用malloc创建从3D数组创建三重指针以传递到函数，该功能将三重指针作为参数并更新其元素

Question

void effect_array(int ***ptr, int rows, int cols, int depth)
{
    for(int i = 0; i < rows; i++)
    {
        for(int j = 0; j < cols; j++)
        {
            for(int k = 0; k < depth; k++)
            {
                ptr[i][j][k] *= 2;
            }
        }
    }
}

在我的尝试中，我成功地为2D/双重指针做了以下操作：

#include <stdio.h>
#include <stdlib.h>

void effect_array2d(int **ptr, int rows, int cols)
{
    for(int i = 0; i < rows; i++)
    {
        for(int j = 0; j < cols; j++)
        {
            ptr[i][j] *= 2;
        }
    }
}

int main()
{
    int arr[2][2] = {
        {1,2},
        {3,4}
    };
    
    int **ptr = (int **)malloc(2 * sizeof(int *));
    for(int i = 0; i < 2; i++) ptr[i] = arr[i];
    
    printf("Array before pass to effect_array2d:\n");
    for(int i = 0; i < 2; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            printf("%d ", arr[i][j]);
        }
        printf("\n");
    }
    printf("\n");
    
    effect_array2d(ptr, 2, 2);
    
    printf("Array after pass to effect_array2d:\n");
    for(int i = 0; i < 2; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            printf("%d ", arr[i][j]);
        }
        printf("\n");
    }
    printf("\n");
    
    free(ptr);
    
    return 0;
}

给出的输出：

Array before pass to effect_array2d:
1 2 
3 4 

Array after pass to effect_array2d:
2 4 
6 8 

当我尝试将相同的方法用于3D数组和三重指针时，但是

#include <stdio.h>
#include <stdlib.h>

void effect_array(int ***ptr, int rows, int cols, int depth)
{
    for(int i = 0; i < rows; i++)
    {
        for(int j = 0; j < cols; j++)
        {
            for(int k = 0; k < depth; k++)
            {
                ptr[i][j][k] *= 2;
            }
        }
    }
}

int main()
{
    int arr[2][2][2] = {
        {
            {1,2},
            {3,4}
        },
        {
            {5,6},
            {7,8}
        }
    };
    
    int ***ptr = (int ***)malloc(2 * sizeof(int **));
    for(int i = 0; i < 2; i++)
    {
        ptr[i] = arr[i];
    }
    
    free(ptr);

    return 0;
}

我遇到了一个警告，说：

malloc3.c: In function ‘main’:
malloc3.c:34:16: warning: assignment to ‘int **’ from incompatible pointer type ‘int (*)[2]’ [-Wincompatible-pointer-types]
   34 |         ptr[i] = arr[i];
      |                ^

确定如何解决这个问题，或者我做错了什么。所有的帮助都将受到极大的赞赏。

我以前曾经看过这个问题，这是一个很艰难的问题，因为C处理3D阵列的差异与2D阵列有所不同。这是我为解决问题所做的工作：

#include <stdio.h>
#include <stdlib.h>

void effect_array(int ***ptr, int rows, int cols, int depth)
{
    for(int i = 0; i < rows; i++)
    {
        for(int j = 0; j < cols; j++)
        {
            for(int k = 0; k < depth; k++)
            {
                ptr[i][j][k] *= 2;
            }
        }
    }
}

int main()
{
    int arr[2][2][2] = {
        {
            {1, 2},
            {3, 4}
        },
        {
            {5, 6},
            {7, 8}
        }
    };

    // Allocate memory for rows
    int ***ptr = (int ***)malloc(2 * sizeof(int **));

    // Allocate memory for columns in each row
    for(int i = 0; i < 2; i++)
    {
        ptr[i] = (int **)malloc(2 * sizeof(int *));
        for(int j = 0; j < 2; j++)
        {
            // Point to the correct part of the original array
            ptr[i][j] = &arr[i][j][0];
        }
    }

    
    printf("Array before:\n");
    for(int i = 0; i < 2; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            for(int k = 0; k < 2; k++)
            {
                printf("%d ", arr[i][j][k]);
            }
            printf("\n");
        }
        printf("\n");
    }

    
    effect_array(ptr, 2, 2, 2);

    
    printf("Array after:\n");
    for(int i = 0; i < 2; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            for(int k = 0; k < 2; k++)
            {
                printf("%d ", arr[i][j][k]);
            }
            printf("\n");
        }
        printf("\n");
    }

    
    for(int i = 0; i < 2; i++)
    {
        free(ptr[i]);
    }
    free(ptr);

    return 0;
}

Answer 1