请问为什么用new建立动态二维数组就会报错，用malloc就不会报错。

Question

请问为什么用注释掉的部分建立动态二维数组就会报错，用malloc就不会报错？

void Matrix()
{
    int i;
    //double** P_ca_x = new double* [Times];
    //double** P_ca_y = new double* [Times];
    //double** P_ca_z = new double* [Times];
    //double** v_ca_x = new double* [Times];
    //double** v_ca_y = new double* [Times];
    //double** v_ca_z = new double* [Times];
    //double** v_ca = new double* [Times];
    P_ca_x = (double**)malloc(sizeof(double*) * Times);
    P_ca_y = (double**)malloc(sizeof(double*) * Times);
    P_ca_z = (double**)malloc(sizeof(double*) * Times);
    v_ca_x = (double**)malloc(sizeof(double*) * Times);
    v_ca_y = (double**)malloc(sizeof(double*) * Times);
    v_ca_z = (double**)malloc(sizeof(double*) * Times);
    v_ca = (double**)malloc(sizeof(double*) * Times);
    for (i = 0; i < Times; i++)
    {
        //P_ca_x[i] = new double[Num];
        //P_ca_y[i] = new double[Num];
        //P_ca_z[i] = new double[Num];
        //v_ca_x[i] = new double[Num];
        //v_ca_y[i] = new double[Num];
        //v_ca_z[i] = new double[Num];
        //v_ca[i] = new double[Num];
        P_ca_x[i] = (double*)malloc(sizeof(double) * Num);
        P_ca_y[i] = (double*)malloc(sizeof(double) * Num);
        P_ca_z[i] = (double*)malloc(sizeof(double) * Num);
        v_ca_x[i] = (double*)malloc(sizeof(double) * Num);
        v_ca_y[i] = (double*)malloc(sizeof(double) * Num);
        v_ca_z[i] = (double*)malloc(sizeof(double) * Num);
        v_ca[i] = (double*)malloc(sizeof(double) * Num);
    }
}
void Delete()
{
    int i;
    for (i = 0; i < Times; i++)
    {
        //delete[] P_ca_x[i];
        //delete[] P_ca_y[i];
        //delete[] P_ca_z[i];
        //delete[] v_ca_x[i];
        //delete[] v_ca_y[i];
        //delete[] v_ca_z[i];
        //delete[] v_ca[i];
        free(P_ca_x[i]);
        free(P_ca_y[i]);
        free(P_ca_z[i]);
        free(v_ca_x[i]);
        free(v_ca_y[i]);
        free(v_ca_z[i]);
        free(v_ca[i]);
    }
    //delete[] P_ca_x;
    //delete[] P_ca_y;
    //delete[] P_ca_z;
    //delete[] v_ca_x;
    //delete[] v_ca_y;
    //delete[] v_ca_z;
    //delete[] v_ca;
    free(P_ca_x);
    free(P_ca_y);
    free(P_ca_z);
    free(v_ca_x);
    free(v_ca_y);
    free(v_ca_z);
    free(v_ca);
}

在对数组赋值时报错，用malloc则不会报错

Answer 1

不会报错把，麻烦把错误信息贴一下。不应该报错的

你先把下面的程序编译一下，如果没报错就不是new的问题

#include <iostream>
#include <string.h>

using namespace std;

int main()
{
    //申请
    double ** a = new double*[5];
    for(int i = 0; i<5; ++i)
    {
        a[i] = new double[6];
        memset(a[i], 0, sizeof(double)*6);
    }

    //赋值
    for(int i = 0; i<5; ++i)
        for(int j = 0; j < 6; ++j)
            a[i][j] = i*j*1.3;

    //输出
    for(int i = 0; i<5; ++i)
    {
        for(int j = 0; j < 6; ++j)
            cout<<a[i][j]<<" ";
        cout<<endl;
    }

    //删除
    for(int i = 0; i<5; ++i)
    {
        delete[] a[i]; 
    }
    delete[] a;
}