c - 返回VLA和使用情况

Question

我有以下功能:

int* create_matrix_2(int rows, int cols)
{
    double (*A)[rows][cols] = malloc(sizeof(int[rows][cols]));

    for (int row = 0; row < rows; row++)
    {
        for (int col = 0; col < cols; col++)
        {
            *A[row][col] = row * cols + col;
        }
    }

    for (int row = 0; row < rows; row++)
    {
        for (int col = 0; col < cols; col++)
        {
            printf("%lf, " , *A[row][col]);
        }
        printf("\n");
    }
    return A;
}

我的问题是:如何从该函数返回 VLA，类型是什么，如何将其编写为函数签名，以及如何在接收返回的数组时使用它？

目前我正在尝试这样做:

int (*matrix2)[height][width] = create_matrix_2(height, width);

for (int row = 0; row < height; row++)
{
    for (int col = 0; col < width; col++)
    {
        printf("%d" , (*matrix2)[row][col]);
    }
    printf("\n");
}

然后运行:gcc 2d_array_test.c -o 2d_array_test.out -std=c99 -O0

但这会导致以下问题:

2d_array_test.c: In function ‘main’:
2d_array_test.c:35:34: warning: initialization from incompatible pointer type [enabled by default]
  int (*matrix2)[height][width] = create_matrix_2(height, width);
                                  ^
2d_array_test.c: In function ‘create_matrix_2’:
2d_array_test.c:105:2: warning: return from incompatible pointer type [enabled by default]
  return A;
  ^

编辑#1:

我尝试使用alk建议的代码，但编译时出现很多错误。这是一个单独的程序，其中包含您建议的带有主函数的代码:http://pastebin.com/R6hKgvM0我收到以下错误:

2d_array_test_new.c: In function ‘main’:
2d_array_test_new.c:18:2: warning: passing argument 3 of ‘create_matrix’ from incompatible pointer type [enabled by default]
  if (-1 == create_matrix(height, width, &matrix))
  ^
2d_array_test_new.c:10:5: note: expected ‘int **’ but argument is of type ‘int (**)[(sizetype)(height)][(sizetype)(width)]’
 int create_matrix(size_t, size_t, int**);
     ^
2d_array_test_new.c: At top level:
2d_array_test_new.c:37:5: error: conflicting types for ‘create_matrix’
 int create_matrix(size_t rows, size_t cols, int(**a)[rows][cols])
     ^
2d_array_test_new.c:10:5: note: previous declaration of ‘create_matrix’ was here
 int create_matrix(size_t, size_t, int**);
     ^
2d_array_test_new.c: In function ‘create_matrix’:
2d_array_test_new.c:45:11: error: ‘EINVAL’ undeclared (first use in this function)
   errno = EINVAL;
           ^
2d_array_test_new.c:45:11: note: each undeclared identifier is reported only once for each function it appears in
2d_array_test_new.c:40:6: warning: variable ‘errno’ set but not used [-Wunused-but-set-variable]
  int errno;
      ^

错误主要似乎与返回类型有关。如何正确编写该数组的类型？

Answer 1

• 引用第一个^第警告:

  warning: initialization from incompatible pointer type
  

  这里

  int (*matrix2)[height][width] = create_matrix_2(height, width);
  

  int (*matrix2)[height][width] 和 int * 根本不一样。

• 引用第二个警告:

  warning: return from incompatible pointer type 
  

  这归因于定义

  double (*A)[rows][cols] = malloc(sizeof(int[rows][cols]));
  

  同时从int * create_matrix_2()返回A。

  int * 和 double (*A)[rows][cols] 也不相同。

我建议你像这样更改函数:

#include <errno.h> /* for errno and EINVAL */
#include <stdlib.h> /* for malloc() */


int create_matrix_2(size_t rows, size_t cols, int(**a)[rows][cols])
{
  int result = 0;

  if (NULL == a)
  {
    result = -1;
    errno = EINVAL;
  }
  else
  {
    (*a) = malloc(sizeof **a);
    if (NULL == (*a))
    {
      result = -1;
    }
    else
    {
      for (size_t row = 0; row < rows; row++)
      {
        for (size_t col = 0; col < cols; col++)
        {
          (**a)[row][col] = row * cols + col;
        }
      }

      for (size_t row = 0; row < rows; row++)
      {
        for (size_t col = 0; col < cols; col++)
        {
          printf("%d, " , (**a)[row][col]);
        }

        printf("\n");
      }
    }
  }

  return result;
}

并这样调用它:

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


int create_matrix_2(size_t rows, size_t cols, int(**a)[rows][cols]);


int main(void)
{
  int result = EXIT_SUCCESS;

  int (*matrix2)[height][width] = NULL;
  if (-1 == create_matrix_2(height, width, &matrix2))
  {
    perror("create_matrix_2() failed");
    result = EXIT_FAILURE;
  }
  else
  {
    for (size_t row = 0; row < height; row++)
    {
      for (size_t col = 0; col < width; col++)
      {
        printf("%d, " , (*matrix2)[row][col]);
      }

      printf("\n");
    }

    free(matrix2);
  }

  return result;
}