【问题标题】:Problem with Memory Allocation while Using Threads使用线程时内存分配问题
【发布时间】:2021-01-23 21:04:14
【问题描述】:

我正在尝试使用使用数据分配的结构。尽管在每个函数中都获得了正确的输出,但当这些函数通过线程运行时,我得到了一个溢出错误。我目前的代码是:

#include <iostream> // cout, endl
#include <thread>   // thread

using namespace std;

// a structure to hold parameters to pass to a thread
struct StatData
{
    // the number of numbers
    int n;
    // the array of numbers
    double *numbers;
};

void average(StatData *data, double *avg) {
    int id;
    int size = data->n;
    double sum = 0.0;

    for(int i = 0; i < size; i++){
        sum += data->numbers[i];
    }
    *avg = sum / size;
}

void minimum(StatData *data, double *min) {
    int id;
    int size = data->n;
    *min = data->numbers[0];

    for(int i = 0; i < size; i++){
        if (*min > data->numbers[i]){
            *min = data->numbers[i];
        }
    }
}

void maximum(StatData *data, double *max) {
    int id;
    int size = data->n;
    *max = data->numbers[0];

    for(int i = 0; i < size; i++){
        if (*max < data->numbers[i]){
            *max = data->numbers[i];
        }
    }
}

int main(int argc, char** argv) {
    // checking if arguments were passed
    if (argc <= 1) {
        cout << "No numbers were passed." << endl;
        cout << "At least one number needs to be entered." << endl;
        return 1;
    }

    // declaring worker threads
    thread avg_t, min_t, max_t;

    // variables for values
    double avg_v;
    double min_v;
    double max_v;

    // initalizing data structure to hold numbers
    StatData data;
    data.n = argc - 1;                  // the amount of numbers passed
    data.numbers = new double[data.n];  // allocating space for nums

    // Filling in the array with numbers using atof
    for (int i = 0; i < data.n; i++) {
        data.numbers[i] = atof(argv[i+1]); // converting string to double
    }

    // creating average thread
    avg_t = thread(average, &data, &avg_v);

    // creating minimum thread
    min_t = thread(minimum, &data, &min_v);

    // creating maximum thread
    max_t = thread(maximum, &data, &max_v);

    // wating for threads to finish
    avg_t.join();
    min_t.join();
    max_t.join();

    printf ("The average value is %d\n", &avg_v);
    printf ("The minimum value is %d\n", &min_v);
    printf ("The maximum value is %d\n", &max_v);

    // freeing up dynamically allocated space
    delete[] data.numbers;
}

当我运行一个值为 47、58 和 6 的 ./test 时,我得到一个打印语句: 平均值为-502425280 最小值为 -502425288 最大值为-502425296

我不确定我哪里出错了导致代码这样做。

【问题讨论】:

  • 这是否取决于线程的使用?作为新用户,请同时关注tour 并阅读How to Ask。哦,顺便说一句,使用std::vector

标签: c++ multithreading memory allocation


【解决方案1】:

您正在打印值的地址(&amp;avg_v 是指向 avg_v 的指针,当打印为 %d 时,将打印指针本身值的一部分)。

要自己打印double 值,只需这样做

    printf ("The average value is %lf\n", avg_v);
    printf ("The minimum value is %lf\n", min_v);
    printf ("The maximum value is %lf\n", max_v);

注意%d 用于打印整数,浮点值使用%f%lf(两者是等价的)。

&amp; 仅在 scanf 中需要,因为 scanf 采用变量地址。 printf 采用实际值,因此无需将地址传递给它(当然,除非您想打印地址)。


解决了这些问题后,可以说通过使用更多 C++ 标准库功能可以使程序变得更简单。仅举几例:std::vectorstd::accumulatestd::min_elementstd::max_elementstd::async(为您创建一个线程并返回一个未来)和 lambda:

#include <iostream>
#include <thread>
#include <vector>
#include <numeric>
#include <algorithm>
#include <future>

int main(int argc, char** argv) {
    if (argc <= 1) {
        std::cout << "No numbers were passed.\n"
            "At least one number needs to be entered.\n";
        return 1;
    }

    std::vector<double> data;
    for (int i = 1; i < argc; i++) {
        data.push_back(atof(argv[i]));
    }

    auto avg_f = std::async([&] {
        return accumulate(data.begin(), data.end(), 0.0) / data.size();
    });
    auto min_f = std::async([&] {
        return *min_element(data.begin(), data.end());
    });
    auto max_f = std::async([&] {
        return *max_element(data.begin(), data.end());
    });

    // the 3 values are now being computed in parallel ...

    double avg_v = avg_f.get(); // get the values from futures (waits as necessary)
    double min_v = min_f.get();
    double max_v = max_f.get();

    std::cout << "The average value is " << avg_v << "\n";
    std::cout << "The minimum value is " << min_v << "\n";
    std::cout << "The maximum value is " << max_v << "\n";
}

【讨论】:

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