正确的方式来初始化一个多维数组的动态指针？ [英] The correct way to initialize a dynamic pointer to a multidimensional array?

问题描述

当我将它们扩展到2个维度和更高的时候，我一直在使用动态指针。例如我想要一个指向二维数组的指针。我知道：

  int A [3] [4]; 
 int（* P）[4] = A; 
  

完全合法（即使我不完全明白为什么）。考虑到：

  int * P = new int [4] 
  

有效，我想象：

  int ** P = new int [5] [7]; 
  

也可以工作，但不是。此代码说明错误：

 错误：类型（*）[7]的值不能用于初始化实体
类型int **
  

通过看到这个新部分成为一个指针我创建了一个7个整数的数组：

  int（* P）[4] = new int [7] ; 
  

这样可以工作，但这不是我想要完成的。通过这样做，我被限制为至少为任何后续维度使用一个常量值，但我希望它在运行时被完全定义，因此是动态的。

解决方案

让我们从一些基本的例子开始

当你说 int * P = new int [4];

1. new int [4]; 调用运算符新函数（）


2. 整数。


3. 返回对此内存的引用。


4. 要捕获此引用，您需要具有与返回引用相同类型的指针， do

int * P = new int [4]; //你创建的数组的整数你应该抓住它整数指针
添加在 - ：

但是当你想创建一个多维数组时，应该这样做：

  int * * p; 
 p = new int * [4]; （int i = 0; i <4; ++ i）{
p [i] = new int [int]指向int类型的指针的动态数组10]; 
 //每个第i个指针现在指向实际int值的动态数组（大小10）
} 
  

这里是试图描述

释放内存

1. 对于一维数组，
   //需要使用delete []运算符，因为我们使用新的[]运算符
   delete [] p; //由p指向的空闲内存;

   


2. 对于2d数组，
   //需要使用delete [因为我们使用新的[]运算符
   for（int i = 0; i <4; ++ i）{
delete [] p [i]删除整数的内部数组;
}

   
   
   

   delete [] p; // delete pointer holding array of pointers;




避免<悬挂指针！

I've been having bad luck with with dynamic pointers when I range them to 2 dimensions and higher. For example I want a pointer to a 2D array. I know that:

int A[3][4];
int (*P)[4] = A;

Is completely legit (even if I don't completely understand why). Taking into consideration that:

int *P = new int[4];

works, I imagined that:

int **P = new int[5][7];

Would also work, but it's not. This code states the error:

Error: A value of type "(*)[7]" cannot be used to initialize an entity of
       type "int **"

By seeing this the new part becomes a pointer to an array of 7 integers I made:

int (*P)[4] = new int[7][4];

And this does work but it's not what I want to accomplish. By doing it like that I'm limited to at least using a constant value for any subsequent dimension, but I want it to be fully defined at run time and therefore "dynamic".

How could I go and make this multidimensional pointer work??

解决方案

Lets starts with some basic examples

When you say int *P = new int[4];

1. new int[4]; calls operator new function()
2. allocates a memory for 4 integers.
3. returns a reference to this memory.
4. to catch this reference you need to have same type of pointer as that of return reference so you do

int *P = new int[4]; //as you created array of integer you should catch the it integer pointer Add On -:

But when you want to create a multidimensional array you should do,

int **p;
p = new int*[4]; // dynamic `array (size 4) of pointers to int`

for (int i = 0; i < 4; ++i) {
  p[i] = new int[10];
  // each i-th pointer is now pointing to dynamic array (size 10) of actual int values
}

Here is what am trying to depict

To free the memory

1. For one dimensional array, // need to use the delete[] operator because we used the new[] operator delete[] p; //free memory pointed by p;

2. For 2d Array, // need to use the delete[] operator because we used the new[] operator for(int i = 0; i < 4; ++i){ delete[] p[i];//deletes an inner array of integer; }

   delete[] p; //delete pointer holding array of pointers;

Avoid memory leakage and dangling pointers!

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