我的静态矩阵的insdead我尝试获得2d动态matrics 我想要做的就是更改init函数，而不是使用定义的高度和宽度 它会初始动态 - 请告诉我如何
void init(int board[][WIDTH], int rows) {
int x, y;
for (y = 0; y < rows; y++)
for (x = 0; x < WIDTH; x++)
board[y][x] = 0;
/* Scatter some live cells: */
board[10][25] = 1;
board[10][26] = 1;
board[10][27] = 1;
board[11][25] = 1;
board[12][26] = 1;
}
int main(void) {
int board[HEIGHT][WIDTH];
init(board, HEIGHT);
..
..
}
这是我想要使用的代码 - 请告诉我正确的分辨
不使用#define WIDTH 50 #define HEIGHT 20
int **matrix_dyn(int n, int m)
{
int i = 0;
int j = 0;
printf ("please enter the horizontal size of the board \n");
scanf ("%d", &n);
printf ("please enter the vertical size of the board \n");
scanf ("%d", &m);
int **board = (int**)malloc(n * sizeof(int*));
printf("please enter the 0's or 1's to fill the matrix \n");
for (i = 0; i <= n; i++)
board[i] = (int*)malloc(m*sizeof(int));
for(i = 0; i <= n; i++)
{
for(j = 0; j <= m; j++)
scanf ("%d", &board[i][j]);
}
return board;
}
这是我的全部代码：
#include <stdio.h>
#define WIDTH 50
#define HEIGHT 20
void init(int board[][WIDTH], int rows) {
int x, y;
for (y = 0; y < rows; y++)
for (x = 0; x < WIDTH; x++)
board[y][x] = 0;
/* Scatter some live cells: */
board[10][25] = 1;
board[10][26] = 1;
board[10][27] = 1;
board[11][25] = 1;
board[12][26] = 1;
}
void print(int board[][WIDTH], int rows, int cols)
{
int x, y;
char c;
for (y = 0; y < rows; y++) {
for (x = 0; x < cols; x++) {
if (board[y][x] == 1)
printf("X");
else
printf(" ");
}
printf("\n");
}
printf("Press any key to continue:\n");
getchar();
}
int count_neighbors(int board[][WIDTH], int rows,
int y, int x)
{
int i, j;
int result = 0;
for (i = -1; i <= 1; i++)
if ((y+i >= 0) && (y+i < rows))
for (j = -1; j <= 1; j++)
if ((x+j >= 0) && (x+j < WIDTH))
if ((i != 0) || (j != 0))
result += board[y+i][x+j];
return result;
}
int step(int board[][WIDTH], int rows) { // now returns a bool
int x, y;
int neighbors[HEIGHT][WIDTH];
int changed = 0; // save changes
for (y = 0; y < rows; y++)
for (x = 0; x < WIDTH; x++)
neighbors[y][x] = count_neighbors(board, rows, y, x);
for (y = 0; y < rows; y++)
for (x = 0; x < WIDTH; x++)
if (board[y][x] == 1) { /* Currently alive */
if (neighbors[y][x] < 2)
{
board[y][x] = 0; /* Death by boredom */
changed = 1; // change happened
}
else if (neighbors[y][x] > 3)
{
board[y][x] = 0; /* Death by overcrowding */
changed = 1; // change happened
}
}
else { /* Currently empty */
if (neighbors[y][x] == 3)
{
board[y][x] = 1;
changed = 1; // change happened
}
}
return changed; // return the status (changed yes/no?)
}
int main(void) {
int board[HEIGHT][WIDTH];
init(board, HEIGHT);
while (1) {
print(board, HEIGHT, WIDTH);
if(step(board, HEIGHT) == 0) // no change
break; // leave the loop
}
return 0;
}
答案 0 :(得分：1)
声明＆amp;像这样分配董事会：
int *board = malloc( n * m * sizeof(int) );
然后，只要您想访问board [x] [y]，请使用以下表达式：
board[y*n+x]
答案 1 :(得分：0)
尝试使用结构，这是我编写的一个简单实现（使用GCC编译，以支持constructor属性。
// IntGrid.h
typedef struct intGrid_t {
int **data;
int rows;
int cols;
} *IntGridRef;
struct {
IntGridRef(* create)(int, int);
void (* print)(IntGridRef);
void (* free)(IntGridRef);
int **(* data)(IntGridRef);
int (* rows)(IntGridRef);
int (* cols)(IntGridRef);
} IntGrid;
// IntGrid.c
IntGridRef _intGrid_create(int rows, int cols);
void _intGrid_print(IntGridRef this);
void _intGrid_free(IntGridRef this);
int **_intGrid_data(IntGridRef this);
int _intGrid_rows(IntGridRef this);
int _intGrid_cols(IntGridRef this);
__attribute__((constructor))
static void intGrid_setup()
{
IntGrid.create = _intGrid_create;
IntGrid.print = _intGrid_print;
IntGrid.free = _intGrid_free;
IntGrid.data = _intGrid_data;
IntGrid.rows = _intGrid_rows;
IntGrid.cols = _intGrid_cols;
}
IntGridRef _intGrid_create(int rows, int cols)
{
IntGridRef this = calloc(1, sizeof(struct intGrid_t));
this->rows = rows;
this->cols = cols;
this->data = calloc(rows, sizeof(int *));
for (int i = 0; i < rows; i++) {
this->data[i] = calloc(cols, sizeof(int *));
}
return this;
}
void _intGrid_print(IntGridRef this)
{
printf("{\n");
for (int i = 0; i < this->rows; i++) {
printf(" { ");
for (int j = 0; j < this->cols; j++) {
printf("%i", this->data[i][j]);
if (j != this->cols - 1)
{
printf(", ");
}
}
printf(" }\n");
}
printf("}\n");
}
void _intGrid_free(IntGridRef this)
{
for (int i = 0; i < this->rows; i++) {
free(this->data[i]);
}
free(this->data);
free(this);
}
int **_intGrid_data(IntGridRef this)
{
return this->data;
}
int _intGrid_rows(IntGridRef this)
{
return this->rows;
}
int _intGrid_cols(IntGridRef this)
{
return this->cols;
}
示例用法：
int main(int argc, char *argv[])
{
IntGridRef grid = IntGrid.create(10, 10);
for (int i = 0; i < IntGrid.rows(grid); i++) {
for (int j = 0; j < IntGrid.cols(grid); j++) {
IntGrid.data(grid)[i][j] = arc4random_uniform(10);
}
}
IntGrid.print(grid);
IntGrid.free(grid);
return 0;
}
答案 2 :(得分：0)
你基本上就在那里，但不是让编译器生成代码来将[x] [y]映射到为board分配的内存中的特定元素，你必须自己进行映射：board [x * h + y]或板[y * w + x]（其中w是宽度，h是高度）;你选择哪个并不重要，只需要保持一致（函数或宏在这里会有所帮助）。