？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！