下面的例子可能会有所帮助，

int main(void)
{
    double **a2d = new double*[5]; 
    /* initializing Number of rows, in this case 5 rows) */
    for (int i = 0; i < 5; i++)
    {
        a2d[i] = new double[3]; /* initializing Number of columns, in this case 3 columns */
    }

    for (int i = 0; i < 5; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            a2d[i][j] = 1; /* Assigning value 1 to all elements */
        }
    }

    for (int i = 0; i < 5; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            cout << a2d[i][j] << endl;  /* Printing all elements to verify all elements have been correctly assigned or not */
        }
    }

    for (int i = 0; i < 5; i++)
        delete[] a2d[i];

    delete[] a2d;


    return 0;
}

这个问题困扰了我15年，所有的解决方案都不能让我满意。如何在内存中连续创建动态多维数组?今天我终于找到了答案。使用下面的代码，你可以做到这一点:

#include <iostream>

int main(int argc, char** argv)
{
    if (argc != 3)
    {
        std::cerr << "You have to specify the two array dimensions" << std::endl;
        return -1;
    }

    int sizeX, sizeY;

    sizeX = std::stoi(argv[1]);
    sizeY = std::stoi(argv[2]);

    if (sizeX <= 0)
    {
        std::cerr << "Invalid dimension x" << std::endl;
        return -1;
    }
    if (sizeY <= 0)
    {
        std::cerr << "Invalid dimension y" << std::endl;
        return -1;
    }

    /******** Create a two dimensional dynamic array in continuous memory ******
     *
     * - Define the pointer holding the array
     * - Allocate memory for the array (linear)
     * - Allocate memory for the pointers inside the array
     * - Assign the pointers inside the array the corresponding addresses
     *   in the linear array
     **************************************************************************/

    // The resulting array
    unsigned int** array2d;

    // Linear memory allocation
    unsigned int* temp = new unsigned int[sizeX * sizeY];

    // These are the important steps:
    // Allocate the pointers inside the array,
    // which will be used to index the linear memory
    array2d = new unsigned int*[sizeY];

    // Let the pointers inside the array point to the correct memory addresses
    for (int i = 0; i < sizeY; ++i)
    {
        array2d[i] = (temp + i * sizeX);
    }



    // Fill the array with ascending numbers
    for (int y = 0; y < sizeY; ++y)
    {
        for (int x = 0; x < sizeX; ++x)
        {
            array2d[y][x] = x + y * sizeX;
        }
    }



    // Code for testing
    // Print the addresses
    for (int y = 0; y < sizeY; ++y)
    {
        for (int x = 0; x < sizeX; ++x)
        {
            std::cout << std::hex << &(array2d[y][x]) << ' ';
        }
    }
    std::cout << "\n\n";

    // Print the array
    for (int y = 0; y < sizeY; ++y)
    {
        std::cout << std::hex << &(array2d[y][0]) << std::dec;
        std::cout << ": ";
        for (int x = 0; x < sizeX; ++x)
        {
            std::cout << array2d[y][x] << ' ';
        }
        std::cout << std::endl;
    }



    // Free memory
    delete[] array2d[0];
    delete[] array2d;
    array2d = nullptr;

    return 0;
}

当你调用sizeX=20和sizeey =15值的程序时，输出将如下:

0x603010 0x603014 0x603018 0x60301c 0x603020 0x603024 0x603028 0x60302c 0x603030 0x603034 0x603038 0x60303c 0x603040 0x603044 0x603048 0x60304c 0x603050 0x603054 0x603058 0x60305c 0x603060 0x603064 0x603068 0x60306c 0x603070 0x603074 0x603078 0x60307c 0x603080 0x603084 0x603088 0x60308c 0x603090 0x603094 0x603098 0x60309c 0x6030a0 0x6030a4 0x6030a8 0x6030ac 0x6030b0 0x6030b4 0x6030b8 0x6030bc 0x6030c0 0x6030c4 0x6030c8 0x6030cc 0x6030d0 0x6030d4 0x6030d8 0x6030dc 0x6030e0 0x6030e4 0x6030e8 0x6030ec 0x6030f0 0x6030f4 0x6030f8 0x6030fc 0x603100 0x603104 0x603108 0x60310c 0x603110 0x603114 0x603118 0x60311c 0x603120 0x603124 0x603128 0x60312c 0x603130 0x603134 0x603138 0x60313c 0x603140 0x603144 0x603148 0x60314c 0x603150 0x603154 0x603158 0x60315c 0x603160 0x603164 0x603168 0x60316c 0x603170 0x603174 0x603178 0x60317c 0x603180 0x603184 0x603188 0x60318c 0x603190 0x603194 0x603198 0x60319c 0x6031a0 0x6031a4 0x6031a8 0x6031ac 0x6031b0 0x6031b4 0x6031b8 0x6031bc 0x6031c0 0x6031c4 0x6031c8 0x6031cc 0x6031d0 0x6031d4 0x6031d8 0x6031dc 0x6031e0 0x6031e4 0x6031e8 0x6031ec 0x6031f0 0x6031f4 0x6031f8 0x6031fc 0x603200 0x603204 0x603208 0x60320c 0x603210 0x603214 0x603218 0x60321c 0x603220 0x603224 0x603228 0x60322c 0x603230 0x603234 0x603238 0x60323c 0x603240 0x603244 0x603248 0x60324c 0x603250 0x603254 0x603258 0x60325c 0x603260 0x603264 0x603268 0x60326c 0x603270 0x603274 0x603278 0x60327c 0x603280 0x603284 0x603288 0x60328c 0x603290 0x603294 0x603298 0x60329c 0x6032a0 0x6032a4 0x6032a8 0x6032ac 0x6032b0 0x6032b4 0x6032b8 0x6032bc 0x6032c0 0x6032c4 0x6032c8 0x6032cc 0x6032d0 0x6032d4 0x6032d8 0x6032dc 0x6032e0 0x6032e4 0x6032e8 0x6032ec 0x6032f0 0x6032f4 0x6032f8 0x6032fc 0x603300 0x603304 0x603308 0x60330c 0x603310 0x603314 0x603318 0x60331c 0x603320 0x603324 0x603328 0x60332c 0x603330 0x603334 0x603338 0x60333c 0x603340 0x603344 0x603348 0x60334c 0x603350 0x603354 0x603358 0x60335c 0x603360 0x603364 0x603368 0x60336c 0x603370 0x603374 0x603378 0x60337c 0x603380 0x603384 0x603388 0x60338c 0x603390 0x603394 0x603398 0x60339c 0x6033a0 0x6033a4 0x6033a8 0x6033ac 0x6033b0 0x6033b4 0x6033b8 0x6033bc 0x6033c0 0x6033c4 0x6033c8 0x6033cc 0x6033d0 0x6033d4 0x6033d8 0x6033dc 0x6033e0 0x6033e4 0x6033e8 0x6033ec 0x6033f0 0x6033f4 0x6033f8 0x6033fc 0x603400 0x603404 0x603408 0x60340c 0x603410 0x603414 0x603418 0x60341c 0x603420 0x603424 0x603428 0x60342c 0x603430 0x603434 0x603438 0x60343c 0x603440 0x603444 0x603448 0x60344c 0x603450 0x603454 0x603458 0x60345c 0x603460 0x603464 0x603468 0x60346c 0x603470 0x603474 0x603478 0x60347c 0x603480 0x603484 0x603488 0x60348c 0x603490 0x603494 0x603498 0x60349c 0x6034a0 0x6034a4 0x6034a8 0x6034ac 0x6034b0 0x6034b4 0x6034b8 0x6034bc 

0x603010: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 
0x603060: 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 
0x6030b0: 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 
0x603100: 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 
0x603150: 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 
0x6031a0: 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 
0x6031f0: 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 
0x603240: 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 
0x603290: 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 
0x6032e0: 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 
0x603330: 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 
0x603380: 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 
0x6033d0: 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 
0x603420: 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 
0x603470: 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299

可以看到，多维数组在内存中是连续的，没有两个内存地址是重叠的。甚至释放数组的例程也比为每一列(或行，取决于如何查看数组)动态分配内存的标准方法简单。由于数组基本上由两个线性数组组成，只有这两个必须(并且可以)被释放。

这种方法可以用相同的概念扩展到二维以上的空间。我在这里不做了，但是当你理解了它背后的思想，它是一个简单的任务。

我希望这段代码能像帮助我一样帮助你。

动态声明2D数组:

    #include<iostream>
    using namespace std;
    int main()
    {
        int x = 3, y = 3;

        int **ptr = new int *[x];

        for(int i = 0; i<y; i++)
        {
            ptr[i] = new int[y];
        }
        srand(time(0));

        for(int j = 0; j<x; j++)
        {
            for(int k = 0; k<y; k++)
            {
                int a = rand()%10;
                ptr[j][k] = a;
                cout<<ptr[j][k]<<" ";
            }
            cout<<endl;
        }
    }

现在，在上面的代码中，我们获取了一个双指针，并为它分配了一个动态内存，并给出了列的值。这里分配的内存仅用于列，现在对于行，我们只需要一个for循环，并为每一行分配一个动态内存。现在我们可以像使用2D数组一样使用指针。在上面的例子中，我们将随机数分配给我们的2D数组(指针)。这都是关于二维数组的DMA。

如果你想声明一个预定义的指针数组:

int **x;
x = new int*[2] { 
        new int[2] { 0, 1 },
        new int[2] { 2, 3 } 
    };

访问:

cout << x[0][0];

尽管这个流行的答案将为您提供所需的索引语法，但它的效率是双重的:在空间和时间上都大而慢。有更好的办法。

为什么答案又大又慢

建议的解决方案是创建一个指针的动态数组，然后将每个指针初始化到它自己的独立动态数组。这种方法的优点是它提供了你习惯的索引语法，所以如果你想找到矩阵在x,y位置的值，你说:

int val = matrix[ x ][ y ];

这是因为矩阵[x]返回一个指向数组的指针，然后用[y]作为索引。分解一下:

int* row = matrix[ x ];
int  val = row[ y ];

方便,是吗?我们喜欢[x][y]语法。

但是这个解决方案有一个很大的缺点，那就是它既胖又慢。

Why?

The reason that it's both fat and slow is actually the same. Each "row" in the matrix is a separately allocated dynamic array. Making a heap allocation is expensive both in time and space. The allocator takes time to make the allocation, sometimes running O(n) algorithms to do it. And the allocator "pads" each of your row arrays with extra bytes for bookkeeping and alignment. That extra space costs...well...extra space. The deallocator will also take extra time when you go to deallocate the matrix, painstakingly free-ing up each individual row allocation. Gets me in a sweat just thinking about it.

它慢还有另一个原因。这些单独的分配往往位于内存的不连续部分。一行的地址可能是1000，另一行的地址可能是100000——你可以理解。这意味着当你在穿越矩阵时，你就像一个狂野的人一样在记忆中跳跃。这往往会导致缓存丢失，从而大大降低处理时间。

所以，如果你绝对必须有你可爱的[x][y]索引语法，使用这个解决方案。如果你想要快速和小巧(如果你不关心这些，为什么要用c++ ?)，你需要一个不同的解决方案。

不同的解决方案

更好的解决方案是将整个矩阵分配为单个动态数组，然后使用自己的(稍微)聪明的索引数学来访问单元格。索引的数学运算非常巧妙;不，这一点也不聪明:这是显而易见的。

class Matrix
{
    ...
    size_t index( int x, int y ) const { return x + m_width * y; }
};

给定这个index()函数(我想象它是一个类的成员，因为它需要知道矩阵的m_width)，您可以访问矩阵数组中的单元格。矩阵数组是这样分配的:

array = new int[ width * height ];

所以在缓慢的，高脂肪的溶液中

array[ x ][ y ]

.．.这是一个快速，小的解决方案:

array[ index( x, y )]

很难过，我知道。但你会习惯的。你的CPU会感谢你的。

Typedef是你的朋友

在回顾并查看了许多其他答案之后，我发现需要进行更深层次的解释，因为许多其他答案要么存在性能问题，要么迫使您使用不寻常的或繁重的语法来声明数组，或访问数组元素(或以上所有问题)。

首先，这个答案假设您在编译时知道数组的尺寸。如果你这样做，那么这是最好的解决方案，因为它将提供最好的性能，并允许您使用标准数组语法来访问数组元素。

The reason this gives the best performance is because it allocates all of the arrays as a contiguous block of memory meaning that you are likely to have less page misses and better spacial locality. Allocating in a loop may cause the individual arrays to end up scattered on multiple non-contiguous pages through the virtual memory space as the allocation loop could be interrupted ( possibly multiple times ) by other threads or processes, or simply due to the discretion of the allocator filling in small, empty memory blocks it happens to have available.

其他好处是声明语法简单，数组访问语法标准。

在c++中使用new:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char **argv) {

typedef double (array5k_t)[5000];

array5k_t *array5k = new array5k_t[5000];

array5k[4999][4999] = 10;
printf("array5k[4999][4999] == %f\n", array5k[4999][4999]);

return 0;
}

或使用calloc的C样式:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char **argv) {

typedef double (*array5k_t)[5000];

array5k_t array5k = calloc(5000, sizeof(double)*5000);

array5k[4999][4999] = 10;
printf("array5k[4999][4999] == %f\n", array5k[4999][4999]);

return 0;
}