c++程序调试了好久总是出错，好像是指针的问题

Question

    初学，想实现n阶线性方程组求解。从txt文件中读入A矩阵和b，求解后将结果输出到另一个txt文件。
    调试过程很奇怪，有时不会有问题，有时会报错。我单步执行时会在main函数return时出现断点，百度了一下，好像跟内存溢出什么的有关，没能看得很明白。请问是哪里出了问题。

 #include <iostream>
#include <fstream>
#include <string>
#include <stdlib.h>
using namespace std;

class equation
{
public:
    int n;
    double ** A = new double *[n];          //系数阵
    double *b = new double[n];              //常数项
    double *x = new double[n];
    void solve();
    void DisplayEquation();
    void DisplayResult();
    void input();
    void output();
    equation()
    {
        n = 0;
    }
    equation(int a)
    {
        n = a;
    }
};

int CountLines()                                                    //统计系数阵阶数
{
    int n = 0;
    string temp;
    fstream f;
    f.open("equation.txt", ios::in);
    if (f.fail())
    {
        cout << "打开文件失败\n";
    }
    else
    {
        while (getline(f, temp))
        {
            if (temp[0] == '#')
                break;
            n++;
        }
    }
    f.close();
    return n;
}

void equation::input()
{
    int t, i = 0, j = 0;
    string temp;
    fstream f;
    f.open("equation.txt", ios::in);
    while (getline(f, temp, '   '))                     //从文件读入系数矩阵到A中
    {
        t = temp.find('\n');
        if (t > -1)
            temp.erase(t, sizeof('\n'));
        if (temp[0] == '#')
            break;
        else
        {
            if (j == n - 1)
            {
                A[i][j] = atof(temp.c_str());
                j = 0;
                i++;
            }
            else
            {
                A[i][j] = atof(temp.c_str());
                j++;
            }
        }
    }
    i = 0;
    while (getline(f, temp, '   ') && !f.eof())         //从文件读入常数项到b中
    {
        t = temp.find('\n');
        if (t > -1)
            temp.erase(t, sizeof('\n'));
        b[i] = atof(temp.c_str());
        i++;
    }
//  f.close();  //这里加这东西居然会报错
}

void equation::DisplayEquation()                            //显示系数矩阵与常数向量
{
    cout << "系数矩阵为:\n";
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        {
            if (j == n - 1)
                cout << A[i][j] << "\t\n";
            else
                cout << A[i][j] << "\t";
        }
    cout << "\n常数项为:\n";

    for (int i = 0; i < n; i++)
        cout << b[i] << "\n";
}

void equation::solve()                                              //求解方程，将解存入x中
{
    double c;
    //================================forward elimination==================================
    for (int k = 0; k < n; k++)
    {
        for (int i = k + 1; i < n; i++)
        {
            c = A[i][k] / A[k][k];
            for (int j = k; j < n; j++)
                A[i][j] = A[i][j] - A[k][j] * c;
            b[i] = b[i] - b[k] * c;
        }
    }
    //==========================back substitution==========================================
    double sum;
    x[n - 1] = b[n - 1] / A[n - 1][n - 1];
    for (int i = n - 2; i >= 0; i--)
    {
        sum = 0;
        for (int j = n - 1; j >= i + 1; j--)
            sum = sum + A[i][j] * x[j];
        x[i] = (b[i] - sum) / A[i][i];
    }
}

void equation::DisplayResult()
{
    cout << "方程的解为：\n";
    for (int i = 0; i < n; i++)
        cout << x[i] << "   ";
    cout << endl;
}

void equation::output()
{
    fstream r;
    r.open("result.txt", ios::out);
    if (r.fail())
    {
        cout << "打开文件失败\n";
        exit(1);
    }
    for (int i = 0; i < n; i++)
    {
        r << "X" << i << "=" << x[i] << "\n";
    }
    r.close();
}

int main()
{
    int N, j = 0;
    string temp, a;
    N = CountLines();
    equation e1(N);
    for (int i = 0; i < N; i++)
        e1.A[i] = new double[N];
    e1.input();
    //  e1.DisplayEquation();
    e1.solve();
//  e1.DisplayEquation();                               //上三角矩阵
//  e1.DisplayResult();
    e1.output();
    return 0;
}

刚注册的账号，没有C币。。。

Answer 1

class equation
{
public:
int n;
double ** A = new double *[n]; //系数阵
double *b = new double[n]; //常数项
double *x = new double[n];
void solve();
void DisplayEquation();
void DisplayResult();
void input();
void output();
equation()
{
n = 0;
}
equation(int a)
{
n = a;
}
};

两点问题：
1）构造函数equation(){n=0;}会生成一个长度为0的数组，编译不会抛错，也会返回一个正常的数组指针，但是如果修改了指针会
导致内存被破坏，恰巧这块内存正在使用，就会抛出异常，也就会崩！
2）建议申请的内存最好在析构函数内释放掉。。。