1

在下面的程序中，当我使用从循环缓冲区的前面访问元素时，*(begin_iterator + index)我遇到了崩溃。但是，如果我使用 访问相同的元素buffer[index]，则会消除崩溃。（请参阅下面的两条注释行）。为什么会这样？

#include <boost/circular_buffer.hpp>
#include <thread>

auto buffer = boost::circular_buffer<int>(5000);

void f1()
{
    const auto N = 500;

    while (true) {

        // Make sure the buffer is never empty
        if (buffer.size() < N+1) {
            continue;
        }

        auto front = buffer.begin();

        volatile auto j = int{};
        for (auto i = 0; i < N; ++i) {
            // j = *(front + i);        // un-commenting this causes a crash
            // j = buffer[i];           // un-commenting this doesn't
        }

        buffer.erase_begin(N);
    }
}
void f2()
{
    while (true) {

        // Make sure the buffer is at most half-full.
        if (buffer.size() > buffer.capacity() / 2) {
            continue;
        }

        static auto k = 0;
        buffer.push_back(k++);
    }
}

int main()
{
    auto t1 = std::thread{f1};
    auto t2 = std::thread{f2};

    t1.join();
}
4

1 回答 1

2

您正在经历未定义的行为，因为您正在读取和修改与多个线程不同步的同一个对象。

以一种或另一种方式对其进行编码可能会消除崩溃，但程序仍然是错误的。

如果我们添加一个互斥锁，那么就不会再出现崩溃了：

#include <boost/circular_buffer.hpp>
#include <thread>
#include <mutex>

boost::circular_buffer<int> buffer(5000);
std::mutex mtx;

void f1()
{
    const auto N = 500;

    while (true) {
        std::lock_guard<std::mutex> lock(mtx);

        // Make sure the buffer is never empty
        if (buffer.size() < N + 1) {
            continue;
        }

        auto front = buffer.begin();

        volatile auto j = int{};
        for (auto i = 0; i < N; ++i) {
             j = *(front + i);        // is OK now
            // j = buffer[i];         
        }

        buffer.erase_begin(N);
    }
}
void f2()
{
    while (true) {
        std::lock_guard<std::mutex> lock(mtx);

        // Make sure the buffer is at most half-full.
        if (buffer.size() > buffer.capacity() / 2) {
            continue;
        }

        static auto k = 0;
        buffer.push_back(k++);
    }
}

int main()
{
    auto t1 = std::thread{ f1 };
    auto t2 = std::thread{ f2 };

    t1.join();
}
于 2021-03-20T15:51:42.087 回答