如何创建多维、动态定义的数组？

Question

我知道如何创建一维动态定义的数组

 string *names = new string[number]; //number specified by the user

然而，当我试图让它成为多维的时候

 string *names = new string[number][number]; //doesn't work

它不起作用。是什么赋予了？我找到了这个http://www.cplusplus.com/forum/beginner/63/，但我完全被他们所说的弄糊涂了。有人愿意解释吗？非常感谢。

Answer 1

我试图从您的链接中为示例提供一些解释：

// To dynamically allocate two-dimensional array we will allocate array of pointers to int first. 
// Each pointer will represent a row in your matrix.
// Next step we allocate enough memory for each row and assign this memory for row pointers.

const int rows = 4; //number of rows in your matrix
const int cols = 4; //number of columns in your matrix

// declaration
int ** a; 

/* allocating memory for pointers to rows. Each row will be a dynamically allocated array of int */
a = new int*[rows];
/* for each row you allocate enough memory to contain cols elements. cols - number of columns*/ 
for(int i = 0; i < rows; i++)
   a[i] = new int[cols];

Answer 2

一维数组（比如说三个元素）的内存布局是这样的，

names ------> [0] [1] [2] 

二维数组（假设有 3 X 3 个元素）看起来像，

names ------> [0] --> [0] [1] [2] 
              [1] --> [0] [1] [2]
              [2] --> [0] [1] [2]
               ^
               |
           this is an array of pointers

即二维数组是指向数组的指针数组，因此您首先需要**名称。

string **names = new string*[number]; // allocating space for array of string pointers

现在，您希望这个字符串指针数组的每个元素都指向一个字符串数组。

因此，您需要这样做

for(int i = 0; i < number, i++) {
   names[i] = new string[number];
}

我希望这有助于更好地理解它。