std::async 导致分段错误（或慢）

Question

我编写了简单的类来求解线性方程组，以并行化我使用 std::async 的矩阵行减法。在 10x10 矩阵大小之后，程序由于分段错误而崩溃。 我的第一个实现是用 std::async(std::launch::async, vecSub, std::ref(mat[c]), std::ref(temp)) 替换 vecSub((mat[c]), (temp)); 但后来发现如果我们不将它分配给任何变量析构函数被调用并且 .get() 被调用，并停止主线程（这就是它的原因减慢了），所以我将其更改为以下实现，现在我遇到了段错误
这是单线程：

class solver
{
    Mat mat;

public:
    //give eqn in the form ax1+ax2+ax3..axN = k (coeffiants only)
    Vec solve(Mat &in)
    {

        mat = in;

        ge(mat);
        return (bs(mat));
    }
    Vec solve(Mat &&in)
    {
        mat = std::move(in);
        ge(mat);
        return (bs(mat));
    }

private:
    void ge(Mat &mat)
    {

        using li = long int;

        for (li p = 0; p < mat[0].size() - 1; p++)
        {
            for (li c = p + 1; c < mat.size(); c++)
            {
                auto x = mat[c][p] / mat[p][p];
                auto temp = mat[p];
                vecMul(x, temp);
                vecSub((mat[c]), (temp));
    
            }
        }
    }
    Vec bs(Mat &mat)
    {
        using li = long int;
        Vec x(mat.size());
        for (li i = mat.size() - 1; i >= 0; i--)
        {
            double s = 0;
            for (li j = i; j < mat[0].size() - 1; j++)
            {
                s += mat[i][j] * x[j];
                x[i] = ((mat[i][mat[0].size() - 1] - s) / (mat[i][i]));
            }
        }
        return x;
    }
    static void vecMul(double a, Vec &b)
    {
        using li = size_t;
        for (li i = 0; i < b.size(); i++)
            b[i] *= a;
    }
    //static
    static void vecAdd(Vec &a, Vec &b)
    {
        using li = size_t;
        assert(a.size() == b.size());
        for (li i = 0; i < a.size(); i++)
            a[i] = a[i] + b[i];
    }
    static void vecSub(Vec &a, Vec &b)
    {
        using li = size_t;
        assert(a.size() == b.size());
        for (li i = 0; i < a.size(); i++)
            a[i] = a[i] - b[i];
    }
};

多线程

class solver
{
    Mat mat;

public:
    //give eqn in the form ax1+ax2+ax3..axN = k (coeffiants only)
    Vec solve(Mat &in)
    {

        mat = in;

        ge(mat);
        return (bs(mat));
    }
    Vec solve(Mat &&in)
    {
        mat = std::move(in);
        ge(mat);
        return (bs(mat));
    }

private:
    void ge(Mat &mat)
    {

        using li = long int;

        for (li p = 0; p < mat[0].size() - 1; p++)
        {
            std::vector<std::future<void>> ts;

            for (li c = p + 1; c < mat.size(); c++)
            {
                auto x = mat[c][p] / mat[p][p];
                auto temp = mat[p];
                vecMul(x, temp);
                ts.push_back(std::async(std::launch::async, vecSub, std::ref(mat[c]), std::ref(temp)));
            }
            for (auto &t : ts)
            {

                t.get();
            }
        }
    }
    Vec bs(Mat &mat)
    {
        using li = long int;
        Vec x(mat.size());
        for (li i = mat.size() - 1; i >= 0; i--)
        {
            double s = 0;
            for (li j = i; j < mat[0].size() - 1; j++)
            {
                s += mat[i][j] * x[j];
                x[i] = ((mat[i][mat[0].size() - 1] - s) / (mat[i][i]));
            }
        }
        return x;
    }
    static void vecMul(double a, Vec &b)
    {
        using li = size_t;
        for (li i = 0; i < b.size(); i++)
            b[i] *= a;
    }
    //static
    static void vecAdd(Vec &a, Vec &b)
    {
        using li = size_t;
        assert(a.size() == b.size());
        for (li i = 0; i < a.size(); i++)
            a[i] = a[i] + b[i];
    }
    static void vecSub(Vec &a, Vec &b)
    {
        using li = size_t;
        // assert(a.size() == b.size());
        for (li i = 0; i < a.size(); i++)
            a[i] = a[i] - b[i];
    }
};

分割发生在

static void vecSub(Vec &a, Vec &b)
    {
        using li = size_t;
        assert(a.size() == b.size());
        for (li i = 0; i < a.size(); i++)
            a[i] = a[i] - b[i];
    }

Answer 1

无论Mat 是什么类型，我都非常确定：

auto temp = mat[p];

不创建引用而是创建副本。就是这个意思

ts.push_back(std::async(std::launch::async, vecSub, std::ref(mat[c]), std::ref(temp)));`

实际上引用了一个临时值temp，该值将在循环结束时被销毁。这是未定义的行为。
这可能会解决问题：

auto& temp = mat[p];