初始化指向多维数组的动态指针的正确方法?

The correct way to initialize a dynamic pointer to a multidimensional array?(初始化指向多维数组的动态指针的正确方法?)

问题描述

当我将动态指针范围扩展到 2 维和更高维度时，我对动态指针的处理一直很糟糕.例如，我想要一个指向二维数组的指针.我知道:

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;

完全合法(即使我不完全理解为什么).考虑到:

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

int *P = new int[4];

作品，我想象的:

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

也可以，但不行.此代码说明错误:

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 **"

通过看到这一点，新部分变成了一个指向我创建的 7 个整数数组的指针:

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

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

这确实有效，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一个常量值，但我希望它在运行时完全定义，因此是动态的".

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".

我怎样才能让这个多维指针工作??

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

推荐答案

让我们从一些基本的例子开始.

Let's start with some basic examples.

当你说int *P = new int[4];

1. new int[4]; 调用 operator new function()
2. 为 4 个整数分配内存.
3. 返回对此内存的引用.

4. 要绑定这个引用，你需要有与返回引用相同类型的指针，所以你这样做

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

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

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
}

这是它的样子:

1. 对于一维数组，

1. For one dimensional array,

 // need to use the delete[] operator because we used the new[] operator
delete[] p; //free memory pointed by p;`

// 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;

避免内存泄漏和悬空指针！

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