Question

运行位于邮件末尾的代码后收到以下错误消息：

在抛出＆＃39; std :: out_of_range＆＃39;的实例后终止调用 what（）：vector :: _ M_range_check：__n（为0）＆gt; = this-＆gt; size（）（为0）

此应用程序已请求Runtime以不寻常的方式终止它。请联系应用程序的支持团队以获取更多信息。

我很抱歉代码的长度。似乎错误来自于我在f函数中调用numerov函数时。如果您能够确定错误是什么，请告诉我？谢谢！

#include <iostream>
#include <cmath>
#include <fstream>
#include <vector>

using namespace std;

int nx = 500, m = 10, ni = 10;
double x1 = 0, x2 = 1, h = (x2 - x1)/nx;
int nr, nl;
vector<double> ul, q, u;


//Method to achieve the evenly spaced Simpson rule
double simpson(vector <double> y, double h)
{
    int n = y.size() - 1;
    double s0 = 0, s1 = 0, s2 = 0;
    for (int i = 1; i < n; i += 2)
    {
        s0 += y.at(i);
        s1 += y.at(i-1);
        s2 += y.at(i+1);
    }
    double s = (s1 + 4*s0 + s2)/3;

//Add the last slice separately for an even n+1
if ((n+1)%2 == 0)
    return h*(s + (5*y.at(n) + 8*y.at(n-1) - y.at(n-2))/12);
else
    return h*2;
}

//Method to perform the Numerov integration
vector <double> numerov(int m, double h, double u0, double u1, double q)
{
    vector<double> u;
    u.push_back(u0);
    u.push_back(u1);
    double g = h*h/12;
    for (int i = 1; i < m+1; i++)
    {
        double c0 = 1 + g*q;
        double c1 = 2 - 10*g*q;
        double c2 = 1 + g*q;
        double d = g*(0);
        u.push_back((c1*u.at(i) - c0*u.at(i-1) + d)/c2);
    }
    return u;
}

//Method to provide the function for the root search
double f(double x)
{
    vector<double> w;
    vector<double> j = numerov(nx + 1, h, 0.0, 0.001, x);
    for (int i = 0; i < 0; i++)
    {
        w.push_back(j.at(i));
    }
    return w.at(0);
}

//Method to carry out the secant search
double secant(int n, double del, double x, double dx)
{
    int k = 0;
    double x1 = x + dx;
    while ((abs(dx) > del) && (k < n))
    {
        double d = f(x1) - f(x);
        double x2 = x1 - f(x1)*(x1 - x)/d;
        x = x1;
        x1 = x2;
        dx = x1 - x;
        k++;
    }
    if (k == n)
        cout << "Convergence not found after " << n << " iterations." <<                 endl;
    return x1;
}


int main()
{
    double del = 1e-6, e = 0, de = 0.1;

    //Find the eigenvalue via the secant method
    e = secant (ni, del, e, de);

    //Find the solution u(x)
    u = numerov(nx + 1, h, 0.0, 0.01, e);

    //Output the wavefunction to a file
    ofstream myfile ("Problem 2.txt");
    if (myfile.is_open())
    {
        myfile << "Input" << "\t" << "u(x)" << endl;
        double x = x1;
        double mh = m*h;
        for (int i = 0; i <= nx; i += m)
        {
            myfile << x << "\t" << u.at(i) << endl;
            x += mh;
        }
        myfile.close();
    }

    return 0;
}

Answer 1

vector<double> w;
for (int i = 0; i < 0; i++)
{
    w.push_back(j.at(i));
}
return w.at(0);

w将不包含任何内容，因为该循环将运行0次。因此，w.at(0)将抛出超出范围的错误。

Answer 2

为什么您认为问题出在numerov函数中？

我在函数f中看到错误？

vector<double> w;
vector<double> j = numerov(nx + 1, h, 0.0, 0.001, x);
for (int i = 0; i < 0; i++)
{
    w.push_back(j.at(i));
}
return w.at(0);

向量w上没有任何内容，您尝试访问元素0。

＆＃39;的std :: out_of_range＆＃39;代码错误

2 个答案: