C++ 錯誤代碼：std::bad_alloc 在 memory 位置

Question

我現在正在使用 Visual Studio，我對收到的錯誤消息有疑問。

它說“memory 位置的 std::bad_alloc”。

我想這意味着我沒有足夠的 memory。 我正在使用 arrays 進行計算，我認為問題出在每個數組的大小上。

首先我使用堆棧而不是堆，所以我動態分配每個數組。 但是，我仍然收到相同的錯誤消息。

如果我減小數組的大小，我可以運行此代碼，但我需要將它與更大的 arrays 一起使用。

如果我使用更好的計算機，我可以運行此代碼嗎？ 我不確定如何減少運行此代碼所需的 memory .... 我應該使用更少的 arrays 嗎？

下面，我把整個代碼，以防萬一......

提前致謝。

include <iostream>
#include <cstdio>
#include <chrono>
#include <vector>
#include <math.h>       // power
#include <cmath>        // abs
#include <fstream>
#include <omp.h>

using namespace std;
using namespace chrono;

// Dynamically allocation with values(double)
void dallo_fn(double**** pMat, int Na, int Nd, int Ny) {
    double*** Mat = new double** [Na];
    for (int i = 0; i < Na; i++) {
        Mat[i] = new double* [Nd];
        for (int j = 0; j < Nd; j++) {
            Mat[i][j] = new double[Ny];
            fill_n(Mat[i][j], Ny, 1);
        }
    }
    *pMat = Mat;
}

// Dynamically allocation without values(double)
void dallo_fn0(double**** pMat, int Na, int Nd, int Ny) {
    double*** Mat = new double** [Na];
    for (int i = 0; i < Na; i++) {
        Mat[i] = new double* [Nd];
        for (int j = 0; j < Nd; j++) {
            Mat[i][j] = new double[Ny];
        }
    }
    *pMat = Mat;
}

// Dynamically allocation without values(int)
void dallo_fn1(int**** pMat, int Na, int Nd, int Ny) {
    int*** Mat = new int** [Na];
    for (int i = 0; i < Na; i++) {
        Mat[i] = new int* [Nd];
        for (int j = 0; j < Nd; j++) {
            Mat[i][j] = new int[Ny];
        }
    }
    *pMat = Mat;
}

// Utility function
double utility(double a, double a_f, double d, double d_f, double y, double sig, double psi, double delta, double R) {
    double C;
    C = y + a - a_f / R - (d_f - (1 - delta) * d);
    double result;
    if (C > 0) {
        result = 1 / (1 - 1 / sig) * pow(pow(C, psi) * pow(d_f, 1 - psi), (1 - 1 / sig));
    }
    else {
        result = -999999;
    }
    return result;
}


int main()
{
#if defined _OPENMP
    omp_set_num_threads(8);
#endif

    double duration;

    // Iteration Parameters
    double tol = 0.000001;
    double itmax = 200;
    double H = 15;

    // Model Parameters and utility function
    double sig = 0.75;
    double beta = 0.95;
    double psi = 0.5;
    double delta = 0.1;
    double R = 1 / beta - 0.00215;

    // =============== 2. Discretizing the state space =========================

    // Size of arrays
    const int Na = 2 * 91;
    const int Nd = 1 * 71;
    const int Ny = 3;

    // Variables for discretization of state space
    const double amin = -2;
    const double amax = 7;
    const double dmin = 0.01;
    const double dmax = 7;
    const double ymin = 0.5;
    const double ymax = 1.5;
    const double Ptrans[3] = { 0.2, 0.6, 0.2 };

    // Discretization of state space
    double ca = (amax - amin) / (Na - 1.0);
    double cd = (dmax - dmin) / (Nd - 1.0);
    double cy = (ymax - ymin) / (Ny - 1.0);

    double* A = new double[Na];
    double* Y = new double[Ny];
    double* D = new double[Nd];

    for (int i = 0; i < Na; i++) {
        A[i] = amin + i * ca;
    }
    for (int i = 0; i < Nd; i++) {
        D[i] = dmin + i * cd;
    }
    for (int i = 0; i < Ny; i++) {
        Y[i] = ymin + i * cy;
    }

    // === 3. Initial guesses, Variable initialization and Transition matrix ===

    // Initial guess for value function
    double*** V;
    dallo_fn(&V, Na, Nd, Ny);
    double*** Vnew;
    dallo_fn(&Vnew, Na, Nd, Ny);

    // Initialization of other variables
    double Val[Na][Nd];
    double Vfuture[Na][Nd];
    double** temphoward = new double* [Na];
    for (int i = 0; i < Na; i++)
    {
        temphoward[i] = new double[Nd];
    }
    
    double*** Vhoward;
    dallo_fn0(&Vhoward, Na, Nd, Ny);
    double*** tempdiff;
    dallo_fn0(&tempdiff, Na, Nd, Ny);
    int*** maxposition_a;
    dallo_fn1(&maxposition_a, Na, Nd, Ny);
    int*** maxposition_d;
    dallo_fn1(&maxposition_d, Na, Nd, Ny);

    double** mg_A_v = new double* [Na];
    for (int i = 0; i < Na; i++)
    {
        mg_A_v[i] = new double[Nd];
    }
    for (int j = 0; j < Nd; j++) {
        for (int i = 0; i < Na; i++) {
            mg_A_v[i][j] = A[i];
        }
    }

    double** mg_D_v = new double* [Na];
    for (int i = 0; i < Na; i++)
    {
        mg_D_v[i] = new double[Nd];
    }
    for (int j = 0; j < Nd; j++) {
        for (int i = 0; i < Na; i++) {
            mg_D_v[i][j] = D[j];
        }
    }

    double***** Uvec = new double**** [Na];
    for (int i = 0; i < Na; i++) {
        Uvec[i] = new double*** [Nd];
        for (int j = 0; j < Nd; j++) {
            Uvec[i][j] = new double** [Ny];
            for (int k = 0; k < Ny; k++) {
                Uvec[i][j][k] = new double* [Na];
                for (int l = 0; l < Na; l++) {
                    Uvec[i][j][k][l] = new double[Nd];
                }
            }
        }
    }

    for (int i = 0; i < Na; i++) {
        for (int j = 0; j < Nd; j++) {
            for (int k = 0; k < Ny; k++) {
                for (int l = 0; l < Na; l++) {
                    for (int m = 0; m < Nd; m++) {
                        Uvec[i][j][k][l][m] = utility(A[i], mg_A_v[l][m], D[j], mg_D_v[l][m], Y[k], sig, psi, delta, R);
                    }
                }
            }
        }
    }

    // Value function iteration
    int it;
    double dif;
    double max;
    it = 0;
    dif = 1;

    // ================ 4. Value function iteration ============================

    while (dif >= tol && it <= itmax) {
        system_clock::time_point start = system_clock::now();
        it = it + 1;
        // V = Vnew;
        for (int i = 0; i < Na; i++) {
            for (int j = 0; j < Nd; j++) {
                for (int k = 0; k < Ny; k++) {
                    V[i][j][k] = Vnew[i][j][k];
                }
            }
        }

        for (int i = 0; i < Na; i++) {
            for (int j = 0; j < Nd; j++) {
                Vfuture[i][j] = 0;
                for (int k = 0; k < Ny; k++) {
                    Vfuture[i][j] += beta * Ptrans[k] * Vnew[i][j][k]; // + beta * Ptrans[1] * Vnew[i][j][1] + beta * Ptrans[2] * Vnew[i][j][2]; // Why is this different from Vfuture[i][j] += beta * Vnew[i][j][k] * Ptrans[k]; with for k
                }
            }
        }
        #pragma omp parallel for private(Val)     
        for (int a = 0; a < Na; a++) {
            for (int b = 0; b < Nd; b++) {
                for (int c = 0; c < Ny; c++) {
                    max = -99999;
                    for (int d = 0; d < Na; d++) {
                        for (int e = 0; e < Nd; e++) {
                            Val[d][e] = Uvec[a][b][c][d][e] + Vfuture[d][e];
                            if (max < Val[d][e]) {
                                max = Val[d][e];
                                maxposition_a[a][b][c] = d;
                                maxposition_d[a][b][c] = e;
                            }
                        }
                    }
                    Vnew[a][b][c] = max;
                }
            }
        }

        // Howard improvement
        for (int h = 0; h < H; h++) {
            for (int i = 0; i < Na; i++) {
                for (int j = 0; j < Nd; j++) {
                    for (int k = 0; k < Ny; k++) {
                        Vhoward[i][j][k] = Vnew[i][j][k];
                    }
                }
            }

            for (int i = 0; i < Na; i++) {
                for (int j = 0; j < Nd; j++) {
                    for (int k = 0; k < Ny; k++) {
                        temphoward[i][j] = beta * Vhoward[maxposition_a[i][j][k]][maxposition_d[i][j][k]][0] * Ptrans[0]
                            + beta * Vhoward[maxposition_a[i][j][k]][maxposition_d[i][j][k]][1] * Ptrans[1]
                            + beta * Vhoward[maxposition_a[i][j][k]][maxposition_d[i][j][k]][2] * Ptrans[2];
                        Vnew[i][j][k] = temphoward[i][j] + Uvec[i][j][k][maxposition_a[i][j][k]][maxposition_d[i][j][k]];
                    }
                }
            }
        }


        // Calculate Diff
        dif = -100000;
        for (int i = 0; i < Na; i++) {
            for (int j = 0; j < Nd; j++) {
                for (int k = 0; k < Ny; k++) {
                    tempdiff[i][j][k] = abs(V[i][j][k] - Vnew[i][j][k]);
                    if (tempdiff[i][j][k] > dif) {
                        dif = tempdiff[i][j][k];
                    }
                }
            }
        }

        system_clock::time_point end = system_clock::now();
        std::chrono::duration<double> sec = end - start;


        cout << dif << endl;
        cout << it << endl;
        cout << sec.count() << endl;
    }

    for (int k = 0; k < Ny; k++) {
        for (int i = 0; i < Na; i++) {
            for (int j = 0; j < Nd; j++) {
                cout << Vnew[i][j][k];
            }
            cout << '\n';
        }
    }
    cout << omp_get_max_threads() << endl;

}

Answer 1

誠然，代碼很漂亮 memory 餓了，但適用於我的 8GB 機器。 請確保您的目標是 x64 平台，否則您會達到 4GB 的固有限制。

Answer 2

是的，只要有足夠的 memory，您就可以運行此代碼。 我剛剛在我的機器上用 64GB 的 RAM 測試過它，它運行得很好。

減少所需 memory 數量的一種方法是使用float s 而不是double s，因為它們占用了一半的大小。

此外，正如其他人在評論中所建議的那樣，您可以將多維結構表示為 1d arrays 並計算數組中的索引，而不是實際嵌套 arrays。 好處是你擺脫了大量的指針。