如何创建多维、动态定义的数组？

Question

我知道如何创建一个单维、动态定义的数组，

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

但是，当我尝试使其成为多维数组时，

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

它不起作用。什么给了？我发现这个http://www.cplusplus.com/forum/beginner/63/ 但我对他们所说的话感到完全困惑。有人愿意解释一下吗？非常感谢。

Answer 1

我试图从您的链接中提供一些解释：

// 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（假设有三个元素）的内存布局看起来像，

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

而二维数组（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，因此你首先需要**名字。

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

现在，您希望该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。