通过引用传递二维数组
我希望能够传递一个二维数组到一个函数,并让它直接在main中更改数组,所以通过引用传递。当我尝试编译时,我在开关盒中得到error: expected expression before {
。 (boardSize = 10,但在编译时未知)通过引用传递二维数组
void fillBoard(int **, int);
int main() {
int **board = malloc(sizeof(int *) * boardSize);
fillBoard(board, boardSize);
}
void fillBoard(int **board) {
int i, *row = malloc(sizeof(int) * boardSize);
for (i=0; i<boardSize; i++) {
board[i] = malloc(sizeof(int) * boardSize);
switch(i) {
case 1: row = {1,0,1,0,1,1,0,0,1,0}; break;
default: row = {0,0,0,0,0,0,0,0,0,0}; break;
}
board[i] = row;
}
}
有很多不同的方法可以做到这一点。关键是要跟踪你在哪里解决问题。您可以使用single
或double
指针来传递并填充board
,这一切取决于您如何跟踪元素。 (而二维数组提供了引用元素的便利,所有值在内存中都是连续的,并且可以通过一维参考和偏移来访问)。
当为numeric
数组分配内存时有一个重要建议。您必须始终初始化数组的全部元素,以防止试图访问或取消引用未初始化的值(未定义的行为)。简单的做法是用calloc
而不是malloc
进行分配。 calloc
分配和将所有值初始化为zero
(NULL
)。
另外请注意,当您不再需要时,需要跟踪您在程序生命周期中分配的内存和free
内存。这将防止内存泄漏的发展。在如下所示的一小段代码中,程序退出时会释放内存。如果这是一些较大代码的一部分,那么当不再需要数据时,您需要释放board
和board2
。
使用原始阵列将是一个例子:
#include <stdio.h>
#include <stdlib.h>
#define boardSize 10
void fillBoard_p (int *a);
void fillBoard_p2p (int **a);
int main() {
int i = 0;
int j = 0;
/* declaring board as an integer pointer */
int *board = calloc (boardSize * boardSize, sizeof (*board));
/* declaring board as a pointer to pointer */
int **board2 = calloc (boardSize, sizeof (*board2));
for (i = 0; i < boardSize; i++) {
board2[i] = calloc (boardSize, sizeof (**board2));
}
fillBoard_p (board);
fillBoard_p2p (board2);
printf ("\nboard as an integer pointer:\n");
for (i = 0; i < boardSize * boardSize; i++) {
if (i % boardSize == 0)
printf ("\n %d", board[i]);
else
printf (" %d", board[i]);
}
printf ("\n");
printf ("\nboard2 as an pointer to integer pointer:\n\n");
for (i = 0; i < boardSize; i++) {
for (j = 0; j < boardSize; j++) {
printf (" %d", board2[i][j]);
}
printf ("\n");
}
printf ("\n");
return 0;
}
void fillBoard_p(int *a) {
// 0=WHITE, 1=BLACK
int i = 0;
int j = 0;
int b [][boardSize] = {
{1,0,1,0,1,1,0,0,1,0},
{1,0,1,1,0,0,1,1,1,0},
{0,0,1,0,1,0,1,0,1,1},
{1,1,0,1,1,0,1,0,0,0},
{0,0,1,0,0,0,1,1,0,1},
{1,1,0,1,1,0,0,1,1,0},
{0,0,1,0,0,1,1,0,1,1},
{0,0,1,0,0,1,0,0,0,0},
{1,1,1,1,0,0,1,1,1,1},
{0,1,0,0,1,1,0,0,0,1}
};
for (i = 0; i < boardSize; i++)
for (j = 0; j < boardSize; j++)
a[i*boardSize+j] = b[i][j];
}
void fillBoard_p2p (int **a) {
// 0=WHITE, 1=BLACK
int i = 0;
int j = 0;
int b [][boardSize] = {
{1,0,1,0,1,1,0,0,1,0},
{1,0,1,1,0,0,1,1,1,0},
{0,0,1,0,1,0,1,0,1,1},
{1,1,0,1,1,0,1,0,0,0},
{0,0,1,0,0,0,1,1,0,1},
{1,1,0,1,1,0,0,1,1,0},
{0,0,1,0,0,1,1,0,1,1},
{0,0,1,0,0,1,0,0,0,0},
{1,1,1,1,0,0,1,1,1,1},
{0,1,0,0,1,1,0,0,0,1}
};
for (i = 0; i < boardSize; i++)
for (j = 0; j < boardSize; j++)
a[i][j] = b[i][j];
}
输出:
$ ./bin/fillboard
board as an integer pointer:
1 0 1 0 1 1 0 0 1 0
1 0 1 1 0 0 1 1 1 0
0 0 1 0 1 0 1 0 1 1
1 1 0 1 1 0 1 0 0 0
0 0 1 0 0 0 1 1 0 1
1 1 0 1 1 0 0 1 1 0
0 0 1 0 0 1 1 0 1 1
0 0 1 0 0 1 0 0 0 0
1 1 1 1 0 0 1 1 1 1
0 1 0 0 1 1 0 0 0 1
board2 as an pointer to integer pointer:
1 0 1 0 1 1 0 0 1 0
1 0 1 1 0 0 1 1 1 0
0 0 1 0 1 0 1 0 1 1
1 1 0 1 1 0 1 0 0 0
0 0 1 0 0 0 1 1 0 1
1 1 0 1 1 0 0 1 1 0
0 0 1 0 0 1 1 0 1 1
0 0 1 0 0 1 0 0 0 0
1 1 1 1 0 0 1 1 1 1
0 1 0 0 1 1 0 0 0 1
另外,由于一个2-D array
被顺序存储在存储器中,则可以利用这一事实的优点和使用memcpy
(在string.h
)来填充传递给你函数的数组。这样可以减少你的函数:
void fillBoard_mc (int *a) {
// 0=WHITE, 1=BLACK
int b [][boardSize] = {
{1,0,1,0,1,1,0,0,1,0},
{1,0,1,1,0,0,1,1,1,0},
{0,0,1,0,1,0,1,0,1,1},
{1,1,0,1,1,0,1,0,0,0},
{0,0,1,0,0,0,1,1,0,1},
{1,1,0,1,1,0,0,1,1,0},
{0,0,1,0,0,1,1,0,1,1},
{0,0,1,0,0,1,0,0,0,0},
{1,1,1,1,0,0,1,1,1,1},
{0,1,0,0,1,1,0,0,0,1}
};
memcpy (a, b, boardSize * boardSize * sizeof (int));
}
要是没有编译器和pointer decay
的特殊性,你可以简单地使用一个静态声明数组,如:
int board[boardSize][boardSize] = {{0}};
传递的地址数组你的函数(成为3星级程序员):
fillBoard (&board);
具有类似的功能:
void fillBoard (int *a[][boardSize]) {
// 0=WHITE, 1=BLACK
int b [][boardSize] = {
{1,0,1,0,1,1,0,0,1,0},
{1,0,1,1,0,0,1,1,1,0},
{0,0,1,0,1,0,1,0,1,1},
{1,1,0,1,1,0,1,0,0,0},
{0,0,1,0,0,0,1,1,0,1},
{1,1,0,1,1,0,0,1,1,0},
{0,0,1,0,0,1,1,0,1,1},
{0,0,1,0,0,1,0,0,0,0},A
{1,1,1,1,0,0,1,1,1,1},
{0,1,0,0,1,1,0,0,0,1}
};
memcpy (a, b, boardSize * boardSize * sizeof (int));
}
由于指针衰减(board[10][10]
=>board[*][10]
),将收到的incompatible pointer type
警告,尽管如预期的函数成功复制存储。在实践中不应该依赖不经过警告编译的代码。
'int **'不等于'int [N] [M]'。一个是指针指针,另一个是数组数组。 – WhozCraig 2014-12-03 17:39:14
http://*.com/a/26877227/971127 – BLUEPIXY 2014-12-03 17:39:16
'void fillBoard(int board [] [boardsize]);' – ooga 2014-12-03 17:41:52