将 2D 数组传递给函数

将 2d 数组传递给函数似乎很简单明了，我们乐意写：

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

#define ROWS 3
#define COLS 2

void fun1(int **, int, int);

int main()
{
  int array_2D[ROWS][COLS] = { {1, 2}, {3, 4}, {5, 6} };
  int n = ROWS;
  int m = COLS;

  fun1(array_2D, n, m);

  return EXIT_SUCCESS;
}

void fun1(int **a, int n, int m)
{
  int i, j;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      printf("array[%d][%d]=%d\n", i, j, a[i][j]);
    }
  }
}

但编译器，版本 4.9.4 中的 GCC，并不是很好。

$ gcc-4.9 -O3 -g3  -W -Wall -Wextra  -std=c11 passarr.c -o passarr
passarr.c: In function ‘main’:
passarr.c:16:8: warning: passing argument 1 of ‘fun1’ from incompatible pointer type
   fun1(array_2D, n, m);
        ^
passarr.c:8:6: note: expected ‘int **’ but argument is of type ‘int (*)[2]’
 void fun1(int **, int, int);

造成这种情况的原因有两个：主要问题是数组不是指针，第二个不便是所谓的指针衰减。将数组传递给函数会将数组衰减为指向数组第一个元素的指针 - 在 2d 数组的情况下，它会衰减到指向第一行的指针，因为 C 数组是按行排序的。

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

#define ROWS 3
#define COLS 2

void fun1(int (*)[COLS], int, int);

int main()
{
  int array_2D[ROWS][COLS] = { {1, 2}, {3, 4}, {5, 6} };
  int n = ROWS;
  int m = COLS;

  fun1(array_2D, n, m);

  return EXIT_SUCCESS;
}

void fun1(int (*a)[COLS], int n, int m)
{
  int i, j;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      printf("array[%d][%d]=%d\n", i, j, a[i][j]);
    }
  }
}

这是必要通过的行数，它们不能被计算出来。

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

#define ROWS 3
#define COLS 2

void fun1(int (*)[COLS], int);

int main()
{
  int array_2D[ROWS][COLS] = { {1, 2}, {3, 4}, {5, 6} };
  int rows = ROWS;

  /* works here because array_2d is still in scope and still an array */
  printf("MAIN: %zu\n",sizeof(array_2D)/sizeof(array_2D[0]));

  fun1(array_2D, rows);

  return EXIT_SUCCESS;
}

void fun1(int (*a)[COLS], int rows)
{
  int i, j;
  int n, m;

  n = rows;
  /* Works, because that information is passed (as "COLS").
     It is also redundant because that value is known at compile time (in "COLS"). */
  m = (int) (sizeof(a[0])/sizeof(a[0][0]));
 
  /* Does not work here because the "decay" in "pointer decay" is meant
     literally--information is lost. */
  printf("FUN1: %zu\n",sizeof(a)/sizeof(a[0]));

  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      printf("array[%d][%d]=%d\n", i, j, a[i][j]);
    }
  }
}

Version = C99

列数是预定义的，因此在编译时固定，但当前 C 标准的前身（即 ISO / IEC 9899：1999，当前是 ISO / IEC 9899：2011）实现了 VLA（TODO：链接它）和虽然目前的标准使它成为可选项，但几乎所有现代 C 编译器都支持它（TODO：检查 MS Visual Studio 现在是否支持它）。

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

/* ALL CHECKS OMMITTED!*/

void fun1(int (*)[], int rows, int cols);

int main(int argc, char **argv)
{
  int rows, cols, i, j;

  if(argc != 3){
     fprintf(stderr,"Usage: %s rows cols\n",argv[0]);
     exit(EXIT_FAILURE);
  }

  rows = atoi(argv[1]);
  cols = atoi(argv[2]);

  int array_2D[rows][cols];

  for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
      array_2D[i][j] = (i + 1) * (j + 1);
      printf("array[%d][%d]=%d\n", i, j, array_2D[i][j]);
    }
  }

  fun1(array_2D, rows, cols);

  exit(EXIT_SUCCESS);
}

void fun1(int (*a)[], int rows, int cols)
{
  int i, j;
  int n, m;

  n = rows;
  /* Does not work anymore, no sizes are specified anymore
  m = (int) (sizeof(a[0])/sizeof(a[0][0])); */
  m = cols;

  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      printf("array[%d][%d]=%d\n", i, j, a[i][j]);
    }
  }
}

这不起作用，编译器抱怨：

$ gcc-4.9 -O3 -g3  -W -Wall -Wextra  -std=c99 passarr.c -o passarr
passarr.c: In function ‘fun1’:
passarr.c:168:7: error: invalid use of array with unspecified bounds
       printf("array[%d][%d]=%d\n", i, j, a[i][j]);

如果我们通过将声明更改为 void fun1(int **a, int rows, int cols) 来故意在函数调用中出错，则会更清楚一些。这导致编译器以不同但同样模糊的方式抱怨

$ gcc-4.9 -O3 -g3  -W -Wall -Wextra  -std=c99 passarr.c -o passarr
passarr.c: In function ‘main’:
passarr.c:208:8: warning: passing argument 1 of ‘fun1’ from incompatible pointer type
   fun1(array_2D, rows, cols);
        ^
passarr.c:185:6: note: expected ‘int **’ but argument is of type ‘int (*)[(sizetype)(cols)]’
 void fun1(int **, int rows, int cols);

我们可以通过多种方式做出反应，其中之一就是忽略所有这些并做一些难以辨认的指针杂耍：

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

/* ALL CHECKS OMMITTED!*/

void fun1(int (*)[], int rows, int cols);

int main(int argc, char **argv)
{
  int rows, cols, i, j;

  if(argc != 3){
     fprintf(stderr,"Usage: %s rows cols\n",argv[0]);
     exit(EXIT_FAILURE);
  }

  rows = atoi(argv[1]);
  cols = atoi(argv[2]);

  int array_2D[rows][cols];
  printf("Make array with %d rows and %d columns\n", rows, cols);
  for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
      array_2D[i][j] = i * cols + j;
      printf("array[%d][%d]=%d\n", i, j, array_2D[i][j]);
    }
  }

  fun1(array_2D, rows, cols);

  exit(EXIT_SUCCESS);
}

void fun1(int (*a)[], int rows, int cols)
{
  int i, j;
  int n, m;

  n = rows;
  m = cols;

  printf("\nPrint array with %d rows and %d columns\n", rows, cols);
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      printf("array[%d][%d]=%d\n", i, j, *( (*a) + (i * cols + j)));
    }
  }
}

或者我们做得对，并将所需信息传递给 fun1。为此，我们需要重新排列 fun1 的参数：列的大小必须在数组声明之前。为了使其更具可读性，保持行数的变量也改变了它的位置，现在是第一个。

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

/* ALL CHECKS OMMITTED!*/

void fun1(int rows, int cols, int (*)[]);

int main(int argc, char **argv)
{
  int rows, cols, i, j;

  if(argc != 3){
     fprintf(stderr,"Usage: %s rows cols\n",argv[0]);
     exit(EXIT_FAILURE);
  }

  rows = atoi(argv[1]);
  cols = atoi(argv[2]);

  int array_2D[rows][cols];
  printf("Make array with %d rows and %d columns\n", rows, cols);
  for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
      array_2D[i][j] = i * cols + j;
      printf("array[%d][%d]=%d\n", i, j, array_2D[i][j]);
    }
  }

  fun1(rows, cols, array_2D);

  exit(EXIT_SUCCESS);
}

void fun1(int rows, int cols, int (*a)[cols])
{
  int i, j;
  int n, m;

  n = rows;
  m = cols;

  printf("\nPrint array with %d rows and %d columns\n", rows, cols);
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      printf("array[%d][%d]=%d\n", i, j, a[i][j]);
    }
  }
}

对于一些认为变量顺序无关紧要的人来说，这看起来很尴尬。这不是什么大问题，只需声明一个指针，让它指向数组。

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

/* ALL CHECKS OMMITTED!*/

void fun1(int rows, int cols, int **);

int main(int argc, char **argv)
{
  int rows, cols, i, j;

  if(argc != 3){
     fprintf(stderr,"Usage: %s rows cols\n",argv[0]);
     exit(EXIT_FAILURE);
  }

  rows = atoi(argv[1]);
  cols = atoi(argv[2]);

  int array_2D[rows][cols];
  printf("Make array with %d rows and %d columns\n", rows, cols);
  for (i = 0; i < rows; i++) {
    for (j = 0; j < cols; j++) {
      array_2D[i][j] = i * cols + j;
      printf("array[%d][%d]=%d\n", i, j, array_2D[i][j]);
    }
  }
  // a "rows" number of pointers to "int". Again a VLA
  int *a[rows];
  // initialize them to point to the individual rows
  for (i = 0; i < rows; i++) {
      a[i] = array_2D[i];
  }

  fun1(rows, cols, a);

  exit(EXIT_SUCCESS);
}

void fun1(int rows, int cols, int **a)
{
  int i, j;
  int n, m;

  n = rows;
  m = cols;

  printf("\nPrint array with %d rows and %d columns\n", rows, cols);
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      printf("array[%d][%d]=%d\n", i, j, a[i][j]);
    }
  }
}