将动态指针初始化为多维数组的正确方法？ [英] The correct way to initialize a dynamic pointer to a multidimensional array?

问题描述

当我将它们定义为2维和更高时，我一直在运用动态指针。例如，我想要一个指向2D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]; 
  

工作，我想像：

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

也可以工作，但不是。此代码指出错误：

 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到我所做的7个整数的数组：

  int（* P）[4] = new int [7] [4] ; 
  

这样做有效，但这不是我想要完成的。通过这样做，我被限制为至少对任何后续维度使用常量值，但是我希望在运行时完全定义它，因此动态。

我如何去做这个多维指针？

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operator new function（）


2. 为4个整数。


3. 返回对此内存的引用。



4. 要绑定此引用，您需要具有相同类型的指针的返回参考，所以你做

     int * P = new int [4]; //当您创建一个整数
    //的数组时，您应该将其分配给一个指针整数
     

对于多维数组，您需要分配一个指针数组，然后用数组的指针填充该数组，如下所示：

  int ** p; 
 p = new int * [5]; （int i = 0; i <5; ++ i）{
p [i] = new int [ 10]; 
 //每个第i个指针现在指向动态数组（大小10）
 //实际int值
} 
  

这是它的样子：

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; // p $; 
     




2. 对于2d Array，

     //需要使用delete []运算符，因为我们使用新的[]运算符
    for（int i = 0; i< ; 5; ++ i）{
    delete [] p [i]; //删除整数的内部数组; 
   } 
    
    delete [] p; //删除指针保持指针数组; 
     

避免内存泄漏和悬挂指针！

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

Let's start 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 bind 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 an array of integer
                        // you should assign it to a pointer-to-integer
   

For a multi-idimensional array, you need to allocate an array of pointers, then fill that array with pointers to arrays, like this:

int **p;
p = new int*[5]; // dynamic `array (size 5) of pointers to int`

for (int i = 0; i < 5; ++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 it looks like:

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 < 5; ++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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