通过将指针传递给 c 中的函数来创建二维数组

Question

因此，我阅读了数十个将二维数组指针传递给函数以在函数中获取/更改该数组的值的示例。但是是否可以在函数内部创建(分配内存)。像这样:

#include <stdio.h>

void createArr(int** arrPtr, int x, int y);

int main() {

    int x, y;       //Dimension
    int i, j;       //Loop indexes
    int** arr;      //2D array pointer
    arr = NULL;
    x=3;
    y=4;

    createArr(arr, x, y);

    for (i = 0; i < x; ++i) {
        for (j = 0; j < y; ++j) {
            printf("%d\n", arr[i][j]);
        }
        printf("\n");
    }
    _getch();    
}

void createArr(int** arrPtr, int x, int y) {
    int i, j;       //Loop indexes
    arrPtr = malloc(x*sizeof(int*));
    for (i = 0; i < x; ++i)
        arrPtr[i] = malloc(y*sizeof(int));

    for (i = 0; i < x; ++i) {
        for (j = 0; j < y; ++j) {
            arrPtr[i][j] = i + j;
        }
    }    
}

Answer 1

忘掉指针到指针吧。它们与二维数组无关。

如何正确操作:How do I correctly set up, access, and free a multidimensional array in C? .

使用指针对指针错误的众多原因之一:Why do I need to use type** to point to type*? .

如何正确执行此操作的示例:

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


void* create_2D_array (size_t x, size_t y)
{
  int (*array)[y] = malloc( sizeof(int[x][y]) );

  for (size_t i = 0; i < x; ++i) 
  {
    for (size_t j = 0; j < y; ++j) 
    {
      array[i][j] = (int)(i + j);
    }
  }

  return array;
}

void print_2D_array (size_t x, size_t y, int array[x][y])
{
  for (size_t i = 0; i < x; ++i) 
  {
    for (size_t j = 0; j < y; ++j) 
    {
      printf("%d ", array[i][j]);
    }
    printf("\n");
  }
}


int main (void)
{
  size_t x = 5;
  size_t y = 3;

  int (*arr_2D)[y];

  arr_2D = create_2D_array(x, y);

  print_2D_array(x, y, arr_2D);

  free(arr_2D);

  return 0;
}