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c++ - 为什么当我使用的线程数比 CPU 的内核数多时，这段代码会变得更快？

转载 作者：IT老高 更新时间：2023-10-28 22:30:19 27 4
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我有一些计时代码用于测量给定代码片段的运行时间:

struct time_data {
    std::chrono::steady_clock::time_point start, end;

    auto get_duration() const {
        return end - start;
    }

    void print_data(std::ostream & out) const {
        out << str();
    }

    std::string str() const {
        static const std::locale loc{ "" };
        std::stringstream ss;
        ss.imbue(loc);
        ss << "Start:    " << std::setw(24) << std::chrono::nanoseconds{ start.time_since_epoch() }.count() << "ns\n";
        ss << "End:      " << std::setw(24) << std::chrono::nanoseconds{ end.time_since_epoch() }.count() << "ns\n";
        ss << "Duration: " << std::setw(24) << std::chrono::nanoseconds{ get_duration() }.count() << "ns\n";
        return ss.str();
    }

    static friend std::ostream & operator<<(std::ostream & out, const time_data & data) {
        return out << data.str();
    }
};

template<typename T>
time_data time_function(T && func) {
    time_data data;
    data.start = std::chrono::steady_clock::now();
    func();
    data.end = std::chrono::steady_clock::now();
    return data;
}

然后，为了利用它，我设计了以下代码来对一组数据进行累加，但不是简单地将数字相加，而是首先找到 Total Stopping Time从应用 Collatz Conjecture而是累积它。

template<typename T>
T accumulation_function(T a, T b) {
    T count = 0;
    while (b > 1) {
        if (b % 2 == 0) b /= 2;
        else b = b * 3 + 1;
        ++count;
    }
    return a + count;
}

template<typename IT>
auto std_sum(IT begin, IT end) {
    auto sum = (*begin - *begin);
    sum = std::accumulate(begin, end, sum, accumulation_function<decltype(sum)>);
    return sum;
}

template<typename IT>
auto single_thread_sum(IT begin, IT end) {
    auto sum = (*begin - *begin);
    IT current = begin;
    while (current != end) {
        sum = accumulation_function(sum, *current);
        ++current;
    }
    return sum;
}

template<typename IT, uint64_t N>
auto N_thread_smart_sum(IT begin, IT end) {
    auto sum = (*begin - *begin);
    std::vector<std::thread> threads;
    std::array<decltype(sum), N> sums;
    auto dist = std::distance(begin, end);
    for (uint64_t i = 0; i < N; i++) {
        threads.emplace_back([=, &sums] {
            IT first = begin;
            IT last = begin;
            auto & tsum = sums[i];
            tsum = 0;
            std::advance(first, i * dist / N);
            std::advance(last, (i + 1) * dist / N);
            while (first != last) {
                tsum = accumulation_function(tsum, *first);
                ++first;
            }
        });
    }
    for (std::thread & thread : threads)
        thread.join();
    for (const auto & s : sums) {
        sum += s;
    }
    return sum;
}

template<typename IT>
auto two_thread_smart_sum(IT begin, IT end) {
    return N_thread_smart_sum<IT, 2>(begin, end);
}

template<typename IT>
auto four_thread_smart_sum(IT begin, IT end) {
    return N_thread_smart_sum<IT, 4>(begin, end);
}

template<typename IT>
auto eight_thread_smart_sum(IT begin, IT end) {
    return N_thread_smart_sum<IT, 8>(begin, end);
}

template<typename IT>
auto sixteen_thread_smart_sum(IT begin, IT end) {
    return N_thread_smart_sum<IT, 16>(begin, end);
}

template<typename IT>
auto thirty_two_thread_smart_sum(IT begin, IT end) {
    return N_thread_smart_sum<IT, 32>(begin, end);
}

template<typename IT>
auto sixty_four_thread_smart_sum(IT begin, IT end) {
    return N_thread_smart_sum<IT, 64>(begin, end);
}

作为引用:运行它的样板 main 代码:

int main() {
    std::vector<int64_t> raw_data;
    auto fill_data = time_function([&raw_data] {
        constexpr uint64_t SIZE = 1'000'000'000ull;
        raw_data.resize(SIZE);
        std::vector<std::thread> threads;
        for (int i = 0; i < 8; i++) {
            threads.emplace_back([i, SIZE, &raw_data] {
                uint64_t begin = i * SIZE / 8;
                uint64_t end = (i + 1) * SIZE / 8;
                for (uint64_t index = begin; index < end; index++) {
                    raw_data[index] = begin % (20 + i);
                }
            });
        }
        for (std::thread & t : threads) 
            t.join();
    });

    int64_t sum;

    std::cout << std::setw(25) << "Fill Data" << std::endl;
    std::cout << fill_data << std::endl;

    auto std_data = time_function([&] {
        sum = std_sum(raw_data.begin(), raw_data.end());
    });

    std::cout << std::setw(25) << "STD Sum: " << sum << std::endl;
    std::cout << std_data << std::endl;

    auto single_data = time_function([&] {
        sum = single_thread_sum(raw_data.begin(), raw_data.end());
    });

    std::cout << std::setw(25) << "Single Sum: " << sum << std::endl;
    std::cout << single_data << std::endl;

    auto smart_2_data = time_function([&] {
        sum = two_thread_smart_sum(raw_data.begin(), raw_data.end());
    });

    std::cout << std::setw(25) << "Two-Thread-Smart Sum: " << sum << std::endl;
    std::cout << smart_2_data << std::endl;

    auto smart_4_data = time_function([&] {
        sum = four_thread_smart_sum(raw_data.begin(), raw_data.end());
    });

    std::cout << std::setw(25) << "Four-Thread-Smart Sum: " << sum << std::endl;
    std::cout << smart_4_data << std::endl;

    auto smart_8_data = time_function([&] {
        sum = eight_thread_smart_sum(raw_data.begin(), raw_data.end());
    });

    std::cout << std::setw(25) << "Eight-Thread-Smart Sum: " << sum << std::endl;
    std::cout << smart_8_data << std::endl;

    auto smart_16_data = time_function([&] {
        sum = sixteen_thread_smart_sum(raw_data.begin(), raw_data.end());
    });

    std::cout << std::setw(25) << "Sixteen-Thread-Smart Sum: " << sum << std::endl;
    std::cout << smart_16_data << std::endl;

    auto smart_32_data = time_function([&] {
        sum = thirty_two_thread_smart_sum(raw_data.begin(), raw_data.end());
    });

    std::cout << std::setw(25) << "Thirty-Two-Thread-Smart Sum: " << sum << std::endl;
    std::cout << smart_32_data << std::endl;

    auto smart_64_data = time_function([&] {
        sum = sixty_four_thread_smart_sum(raw_data.begin(), raw_data.end());
    });

    std::cout << std::setw(25) << "Sixty-Four-Thread-Smart Sum: " << sum << std::endl;
    std::cout << smart_64_data << std::endl;
    return 0;
}

这是程序的输出:

                Fill Data
Start:          16,295,979,890,252ns
End:            16,300,523,805,484ns
Duration:            4,543,915,232ns

                STD Sum: 7750000000
Start:          16,300,550,212,791ns
End:            16,313,216,607,890ns
Duration:           12,666,395,099ns

             Single Sum: 7750000000
Start:          16,313,216,694,522ns
End:            16,325,774,379,684ns
Duration:           12,557,685,162ns

   Two-Thread-Smart Sum: 7750000000
Start:          16,325,774,466,014ns
End:            16,334,441,032,868ns
Duration:            8,666,566,854ns

  Four-Thread-Smart Sum: 7750000000
Start:          16,334,441,137,913ns
End:            16,342,188,642,721ns
Duration:            7,747,504,808ns

 Eight-Thread-Smart Sum: 7750000000
Start:          16,342,188,770,706ns
End:            16,347,850,908,471ns
Duration:            5,662,137,765ns

Sixteen-Thread-Smart Sum: 7750000000
Start:          16,347,850,961,597ns
End:            16,352,187,838,584ns
Duration:            4,336,876,987ns

Thirty-Two-Thread-Smart Sum: 7750000000
Start:          16,352,187,891,710ns
End:            16,356,111,411,220ns
Duration:            3,923,519,510ns

Sixty-Four-Thread-Smart Sum: 7750000000
Start:          16,356,111,471,288ns
End:            16,359,988,028,812ns
Duration:            3,876,557,524ns

前几个结果并不令人惊讶:我自己编写的累加代码与 std::accumulate 函数的工作方式大致相同(我已经看到两者都出现在连续运行，这意味着它们的实现可能是相似的)。当我转到两个线程和四个线程时，代码变得更快。这是有道理的，因为我使用的是 Intel 4 核处理器。

但超出这一点的结果令人困惑。我的 CPU 只有 4 个内核(如果您考虑超线程，则为 8 个)，但即使我原谅超线程在 8 个线程上提供了边际性能提升，将数量增加到 16、32 和 64 个线程都会产生额外的性能提升。为什么是这样？当我早已用尽 CPU 可以物理并发运行的线程数时，额外的线程如何产生额外的性能提升？

注意:这与 this linked question 不同。 ，因为我正在处理特定的用例和代码，而链接的问题是一般性的。

编辑:

我对代码进行了调整，以便在创建线程时，将它们的优先级设置为 Windows 允许的最高优先级。总的来说，结果似乎是一样的。

                Fill Data
Start:          18,950,798,175,057ns
End:            18,955,085,850,007ns
Duration:            4,287,674,950ns

                STD Sum: 7750000000
Start:          18,955,086,975,013ns
End:            18,967,581,061,562ns
Duration:           12,494,086,549ns

             Single Sum: 7750000000
Start:          18,967,581,136,724ns
End:            18,980,127,355,698ns
Duration:           12,546,218,974ns

   Two-Thread-Smart Sum: 7750000000
Start:          18,980,127,426,332ns
End:            18,988,619,042,214ns
Duration:            8,491,615,882ns

  Four-Thread-Smart Sum: 7750000000
Start:          18,988,619,135,487ns
End:            18,996,215,684,824ns
Duration:            7,596,549,337ns

 Eight-Thread-Smart Sum: 7750000000
Start:          18,996,215,763,004ns
End:            19,002,055,977,730ns
Duration:            5,840,214,726ns

Sixteen-Thread-Smart Sum: 7750000000
Start:          19,002,056,055,608ns
End:            19,006,282,772,254ns
Duration:            4,226,716,646ns

Thirty-Two-Thread-Smart Sum: 7750000000
Start:          19,006,282,840,774ns
End:            19,010,539,676,914ns
Duration:            4,256,836,140ns

Sixty-Four-Thread-Smart Sum: 7750000000
Start:          19,010,539,758,113ns
End:            19,014,450,311,829ns
Duration:            3,910,553,716ns

双重编辑组合:

我编辑了程序以确保 tsum 变量是 lambda 内部的局部变量，而不是对 lambda 外部的引用。多线程代码的速度明显加快，但仍表现出相同的行为。

                Fill Data
Start:          21,740,931,802,656ns
End:            21,745,429,501,480ns
Duration:            4,497,698,824ns

                STD Sum: 7750000000
Start:          21,745,430,637,655ns
End:            21,758,206,612,632ns
Duration:           12,775,974,977ns

             Single Sum: 7750000000
Start:          21,758,206,681,153ns
End:            21,771,260,233,797ns
Duration:           13,053,552,644ns

   Two-Thread-Smart Sum: 7750000000
Start:          21,771,260,287,828ns
End:            21,777,845,764,595ns
Duration:            6,585,476,767ns

  Four-Thread-Smart Sum: 7750000000
Start:          21,777,845,831,002ns
End:            21,784,011,543,159ns
Duration:            6,165,712,157ns

 Eight-Thread-Smart Sum: 7750000000
Start:          21,784,011,628,584ns
End:            21,788,846,061,014ns
Duration:            4,834,432,430ns

Sixteen-Thread-Smart Sum: 7750000000
Start:          21,788,846,127,422ns
End:            21,791,921,652,246ns
Duration:            3,075,524,824ns

Thirty-Two-Thread-Smart Sum: 7750000000
Start:          21,791,921,747,330ns
End:            21,794,980,832,033ns
Duration:            3,059,084,703ns

Sixty-Four-Thread-Smart Sum: 7750000000
Start:          21,794,980,901,761ns
End:            21,797,937,401,426ns
Duration:            2,956,499,665ns

三重编辑 WOMBO-COMBO:

我再次运行了相同的测试，但相反。结果非常相似:

                Fill Data
Start:          22,845,472,001,475ns
End:            22,850,746,219,215ns
Duration:            5,274,217,740ns

Sixty-Four-Thread-Smart Sum: 7750000000
Start:          22,850,747,328,223ns
End:            22,853,951,903,952ns
Duration:            3,204,575,729ns

Thirty-Two-Thread-Smart Sum: 7750000000
Start:          22,853,951,981,830ns
End:            22,857,239,237,507ns
Duration:            3,287,255,677ns

Sixteen-Thread-Smart Sum: 7750000000
Start:          22,857,239,307,839ns
End:            22,860,389,285,305ns
Duration:            3,149,977,466ns

 Eight-Thread-Smart Sum: 7750000000
Start:          22,860,389,353,524ns
End:            22,864,828,560,484ns
Duration:            4,439,206,960ns

  Four-Thread-Smart Sum: 7750000000
Start:          22,864,828,633,834ns
End:            22,870,640,982,096ns
Duration:            5,812,348,262ns

   Two-Thread-Smart Sum: 7750000000
Start:          22,870,641,056,956ns
End:            22,877,032,284,384ns
Duration:            6,391,227,428ns

             Single Sum: 7750000000
Start:          22,877,032,367,997ns
End:            22,889,501,695,960ns
Duration:           12,469,327,963ns

                STD Sum: 7750000000
Start:          22,889,501,770,820ns
End:            22,902,014,633,229ns
Duration:           12,512,862,409ns

最佳答案

您所有的 tsum 写入都写入同一个 array，这将占用内存中的连续地址。这将导致所有这些写入的缓存争用问题。拥有更多线程会将这些写入分散到不同的缓存行，因此 CPU 内核花费更少的时间来使缓存行无效和重新加载。

尝试累积到本地 tsum 变量(不是对 sums 的引用)，然后在循环时将其写入 sums[i]完成了。

关于c++ - 为什么当我使用的线程数比 CPU 的内核数多时，这段代码会变得更快？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/40028948/

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